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Photo courtesy: Gil Talbot Photography
Complete Game Notes (PDF)
Live Statistics (at Dartmouth)
Live Video (at Dartmouth - USCHO.com Game of the Week)
Live Statistics (vs. Providence)
Live Video (vs. Providence - with subscription)
ECAC Hockey Standings
Ivy League Standings
Purchase Women's Beanpot Tickets!
The Particulars
The seventh-ranked Harvard women's hockey team continues the stretch run as the team visits Dartmouth Friday afternoon before hosting No. 9 Providence in a non-conference matchup Saturday afternoon at Bright Hockey Center. The Crimson currently stands third in the ECAC Hockey standings with a 8-5-2 record and 18 conference points. Friday's game at Dartmouth has been designated as the USCHO.com Game of the Week.
Follow From Home
GoCrimson.com will provide live statistics for this weekend's games, all home games and most road contests this season. Live audio and video of Friday's game will be available on USCHO.com. Live audio and video for Saturday's contest will be available at GoCrimson.com. Matt Sawyer and Glenn Conway will have the call for Saturday's game.
Series History
Harvard and Dartmouth have played 73 times, with the Big Green owning a 39-32-2 lead in the all-time series, which began in 1978-79. The Crimson has won five of the last six games against its ECAC Hockey and Ivy League rival, including a 3-2 victory over the Big Green at Bright Hockey Center earlier this season on Nov. 25. Liza Ryabkina netted the game-winning goal with 3:25 remaining in the game, while Christina Kessler picked up the victory with a 27-save effort.
Providence sports a 22-13-3 record against Harvard in 38 games, but the Crimson is 6-0-1 in the last seven games between the schools. The Crimson and Friars last played on Dec. 8, 2007, a 4-1 victory for Harvard in Providence, R.I. Sarah Vaillancourt '08-09, Jenny Brine '09 and Liza Ryabkina each had a goal and two assists to guide the offense, while Christina Kessler stopped 17 shots for the win.
Last Time Out
After shutting out Union, 4-0, Friday, Laura Bellamy recorded a career-high 27 saves in a 4-2 loss at the hands of Rensselaer Saturday afternoon. Kate Buesser and Cori Bassett each netted a goal for the Crimson.
Where's Your Beanpot?
Harvard will host the 2010 Beanpot at Bright Hockey Center. The first round, featuring the Crimson vs. Boston College at 8 p.m. and Boston University vs. Northeastern at 5 p.m. will take place Feb. 2, while the second round will be held Feb. 9. Harvard has won 12 Beanpots, while Northeastern has the lead with 14. Boston College has three titles, and Boston University has won three of the tournaments, which began in 1979.
Harvard Against Hockey East
Harvard owns a 1-0-1 record against teams from Hockey East this season. The Crimson tied New Hampshire, 1-1, Nov. 18 and then defeated Connecticut by a score of 2-1, Dec. 8. After facing the Friars, Harvard will play at least two more Hockey East opponents this season, as the Crimson will skate against Boston College in the opening round of the Beanpot Feb. 2, and then will play either Northeastern or Boston University in the second round Feb. 9.
Clean Sheets
With its 4-0 victory against Union Jan. 22, Harvard now has recorded five shutouts this season. Laura Bellamy earned her first career clean sheet against the Dutchwomen, making 13 saves. Four of the Crimson's five shutouts have come at Bright Hockey Center.
Christina Kessler has recorded 25 shutouts during her career at at Harvard, a program record.
Crimson Leading The League
Kate Buesser has an ECAC Hockey-best 15 assists in league games and stands second in points with 22. Rookie Jillian Dempsey (8-10-18) ranks fourth in conference points among all players and first among freshmen in the league. Defender Leanna Coskren, Cori Bassett and Dempsey are tied for fifth with five power-play points in ECAC Hockey contests. Coskren is also tied for first with three game-winning goals during conference action. Bassett, meanwhile, stands third in defenseman scoring with 11 points on four goals and seven assists.
The World's Best
A trio of former Harvard standout skaters have been named to the 2010 U.S. Women's National Team, which will compete at the 2010 Vancouver Olympics. Angela Ruggiero '02-04, the Patty Kazmaier recipient in 2004, Julie Chu '06-07, the winner of the Patty Kazmaier trophy in 2007, and former All-American Caitlin Cahow '07-08 were named to the American team. On Team Canada, Harvard will be represented by two-time Patty Kazmaier Award winner Jennifer Botterill '02-03, and Sarah Vaillancourt '08-09, who was named the top women's college hockey player in 2008. Overall, Harvard has had nine current and former women's hockey players represent their countries in the Olympics since 1996.
When The Game Is On The Line
Seven different Harvard players have scored a game-winning goal this season -- Kaitlin Spurling, Liza Ryabkina, Jillian Dempsey, Leanna Coskren, Anna McDonald, Cori Bassett and Kathryn Farni. Coskren leads the team and ECAC Hockey with three game-winning tallies, while Ryabkina has notched two this season.
Scoring In A Bunch
As a team, Harvard tanks second in ECAC Hockey with 3.20 goals per game, having scored a league-best 48 tallies in 15 conference games entering the weekend.
A Balanced Attack
The Crimson has had 13 different players score at least one goal this year, as the team has netted a total of 46 goals. Sixteen players have at least one point to their credit and Harvard had 15 different players light the lamp last season as well, scoring a total of 95 goals.
Around The Ivy League
The Crimson is unbeaten in five straight games against Ivy League opponents (3-0-2) after tying Cornell Jan. 15. The Crimson stands at 3-2-2 against Ancient Eight opponents, good for third place in the conference standings. Cornell remains in first with a 5-0-2 Ivy League conference record.
License to Kill
The Crimson boasts the nation's ninth-best penalty kill unit, stopping 68 of 77 opponent power plays (.883). Harvard also has a shorthanded goal to its credit, coming off the stick of Jillian Dempsey in a 4-0 win over Union Jan. 22.
Harvard Coach Katey Stone
Katey Stone is in her 16th season as the Landry Family Head Coach for Harvard Women's Ice Hockey, and she brings a 329-141-26 (.690) record into Friday's game. Stone, who ranks second all-time in Division I victories, has led Harvard to all seven of its NCAA tournament appearances. Stone is also the active leader in victories among Division I women's hockey coaches.
Stone is currently eight wins shy of tying the all-time Division I record, held by Laura Halldorson (337 victories).
First Thing's First
The Crimson was 18-0-2 when scoring first and 1-10-1 when the opponent lit the lamp first last season. So far in 2009-10, Harvard is 8-1-2 when netting the game's first goal and 2-4-2 when the opposition scores first.
It's All Over
Harvard is 102-1-5 since the start of the 2004-05 season when leading after two periods. The Crimson is 8-0-1 in this category this season.
Just Win, Christina!
Senior goaltender Christina Kessler is 64-16-6 (.779) in her career in net, placing her first on Harvard's all-time goalie wins, as she passed Ali Boe '06 with a 5-1 victory against Colgate Jan. 16. Kessler currently holds the NCAA career record with a .9413 save percentage, as she has stopped 1,863 of 1,979 shots faced during her career. Jessie Vetter (Wisconsin 2006-09) owned a .941 save percentage to set the record (2,175 of 2,312, .9407).
Welcome To The Show
Harvard freshmen have had the scoring touch this season, led by Jillian Dempsey who is tied for the team with eight goals and is second with 11 assists and 18 points. Kaitlin Spurling (4-3-7), Josephine Pucci (2-5-7) have also been scoring threats.
Storied Program
With its 3-0 victory over St. Lawrence Nov. 7, Harvard earned the 500th victory in program history. The Crimson owns a 508-292-42 (.628)record over its storied 31-year history.
A Stone Wall
The Crimson earned back-to-back shutouts against Minnesota Dec. 4-5, marking the first time that the Golden Gopher were held scoreless since Feb. 2, 2007. The Crimson had not recorded consecutive shutouts since holding Cornell without a goal during the ECAC Hockey quarterfinals Feb. 27-28, 2009.
To Be The Best ...
With its 4-4 tie against No. 10 Cornell Jan. 15, the Crimson stands at 3-2-3 against ranked foes this year. Harvard was 2-6-1 against top 10 opponents last season. The Crimson will play No. 9 Providence Jan. 30.
Bright Spot
The Crimson was 9-4-2 record at Bright Hockey Center last season, while posting a 9-5-1 mark on the road. In 13 games this season, Harvard owns a 7-3-3 record on home ice, but is 3-2-1 in road contests.
Red Lights
Harvard ranks fourth in the nation in scoring defense (1.53 goals allowed per game), fifth in scoring margin (+1.21), 11 on the penalty kill (68 of 77, .883), 15th in scoring offense (2.74 goals scored per game) and 27th on the power play (11of 93, .118).


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The Architects' Journal's new iPad edition is now in the App Store with a free sample issue of the AJ, single issues for sale, and free access to subscribers until April 2013
The Architects' Journal (AJ), the best-selling weekly architecture publication since 1895, has announced the launch of its iPad edition, now available for download free in the App Store. The app is free to existing subscribers until April 2013.
Featuring the weekly magazine with interactive drawings and bonus content, the app includes a free sample issue of the AJ, and offers the ability to purchase single issues or subscribe to the weekly iPad edition on a monthly or annual basis.
The launch is a historic moment for the print title. The AJ is the first weekly architecture title to use the Adobe DPS platform favoured by magazines such as Wired, which enabled the creation of the interactive floorplans and technical drawings, pinch and zoom, bonus photography, video and maps, as well as deep links to the AJ Buildings Library and The AJ website.
Introducing the app in her column in this week's issue of the magazine, Christine Murray, AJ editor, said the app builds upon the title's long tradition in print:
'This is a landmark moment in more than a century of publishing for the AJ. Brought to you by the same team that designs the AJ in print, and built with the same software used to create iPad magazines such as Wired, we've worked hard to craft an edition that is in keeping with the AJ's tradition and style, while making the most of the additional features offered by this new platform.
The building studies include interactive floorplans and technical drawings, maps and video
Murray added: 'The navigation follows a few simple rules. As for interactivity, we've lavished the most attention, rightly, on the building studies, which include interactive floorplans and technical drawings, bonus photography, maps and video. We believe we've created a great new way for architects to experience the AJ every week."
To download the free app, which includes a free sample issue for non-subscribers, visit the App Store
The AJ is now on iPad - download it free today
Readers' comments (5)
I have been, trying to use you Ipad app. a great thing however it is way too slow to download and navigation through the issues is slightly awkward. There is definitely room for improvement....
Thanks for your feedback Etain, we're working to improve download times and navigation, so your comments are valuable to us. Please get in touch with any further thoughts on how your experience could be improved.
Christine Murray
I keep getting an error message when tapping the "tap here" icon on aj's app. Haven't received a downloaded issue since 28/3/13.
We're having trouble with our authentication - but the wizards are on it and it should be fixed soon. Will post a note when they've cracked it.
I've been told authentication is back online - please try again and sorry for the inconvenience!
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subset: commoncrawl


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Sobaos pasiegos
Turismo de Cantabria
Sobaos pasiegos
Ingredients for 8 people: 250g of butter 250g of flour 250g of sugar 1 packet of powdered yeast 1 spoonfuls of rum 3 eggs salt.
Mix the softened butter with the beaten eggs, the rum, the sugar and a pinch of salt in a bowl. Meanwhile, put the flour and the powdered yeast in a separate receptacle; then add the mixture to the other ingredients. When everything is in, stir until it forms an even mixture, without lumps. Pour the mixture into paper cases or on to a baking tray greased with butter until it is about half the size it will reach when it rises with baking. Put in the oven at about 180ºC for about 15 minutes.
More information
Recipe information
- Category: Dessert
- Cooking time: 1 hour
- Price: Low
- Season: All year round
Nutritional information
- Energy: High
- Cholesterol: Average


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Regulators cleared Nasdaq OMX Group Inc.'s plan to pay US$62-million to compensate brokers for its mishandling of Facebook Inc.'s public debut, dealing a defeat to Wall Street firms that say they lost many times that amount.
The Securities and Exchange Commission approved Nasdaq's request to change its rules and expand the compensation pool for member firms in the May 18 initial public offering. The funds will go to traders who lost money after a design flaw in the exchange's computers delayed Facebook's open and left traders confused about how many shares they owned.
Nasdaq's proposal was opposed by Citigroup Inc. and UBS AG, which said in letters urging the SEC to reject it that losses within their market-making units exceeded US$62-million. Nasdaq, balancing its role as an organization with legal immunity for technology breakdowns with its obligations to members, said the pool covers "objective, discernible" losses suffered by brokers.
While agreeing the proposal won't compensate all purported losses, the SEC said it provides "significantly more compensation for eligible claims, outside of litigation, than would otherwise be available," according to its order. "Accordingly, approval of the proposed rule change will make more funds available to compensate investors and Nasdaq members under Nasdaq rules, which the commission believes is in the public interest," it wrote.
Scott Helfman, a spokesman for Citigroup, declined to comment on the settlement. Megan Stinson, a New York-based spokeswoman for UBS, did not immediately respond to a request for comment.
Claim Ceiling
Under existing rules, Nasdaq's liability for losses related to computer malfunctions is US$3-million, and may have been as low as US$500,000 in the Facebook case, the SEC said in its order.
"We're pleased that the Securities and Exchange Commission has approved our accommodation plan, which will enable our customers and members and market participants to receive appropriate restitution as FINRA promptly begins processing claims," Joseph Christinat, a spokesman for Nasdaq OMX, said by phone today.
IPO Cross
The pricing of the first public transaction on May 18, a trade known as the IPO cross, took a half hour longer than Nasdaq planned because of technical malfunctions. In May, Nasdaq OMX Chief Executive Officer Robert Greifeld acknowledged "poor design" in software put the opening auction into a loop that delayed its completion.
Nasdaq's handling of the Facebook IPO may still end up in court. In approving the rule change needed to accommodate the payouts, the SEC said the question of whether Nasdaq is entitled to regulatory immunity in its handling of the offering is outside the scope of the decision. Nasdaq has made releasing it from legal liability a condition for receiving compensation.
That immunity argument was raised in a letter sent to the commission in August by Citigroup. Decisions made by the second- largest U.S. equity trading venue in the IPO were aimed at protecting profits rather than member firms, the company said.
"Market participants suffered hundreds of millions of dollars of losses as a result of Nasdaq's profit-driven conduct prior to and during the Facebook IPO, not as a result of protected regulatory activity by Nasdaq, or routine system failures," Citigroup wrote in August. "Nasdaq should not be permitted to hide behind regulatory immunity."
'Unprecedented Failures'
The proposed plan is "inadequate to address the magnitude of Nasdaq's unprecedented failures," UBS told the SEC in a letter on Aug. 22. The bank entered multiple orders for Facebook shares because it didn't receive confirmations, leading to losses of more than US$350-million, UBS said. It asked the SEC to work with Nasdaq to reformulate the proposal to increase the amount paid and cover a broader range of trading losses.
Citigroup, Chicago-based Citadel, UBS in Zurich and Knight Capital Group Inc. in Jersey City, New Jersey, operate equity wholesaling groups, brokers that execute orders for individual investors sent by securities firms such as Charles Schwab Corp., TD Ameritrade Holding Corp. and Fidelity Investments. Combined, the firms claim they lost almost US$500-million in the offering.
Citadel and Knight previously said they supported Nasdaq's repayment proposal.
Protect Investors
The SEC confined its decision to issues of exchange rulemaking and said Nasdaq's proposal was consistent with laws that require regulations to prevent fraud, remove impediments to a free and open market, protect investors and the public interest, and deter discrimination among customers, issuers, brokers and dealers.
"Commenters have raised a number of concerns about the proposed rule change, many contending that it is not a fair or equitable approach to compensating market participants," the SEC wrote in its order. "Nasdaq has explained, however, that it did not design the proposed rule change to compensate all claims of loss suffered by market participants," it said.
"Rather, Nasdaq, in the accommodation proposal, is proposing to change a Nasdaq rule that in its current form strictly limits the amount of compensation that may be paid to users of the Nasdaq Market Center," it wrote. "In considering whether to approve the proposed rule change, the commission takes into account the existing circumstances and the manner in which the Nasdaq rule would operate if the commission disapproved the proposed rule change."
Delayed Orders
Delays and malfunctions on the Nasdaq Stock Market were the first signs of trouble in the Facebook IPO that burned investors and prompted lawsuits against the company, its exchange and the underwriters. The stock has fallen more than 30% from the price set by underwriters. It slipped 0.9% to US$25.51 as of 10:18 a.m. in New York.
Facebook was sold by underwriters at US$38 on May 17. The pricing of the first public transaction, a trade known as the IPO cross, took place at 11:30 a.m. New York time the next morning, a half hour later than Nasdaq planned. About 30 minutes after that, the market owner reported a delay confirming trades from the opening auction with the brokers that placed orders.
The transaction reports, normally distributed immediately, were sent at 1:50 p.m., leaving market makers, brokers and their customers uncertain about whether orders submitted into Nasdaq's opening cross had been executed. Traders and individuals couldn't sell to limit losses until they received the reports and knew how many shares they held.
The payout proposal covered what's "directly attributable to the system issues experienced by Nasdaq," the exchange operator told the SEC in a Dec. 7 letter.


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The Piano Technicians Guild (PTG) is the world's premier source of expertise in piano service and technology. The mission of the Piano Technicians Guild is to promote the highest possible standards of piano service by providing members with opportunities for professional development, by recognizing technical competence through examinations and by advancing the interests of its members.
Our rigorous technical examinations are used to certify piano technicians as Registered Piano Technicians (RPT). The RPT credential shows that the technician is a professional who can competently perform normal piano tuning, service and repair. We encourage you to find an RPT to service your piano!
Our Resources area has information for tuners, technicians, teachers, piano owners and pianists. Learn how to care for your piano. Download free learning tools for piano students and teachers from the Piano Learning Center and find tips for buying a piano. Explore a career in piano technology and learn more about the history and design of pianos.
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WebSundew is a powerful web scraping tool that extracts data from the web pages with high productivity and speed. WebSundew enables users to automate the whole process of extracting and storing information from the web sites. You can capture large quantities of bad-structured data in minutes at any time in any place and save results in any format. Our customers use WebSundew to collect and analyze the wide range of data that exists on the Internet related to their industry.
WebSundew Services were designed for the users who are too busy to deal with the soft and for the organizations which do not have a complex IT infrastructure of their own. Our extraction services staff can set up a data extraction agent which you can run on your computer or we can extract data from the given web site. You just need to submit a service request and we will do the entire job for you.
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Pictures relating to Farina, SA
Not the 'Farina' you were after? Try the other Farina (the regional area) in South Australia.
Create a custom map: Click here to create a custom map including 'Farina'
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Do you know what the fastest creature on two legs is? Weirdly enough, it's not the road runner! We're busting up these cartoon myths portrayed as fact in our new Fun Fact Friday post.
Myth: Road Runners are super fast.
Facts: According to 10,000 Birds the Road Runner can run up to 20 mph, still faster than most birds, but coyotes can reach speeds of 43 mph. However, Road Runners are masters at ducking and weaving AND can also fly. So chances are, a coyote would have a pretty rough time actually catching one for dinner. Silly coyotes.
Myth: The Tasmanian Devil tears through the lush, thick jungles of Tasmania causing stampedes and devouring all manners of species including "tigers, lions, elephants, buffaloes, donkeys, hippopotamuses, rhinoceroses, giraffes, octopuses, moose, and ducks."
Fact:The Tasmanian devil, despite what Looney Tunes depicts, doesn't travel via mini-tornadoes through jungle, snarling and spitting, but IS "notoriously cantankerous" and "will fly into a maniacal rage when threatened." These marsupials are in fact strictly carnivorous, according to National Geographic, and feed on small mammals, snakes, fish, and insects.
Myth: Popeye the Sailor Man's source of strength and huge muscles came from guzzling spinach.
Fact: Spinach is still good for you, but maybe not as good as you think. This myth is first in a line of what scientists call "Super Myths." In 1870, a German study proclaimed spinach to have as much iron as a steak. Almost a hundred years later it was realized that this error was due to a misplaced decimal point making it appear spinach is extremely iron rich and caused a few generations of confusion.
By the way, the fastest animal on two legs is an ostrich. Happy Friday!
Looking for some fun stuff to do? Check Out
10 (Inexpensive) Weekend Activities
DIY Dorm: String Up Photo Display
Hot Spots for a Road Trip: Spring 2013


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Data interpretation
Mplus Discussion
Growth Modeling of Longitudinal Data
posted on Tuesday, January 16, 2001 - 3:47 pm
In a time-invariant conditional LTM, how do you interpret the significant effects of the predictors on the slope (in this case, an overall downward trend in health over time).
Education: estimate .074 (se .056)
Energy: estimate -.231 (se .054)
For a 1 unit increase in education, you see...
For a 1 unit increase in energy, you see...
Thanks in advance.
Bengt O. Muthen
posted on Wednesday, January 17, 2001 - 9:22 am
Can you describe your model more fully? Does LTM stand for latent transition or latent trait modeling? Do you have dichotomous dependent variables?
posted on Tuesday, January 23, 2001 - 9:31 am
The model has four time-points with a categorical dependent health variable (measured on a 5-point scale). The model is time-invariant with baseline predictors (e.g. education) predicticing the intercept and slope of the model. The education and energy variables are categorical, with higher numbers representing higher levels of education and energy to participate in daily activities.
I used the acronym LTM to mean latent trajectory modeling.
Bengt O. Muthen
posted on Tuesday, January 23, 2001 - 11:47 am
With categorical repeated measures, the interpretation of the effects of time-invariant covariates on the slope growth factor can be expressed in several different ways. First, the coefficient (unstandardized) can be simply interpreted as in regular regression in terms of change in the slope for a unit change in a covariate (holding other covariates constant). This may not carry much meaning because the scale of the slope is arbitrary. Second, one can consider the standardized coefficient, in which case the change in the slope is expressed in slope standard deviations. This still doesn't mean much given that the outcome is categorical. Third, one can express the ultimate effect of the change in the covariate on the outcome variable probabilities. This may give a more "down to earth" interpretation. For instance, you can compute the outcome probabilities for some chosen values of your covariates. You do this by first computing the mean values of the slope given the chosen covariate values and then computing the outcome variable probabilities for these mean values.
posted on Thursday, June 28, 2001 - 2:00 pm
How do you calcute the mean value of the slope given the chosen covariate value and then compute the outcome variable probabilities? Can this be done directly in MPLUS or hand-calculated. Thanks for your assistance.
posted on Sunday, July 01, 2001 - 12:10 pm
I assume that you have a categorical outcome and that by slope you mean the slope growth factor. The mean value of the slope is obtained in TECH4 and is s_m = a+g*x for covariate value x, where a is the intercept of the slope factor and g is the regression coefficient for the slope regressed on x. The probability has to be computed by hand, for instance with unit scale factors delta (see User's Guide), you have for a binary y scored 0/1,
P (y=1 |x) = F(-tau + s_m*x_t),
where F is the normal distribution function, tau is the threshold parameter held equal across time, and x_t are the time scores (the slope loadings).
Louise Sullivan
posted on Wednesday, November 10, 2004 - 11:17 am
I'm looking at social mobility across 4 different time points and according to the BIC value the best fitting unrestricted latent class growth analysis is a 7 class model. Seven different classes is not substantively useful and I notice from your June 2000 paper (Muthen and Muthen 'Integrating Person-Centered and Variable-Centered Analyses: Growth Mixture Modeling with Latent Variables' in Alchoholism: Clinical and Experimental Research) that you outline other criteria on page 887 for assessing how many latent classes to use. Please could you explain how high the average posterior probabilites should be - for my 7 class model I've got cross-classification figures as low as 0.683, 0.610 and 0.662. I would prefer to use a 4 class model which has a higher BIC (26224 compared to 26165) but better cross-classification values ie between 0.764 and 0.91. Am I justified in using a 4 class model?
posted on Sunday, November 14, 2004 - 12:35 pm
The posterior probabilities tell you how useful the model is, but not how many classes fit the data best. You can consider other fit statistics. For example, several simulation studies indicate that Mplus' sample-size adjusted BIC is better than BIC. Also, the Lo-Mendell-Rubin test in Mplus' Tech11 can be used. Ultimately, the usefulness of the model is a key consideration besides statistical fit indices, e.g. predictive performance.
Marion Sheenan
posted on Saturday, August 13, 2005 - 3:38 pm
I want to perform hypothesis testing on the individual parameters in my model. I know that I can use est/error but should I use a T or Z distribution?
Linda K. Muthen
posted on Sunday, August 14, 2005 - 11:14 am
The estimate divided by the standard error shown in the Mplus output follows an approximate z distribution.
Gwen Marchand
posted on Saturday, February 02, 2008 - 4:02 pm
I am relatively new to growth modeling and have managed to confuse myself. I have a simple question regarding interpretation of coefficients.
I understand that including a time-invariant covariate into a model influences the latent slope and intercept, so that estimates listed under the "intercept" section of the output for the slope and intercept account for the influence of the covariate on the latent factors. In my case, the slope mean in the tech4 output is negative, but I have a positive coefficient estimate for my slope in the intercept section of the output (slope estimate = .113). I'm currently exploring why that may have occurred. But in the meantime, the covariate has a negative association with the slope. (-.05).
Would I interpret this so that higher scores on the covariate at time 1 are associate with more slowly increasing slopes (based on the slope coefficent)?
Thank you in advance.
Linda K. Muthen
posted on Monday, February 04, 2008 - 9:12 am
If s is the slope growth factor,
mean (s) = a + b mean (x)
When x is zero, the mean (s) is equal to the intercept (s), that is, a.
In your case, a is positive and b is negative, so the mean of x must be a positive value large enough to cause the product to be negative and larger than a resulting in a negative mean of s.
The interpretation is that as x increases the slope is a larger negative value.
posted on Saturday, February 14, 2009 - 8:50 am
First, I'd like to thank you for making this forum available, it is such a great help!
I am attempting to revise a paper and have some questions related to interpreting the correlation between intercept and growth factors. The LGM focuses on symptoms from time 1 to time 7.
1. Given a positive intercept mean (0.82), a negative linear slope mean (-0.16), and a positive quadratic slope mean (0.02), how do you interpret a negative correlation between the slope and intercept? Is it that the higher individuals are on symptoms at time 1, the slower the rate of decline in symptoms?
2. If the variance of the quadratic factor is fixed to 0, is it necessary to include it in your interpretation or to include a correlation between the intercept or linear slope and quadratic factor?
3. Given a positive intercept mean (0.97), a negative linear slope mean (-0.13), and a positive quadratic slope mean (0.02), how do you interpret:
a) a positive (but nonsignificant) correlation between the intercept and quadratic slope
b) a negative correlation between the linear and quadratic slope?
Thanks very much in advance!
Bengt O. Muthen
posted on Sunday, February 15, 2009 - 11:17 am
1. If you center at time 1, then the higher an individual is at time 1, the lower his/her slope - that is, the steeper the decline. I am referring to the correlation between the intercept and the linear slope (but see also the caveat in 3 below).
2. With Var(q)=0 you don't have covariances between q and other growth factors. You still have the mean of q to explain.
3. With a quadratic growth model the linear and quadratic terms are partly confounded and are not easy to give separate interpretations for (this is why orthogonal polynomials are sometimes used). Vaguely speaking, with centering at time 1 the linear slope has the biggest influence in the beginning of the growth and the quadratic the end of the growth. Because of the confounding, I would not go into interpretations of correlations among growth factors in a quadratic model. With that caveat,
a) if it were significant this would probably mean that a person with a high intercept also has a high upturn towards the end.
b) when the initial decline is steeper, the ending upturn is higher.
posted on Tuesday, February 17, 2009 - 3:20 pm
I conducted a conditional two-group LGM and I'm having some trouble wrapping my head around interpreting the effect of two predictors on the slope functions.
In the first group, which consists of only an intercept (intcpt=.315) and linear factor (intercept = .023), how do you suggest I interpret the following:
1. a positive path coefficient (0.032) for the regression of the slope on predictor 1.
2. a negative path coefficient of -0.034 for the regression of the slope on predictor 2.
In the second group, which consists of an intercept (int= 0.844), linear factor (int=-0.015), and a quadratic factor (int=-0.004), how do you suggest I interpret the following:
1. a negative path coefficient of -0.062 for the regression of the linear slope on predictor 2.
2. a positive path coefficient of 0.012 for the regression of the quadratic slope on predictor 2.
Thanks for your assistance!
Bengt O. Muthen
posted on Thursday, February 19, 2009 - 9:21 am
Use the rules for interpreting coefficients in linear regression. As in that case, these path coefficients are partial regression coefficients, so giving the effect on the DV as the predictor changes 1 unit while holding the other predictors constant.
posted on Thursday, February 19, 2009 - 11:52 am
Thanks for your help. However, I still am not clear on whether the predictor is predicting a faster or slower rate of change.
Bengt O. Muthen
posted on Friday, February 20, 2009 - 5:01 am
Since you mention predictor (singular form) I assume you refer to the question you have for the second group, regarding the quadratic model. Is that right?
posted on Friday, February 20, 2009 - 5:43 am
Yes, that is correct. I think (but please correct me if I'm wrong!) that for the group with the negative linear trend (the first group), predictor 1 (with a positive coefficient) predicts a slower decline and predictor 2 (with a negative coefficient) predicts a faster decline
However, I am completely confused as to how to interpret the effect of the predictor on the quadratic group.
Thanks again in advance!
Linda K. Muthen
posted on Friday, February 20, 2009 - 6:24 am
See the following post from Sunday, February 15:
3. With a quadratic growth model the linear and quadratic terms are partly confounded and are not easy to give separate interpretations for (this is why orthogonal polynomials are sometimes used). Vaguely speaking, with centering at time 1 the linear slope has the biggest influence in the beginning of the growth and the quadratic the end of the growth. Because of the confounding, I would not go into interpretations of correlations among growth factors in a quadratic model. With that caveat,
a) if it were significant this would probably mean that a person with a high intercept also has a high upturn towards the end.
b) when the initial decline is steeper, the ending upturn is higher.
posted on Friday, February 20, 2009 - 9:05 am
Thanks - is this also true for the effect of covariates on linear and quadratic slope?
Linda K. Muthen
posted on Saturday, February 21, 2009 - 9:35 am
Ingrid Holsen
posted on Thursday, February 17, 2011 - 7:45 am
I am really workinng on this to get it right, but I am now confused. I have several covariates predicting latent growth in body image measured at several time points between ages 13 and 30.
(I am here using the �model results�,is STDYX preferable?)
To use the covariate close parent-adolescent relationship as an example; for boys there is a positive estimate at initial level at age 13 (0.14) (understandable!), negative sign. estimate for slope (-0.25), and a positive estimate for q (0.14). Do I interpete the s and q as that parent adolescent relationship influences body image growth to a less degree (during adolescence) for so to be of more importance again in early adulthood (q)? The body image curve for boys are increasing between the ages 13 and 18, so leveling off and decresing some at ages 21 and 23 .(so an increase again up to age 30).
Ingrid Holsen
posted on Thursday, February 17, 2011 - 12:58 pm
Hi again,
We are treating close adolescent relationship and peer relationship as �time invariant covariates�from time 1. I have BMI as a time varying covariate at six points in time. When we include BMI in the model the effects of the time invariant covariates for girls on slope and quadratic growth disappears, while almost no difference for boys. Is there a way that we in one model (one step) can reveal this effect. Now we run it twice.
Thanks in advance (for both my two posts)
Bengt O. Muthen
posted on Thursday, February 17, 2011 - 3:52 pm
Regarding your first post, it is difficult to separately interpret effects on linear and quadratic slopes. This is why sometime "orthogonal polynomials" are used. The effect on the intercept is straightforward, however, and one approach to this issue is shown in
Muth�n, B. & Muth�n, L. (2000). The development of heavy drinking and alcohol-related problems from ages 18 to 37 in a U.S. national sample. Journal of Studies on Alcohol, 61, 290-300.
which is on our web site under Papers.
Regarding the choice of standardization, see the UG.
Bengt O. Muthen
posted on Thursday, February 17, 2011 - 3:56 pm
When you say run it twice, I think you don't mean for boys and girls but with and without BMI. If so, it seems difficult to capture the changing gender role in one model. The growth is different with BMI as a tvc. I wonder if having BMI as a parallel growth process instead of as a tvc would be useful.
Olli Kiviruusu
posted on Wednesday, February 23, 2011 - 7:27 am
I'm analysing a latent growth curve model with four timepoints and time-invariant and time-varying covariates. I'm interested in the total effect of TVCs on the growth factors, especially on the mean (or intercept) of the slope factor i.e. does the mean growth rate change significantly after TVCs are specified. In the model without the TVCs the intercept of the slope factor is .35 and in the model with the TVCs .45 indicating that growth rate would be higher if the effects of TVCs were removed from the equation. How can I assess the significance of this change? Is it okay to constrain the intercept of the slope factor in the model with TVCs to the value it had in the model without TVCs and then analyse the chi-square change in model fit? Or is there a better/correct way to do this?
Thanks in advance.
Bengt O. Muthen
posted on Wednesday, February 23, 2011 - 5:07 pm
No, that doesn't sound correct. As a first step you want to think about how to make the question well defined. What is the intercept/mean of the slope growth factor when the model includes the TVCs - does it mean the same thing as when TVCs are not included? When included, does your model let the TVCs influence the slope growth factor? If not, doesn't the slope refer to the development of the Ys at zero values of the TVCs? Which raises the question, are the TVCs centered (sample means subtracted)?
Olli Kiviruusu
posted on Thursday, February 24, 2011 - 6:44 am
My model looks like this:
! Non-linear crowth curve;
ylevel yslope |
! TICs;
ylevel on tica ticb;
yslope on tica ticb;
! TVCs/concurrent effects;
y1 on tvc1;
y2 on tvc2;
y3 on tvc3;
y4 on tvc4;
! TVCs/lagged effects;
y2 on tvc1;
y3 on tvc1 tvc2;
y4 on tvc1 tvc2 tvc3;
Y is a personality variable and TVCs are the number of certain types of events. Regressions of Ys on TVCs show small, but significant negative effects. If I understand your point right, to me, the essential meaning of Ys and hence (I think) its growth parameters is the same whether TVCs are specified or not. I can do the centering of TVCs, but the interpretation of growth parameters at zero number of events as the TVCs now stands is also well motivated. TVCs do not directly influence the growth factors - actually regressing the slope factor on the TVCs would in a way be the easiest solution to my problem, but I don't think that it is allowed here, or is it?
Thanks again.
Bengt O. Muthen
posted on Thursday, February 24, 2011 - 10:37 am
Say that the TVC means decline linearly over time. Then the direct negative effects onto the Ys will help pull down the Y means instead of the slope mean being the only source affecting the Y means. This affects the interpretation of the slope mean changing across the two models.
You can regress the slope on TVCs. For instance, TVC1 happens before the slope affects the change from time 1 to time 2. TVC1 might also be correlated with the intercept.
These points illustrate the complexity of models with TVCs.
Olli Kiviruusu
posted on Friday, February 25, 2011 - 9:12 am
Thanks for your help.
I regressed the slope (and intercept) factors on TVC1 (and on TVC2 in a three timepoint model) and there were no significant effects on the slope. These models seem to me less than perfect however as TVCs 3 and 4 can't (I think) be used in them.
If there is no way to assess the joint effect of all TVCs on the growth factors I guess I need to consider some other models than LGC. Any suggestions?
Bengt O. Muthen
posted on Friday, February 25, 2011 - 1:28 pm
You might want to take a look at how intercept changes can be modeled by TVCs - see slides 157-159 of the Topic 3 handout of 05/17/2010.
You can also formulate a growth model for the TVC process and do parallel growth modeling where the TVC growth factors influence the growth factors for the Y process.
Elizabeth Adams
posted on Tuesday, March 08, 2011 - 3:15 pm
I am modeling gender and race centrality as predictors of change in cross race contact.
In terms of the analysis interpretation we are unsure of how to interpret the output from MPlus (gender is 0=female and 1=Male)).
Bengt O. Muthen
posted on Tuesday, March 08, 2011 - 6:42 pm
You interpret these slopes just like you would in a regular linear regression with a continuous dependent variable - that is, if INT was observed and if SLOPE was observed.
Ingrid Holsen
posted on Saturday, May 07, 2011 - 2:53 am
I have a question regarding the interpretion of BMI as a time varying covariate. It is a latent growth curve, i s and q, outcome; body image at 6 ages from 13 to 30, also background variables. The time varying covariate BMI has a sign neg estimate for males at age 13 (-.18) and 30 (-.27), but a sign. pos at age 21 (.04, p<0.01). Females have a sign pos at age 21 and a neg at age 30. I have checked this several times now, it seems correct.
So BMI has an additional effect on body image at this ages; at ages 13 and 30 boys� relatively high BMI led to further decline in body satisfaction, while at age 21 the opposite occured?
Or am I interpreting the estimates with time varying covariates wrong here? How is it best to express it?
Many thanks!
Linda K. Muthen
posted on Saturday, May 07, 2011 - 8:12 am
See the following book whcih has a section on the interpreation of time-varying covariates:
Bollen, K.A. and Curran, Patrick, J. Latent Curve Models: A Structural Equation Modeling Perspective. Wiley 2006.
Ingrid Holsen
posted on Sunday, May 08, 2011 - 2:42 am
Thanks for quick reply! My worry is
A) that the results (see above) might be incorrect. Based on previous research I just cannot see how a positive prediction age 21 (BMI (tvc)/body satisf.) can be correct, particularly not for girls. Also the correlations are negative, around -.30.
B)The fit measures for the model are not that good, cfi .93, RMSEA 0.04, SRMR 0.08. I have looked at mod. indic.: When I include a path Q on bmi30 the fit is much better, cfi .97, RMSEA 0.03 and SRMR 0.05.. Chi square much lower too (but still sign, sample is 1082). It makes sense to me that BMI30 predicts Q; males (-.20), females (-.39). (curve in body image adulthood levels off), but can I do that? Then BMI age21 are no longer positive! and not significant, all effects go through Q.
Linda K. Muthen
posted on Sunday, May 08, 2011 - 6:20 am
A) This I cannot comment on without more information than can be handled on Mplus Discussion.
B) If the model doesn't fit, any interpretation of the results is invalid.
Ingrid Holsen
posted on Sunday, May 08, 2011 - 6:34 am
Ok, we are struggling with this. Can I post more information here or can we do it another way?
posted on Monday, August 22, 2011 - 2:38 am
I'm analyzing a quadratic GMM with four timepoints and covariates. One covariate has a significant influence on the linear slope factor, but insignificant influence on the quadratic factor. How can I interpret this? Does this mean that the covariate only affects the change between T1 and T2 and after this there is no influence?
Thanks a lot
Bengt O. Muthen
posted on Monday, August 22, 2011 - 10:07 am
Not quite. Telling effects on the linear and quadratic parts growth factors is difficult because those two factors interact. The covariate that influnces the linear factor significantly continues to have an influence after T2 because the linear slope continues to have an influence beyond T2. But beyond that it is hard to parse out the influences via the two growth factors.
Karen Offermans
posted on Friday, September 09, 2011 - 1:22 am
Dear Linda and Bengt,
I am analyzing the following longitudinal growth model on a continuous variable (perceived alcohol availability), including multiple group analyses on age:
GROUPING is age (13=13 14=14 15=15);
iPalav sPAlav |
iPAlav on group1;
sPAlav on group1;
I get the following warning in the output:
However I do not know what to check in the technical output 4 and what I can conclude from this??
I hope you can help me further.
Karen Offermans
posted on Friday, September 09, 2011 - 1:55 am
Just to give some more information in relation to my previous question;
In the tech 4 outcome concerning 14 year olds Mplus doesn't give the correlations between sPalav and the other latent variables (stated as 999) or itself. Furthermore, in the estimated covariance matrix for latent variables I can see a negative covariance between spalav and spalav of -.107.
Is there anything we can do to solve this problem?
Looking forward to your reply.
Linda K. Muthen
posted on Friday, September 09, 2011 - 7:01 am
It sounds like spalav has a negative variance. This makes the model inadmissible. You would need to change the model.
posted on Monday, April 09, 2012 - 4:55 am
I have two questions about this formula for calculating outcome variable probabilities (binary y scored 0/1) at different levels of a covariate over time, in categorical growth models:
"P (y=1 |x) = F(-tau + s_m*x_t),
where F is the normal distribution function, tau is the threshold parameter held equal across time, and x_t are the time scores (the slope loadings).
The mean value of the slope is obtained in TECH4 and is s_m = a+g*x for covariate value x, where a is the intercept of the slope factor and g is the regression coefficient for the slope regressed on x"
1. The mean value of the slope obtained in TECH4 is different from the the intercept of the slope factor. If the intercept of the slope factor is used in the formula, why is the mean value of the slope obtained in TECH4 relevant?
2. Is this formula the same for logit and probit coefficients? If different, how should it be modified?
Linda K. Muthen
posted on Monday, April 09, 2012 - 9:12 am
1. The mean and intercept are different parameters. The mean of y, y_bar, is
y_bar = a + b*x_bar;
The intercept is
a = y_bar - x_bar.
2. See Chapter 14 of the user's guide. There is a section on probit and another on logit.
posted on Monday, June 11, 2012 - 6:48 am
I have a question regarding the interpretation of my growth model. I am looking at depression (outcome) across four time points and assessing the influence of time-variant and invariant predictors on the slope and intercept of the depression curve. The UGM suggests that depression decreases from mid adolescence through young adulthood (I used a freed-factor loading approach). I am a little confused as to how to interpret the regression coefficients because most of the examples have positive slopes. Here are my questions:
1. if the estimate for gender (boys) predicting the slope of depression is positive, does this mean that boys have a slower rate of decline in depression over time?
2. The time-varying covariate also has a declining slope. The regression estimate for the effect of the intercept of the TVC on the slope of depression is positive (.085). Does this mean that those people with higher levels on the TVC have a slower rate of decline in depression?
Linda K. Muthen
posted on Monday, June 11, 2012 - 4:56 pm
1. Yes.
2. Yes.
posted on Thursday, November 01, 2012 - 7:02 am
Suppose I have a parallel process growth model with categorical outcomes. The intercepts and slopes are regressed on covariates, and the intercepts and slopes are correlated.
For each covariate, I have calculated probabilities using the formulae in the discussion above: (1) s_m = a+g*x for covariate value x (2) F(-tau + s_m*x_t).
How can this formula be modified to estimate the probability at the first time point that the one outcome is already present, when the other outcome is already present at baseline? The two intercepts are correlated, so I want to illustrate that someone already having outcome 1 is more likely to already have outcome 2 at baseline. The correlation between the intercepts is captured by a covariance rather than by a regression coefficient.
Bengt O. Muthen
posted on Thursday, November 01, 2012 - 8:59 pm
You say
P (y=1 |x) = F(-tau + s_m*x_t)
but that is P(y=1 | x, s=s_m). If s is a random effect, to get P(y=1 | x) you have to integrate over s.
P(y1=1, y2=1 | x) requires bivariate integration over s1, s2.
Cameron McIntosh
posted on Wednesday, December 05, 2012 - 6:40 pm
I am estimating longitudinal models with binary and ordinal observed outcomes using the multilevel features (TWOLEVEL RANDOM) in Mplus (as opposed to the SEM/LGCM approach). I know that with SEM/LGCM, the regressions of growth factors on covariates are linear regressions (as the intercept and slope are continuous latent variables with arbitrary metrics) and the factor loadings for the intercept and slope are fixed logit or probit coefficients. One can, however, get predicted probabilities for the observed categorical indicators of growth by combining the appropriate parameter estimates.
But in the case of MLM, where the intercept and slope are estimated by directly regressing the categorical outcome on an observed ordinal time variable using stacked (long) data (rather than creating a measurement model for the growth factors) and a logit or probit link, can the estimated growth parameters themselves be more directly interpreted on the logit (or probit) scale... following which one could simply convert these to odds ratios and thus predicted probabilities? Is this correct or am I missing something?
Bengt O. Muthen
posted on Wednesday, December 05, 2012 - 8:29 pm
I think this is the right way to look at it. This is like UG ex 9.16 but with categorical outcomes. Here x1 and x2 influence s and y on the Between (subject) level) which in turn influence the outcome at each time point as the figure implies. So x1 and x2 ultimately have a logit/probit influence on the outcomes.
Dustin Pardini
posted on Thursday, May 16, 2013 - 9:41 am
I am using a simple linear growth curve to predict a distal outcome. In explaining the contribution of the slope as a predictor, in addition to interpreting the estimates provided, is it practical to convey this information by using R2? Specifically, would it be feasible to run the model with only the intercept or slope being used as a predictor (ex.- y on I), then the same model but with both I and S as predictors (y on I S) and report the differences in R2 between these two models?
Bengt O. Muthen
posted on Thursday, May 16, 2013 - 11:28 am
That doesn't seem unreasonable, as long as i and s are not too highly correlated.
B Chenoworth
posted on Wednesday, June 12, 2013 - 2:11 am
I am estimating a linear latent growth curve model across three time points using ordinal-categorical data. Specifically the response scale of the outcome variable is a 4-point likert scale (none, 1-2 days, 3-5 days, 6-7 days). As such I have used WLSMV to estimate the model.
In the output, the mean of the intercept is fixed at 0 and the mean of the slope is estimated. The output tells me that the mean of the slope is -0.3 (p<.05). So the outcome variable decreases by 0.3 points between each time point.
Reviewers of my work continue to ask me, what does this mean in term of how much change occurred. Because the outcome variable is ordinal-categorical I am finding it difficult to answer this.
Would this best be answered in terms of calculating an effect size for the slope, and if so, how would I do this?
Thank you in advance for your help.
Linda K. Muthen
posted on Wednesday, June 12, 2013 - 1:13 pm
I think looking at a plot of probabilities would be helpful. See the SERIES option of the PLOT command.
B Chenoworth
posted on Wednesday, June 12, 2013 - 3:15 pm
Thank you for your response Linda.
Just to follow up, it is possible to calculate an effect size for the slope, in order to report that the decrease represented a small, medium, or large change?
Bengt O. Muthen
posted on Wednesday, June 12, 2013 - 3:39 pm
It is possible, but does not convey how that impacts the probabilities of the observed variables.
Ivana Igic
posted on Wednesday, August 14, 2013 - 3:17 am
Dear Drs. Muthen,
I�m running a 3-step GMM (Mplus Web Notes: No. 15). In the first step after I have tested different models I got a 5 class curvilinear solution as the most suitable.In the 3-step I have predicted the distal outcome in T5, while I have controlled for the T1 value of distal outcome.
1.Is the intercept value of distal outcome within the class the mean value of the distal outcome per class?
2. The values of distal outcome are 1-6, what is wrong if I got the negative values for intercept or the values higher than 6 for the intercept per class?
3.I also analyzed the same data in SPSS using ANCOVA and I got very different values for the estimated mean value of distal outcome per class.
4.I used Wald test for distal outcome means comparison as suggested but this doesn�t work. Did I do something wrong?
[t5_y ] (m1);
[t5_y ] (m2);
[t5_y ] (m3);
[t5_y ] (m4);
[t5_y ] (m5);
Model test:
Thank you very much for your help.
Linda K. Muthen
posted on Wednesday, August 14, 2013 - 9:17 am
1-3. The intercepts not means are being estimated.
4. Remove
The other tests imply those tests.
Ivana Igic
posted on Wednesday, August 14, 2013 - 10:49 am
Thank you very much for answering me!
1. I want to compare the value of distal outcome within diff. classes, how should I then interpret the intercept values?
I want to be able to say that people within one class feel better/worse (my distal outcome) compare to people in other classes and to test the significance of these differences using the wald test.
2.The model test is still not working.
Thank you very much for your help and have a nice day.
Linda K. Muthen
posted on Thursday, August 15, 2013 - 8:45 am
1. The intercept is the mean controlled for the covariate.
2. Please send the output and your license number to
Maike Theimann
posted on Wednesday, August 28, 2013 - 12:37 am
my question relates to conditional linear latent growth curve models.
I have an unconditional model which shows that the endogenous variable declines over time. If I regress the slope Factor on an exogenous variable, its effect on the slope factor is negative.
Unconditional model:
Unstandardized Means
Conditional Model:
As you see there is a negative effect of AGSLB on the slope Factor. Does this mean that if AGSLB increases by one, the curve of the endogenous variable will move (0.194 units) towards zero?
And in general does a positive effect on the Slope-Factor mean that if the exogenous variable increases, the curve of the endogenous variable will move more into the direction it has in the unconditional model and a negative Effect that it will move more towards zero, indifferent of the curve�s shape in the unconditional model?
Bengt O. Muthen
posted on Wednesday, August 28, 2013 - 6:20 pm
A covariate that has a negative effect on a slope is interpreted as follows. As the covariate value increases, the slope value decreases. It doesn't matter if the slope mean is negative or positive. If the mean is negative, increasing covariate value beyond its mean makes it even more negative.
Martina Narayanan
posted on Tuesday, August 12, 2014 - 3:54 am
I am running a linear growth curve model. The mean slope (-0.003) is not signifcant and negative, but the variance of the slope (0.002) is significant. I am testing how the slope predicts an outcome variable, and I find a significant and positive unstandardized regression coefficient (0.245) for slope predicting that outcome.
How do I interpret this finding?
I know a positive regression coefficient means, that with an increasing slope, the outcome variable increases. But since my slope is slightly negative to begin with, does that mean the more negative my slope, the more increased the outcome? Does this logic apply even when my slope is not significant to begin with (it may be sort of random that it is slightly negative)?
Linda K. Muthen
posted on Tuesday, August 12, 2014 - 10:34 am
There is variability around your slope mean. It can increase by going from, for example -1 to -.5. This increase is associated with an increase in the outcome.
Martina Narayanan
posted on Tuesday, August 12, 2014 - 1:08 pm
Thank you Linda. Just a follow up question to make sure I understand you right:
When you talk about a change in a negative slope from -1 to -.5 you would refer to that as an increase in the slope? Because I thought this would be called a decrease as the absolute value of the slope is decreasing. Could you once again explain what we call an increase or decrease in the case of a negative slope? I believe, this difference is quite important for how we interpret the regression coefficient from the slope to the outcome.
Thank you, Martina
Linda K. Muthen
posted on Tuesday, August 12, 2014 - 2:58 pm
We aren't talking absolute values. We are talking the real values.
Nick Shryane
posted on Thursday, June 25, 2015 - 9:11 am
I have five longitudinal binary outcomes. I�m fitting linear growth curve models and I want to check my understanding of the output by computing the model predicted probability of the outcome at the first time point.
Fitting using probit/WLS, I�m using the equation from chapter 14 of the manual (v7, p.492):
Pr(u=1|x_t) = F(-tau + slope*x_t),
where F is the normal distribution function, tau is the threshold parameter (same for all time points), and x_t are the time scores (the slope loadings).
So, for x_t = 0 the probability of u is given just by �tau. Fitting the model, I obtain tau = 0.721, which is Pr=0.235, which fits the observed proportion pretty well (0.225).
My difficulties start when I fit the model by logit/MLR. When I fit the same model I get a value of 3.67 for the thresholds. To compute a probability I use equation (1) (p.493), with tau = -a,
Pr(u=1|x_t) = 1/(1+exp(tau - slope*x_t)),
so again with x_t = 0 it should be 1/(1+exp(tau). But this gives me Pr = 0.023 for the fitted threshold (3.67).
When I use plot3 to graph the predicted probabilities they come out fine (i.e. around Pr = 0.235), so I presume I�m doing something wrong.
Any help would be appreciated.
Bengt O. Muthen
posted on Thursday, June 25, 2015 - 6:33 pm
First, do you have random effect, that is, growth factors with variances? If so, the computations have to take that into account by numerical integration.
Nick Shryane
posted on Friday, June 26, 2015 - 5:39 am
Thanks for getting back to me. Yes, the model has latent growth intercept and slope factors. The model statement is:
i s |
I'm interested in the predicted probability for u1.
I thought that, for time 1 (u1), the contribution of the latent growth factors would be zero.
This assumption appeared to hold (i.e. the equation above produced the correct answer) when estimating the model by probit/WLSMV but not when using logit/MLR.
Nick Shryane
posted on Friday, June 26, 2015 - 6:23 am
A quick update to my query.
Thanks for your mention of the latent variable variances - when I fit the logit/MLR model with these constrained to zero the threshold on its own gives the predicted probability I was expecting (logit 1.043, Pr(0.26)).
But I'm still curious as to why this should matter for predicting the probability using MLR but not for WLSMV.
Bengt O. Muthen
posted on Saturday, June 27, 2015 - 4:41 pm
The WLSMV model uses a probit link together with the normality for the growth factors and this results in a normal y* which gives an explicit form for the probability in terms of the normal dist fcn.
The ML model uses a logit link together with the normality for the growth factors which does not give an explicit form but needs numerical integration.
Nick Shryane
posted on Monday, June 29, 2015 - 7:36 am
Aha. Thanks for this. One final question in this case - what is the interpretation of the item threshold parameter in such a logit model?
I have always assumed it was the logit of the probability of an item response for a case located at zero on the latent intercept and slope parameters, but it seems that this isn't the case.
Bengt O. Muthen
posted on Monday, June 29, 2015 - 11:22 am
When you condition on zero latent variable values, the threshold interpretation is as you say (although with logit that reverses the sign of the threshold). It is when you don't condition that you need the numerical integration.
Kelly Murphy
posted on Wednesday, August 05, 2015 - 11:15 am
I estimated a parallel process model/dual domain model (two latent growth curve models at once), and am having trouble interpreting the covariance between the two slopes. How do I interpret a negative covariance between a negative slope and a positive slope?
Thank you!
Bengt O. Muthen
posted on Thursday, August 06, 2015 - 7:25 am
When the positive slope increases the negative slope decreases - becomes more negative.
Kelly Murphy
posted on Thursday, August 06, 2015 - 2:06 pm
Thank you so much for taking the time to respond to my question, I sincerely appreciate it.
So would a positive covariance between a negative slope and a positive slope mean that when the positive slope increases, the negative slope decreases less?
Bengt O. Muthen
posted on Friday, August 07, 2015 - 5:44 pm
Marike Deutz
posted on Thursday, March 03, 2016 - 2:22 am
Dear Linda and Bengt Muth�n,
I have a cohort-sequential LGM with distal outcomes and I have some trouble interpreting the results. The linear slope is positive (.255) and the quadratic slope is negative (-.191). I understood from the forum to always interpret them together. I plotted the growth curve and there is an increase first and then a decrease.
1) the linear and quadratic slope are negatively related. Can I interpret this relation? It seems obvious that they are related as a linear increase is followed by quadratic decrease?
2) The linear slope positively and the quadratic slope negatively predicted 3 of my outcomes. Does that mean that a developmental course with a steeper increase and a steeper decrease, so a higher overall curve, predicts my outcomes?
3) would you say the intercept is the initial level? Or a constant determined by all data waves?
Thank you!
Bengt O. Muthen
posted on Thursday, March 03, 2016 - 6:55 pm
Consider centering the time scores to reduced the s, q correlation.
Some answers are in the paper on our website:
Muth�n, B. & Muth�n, L. (2000). The development of heavy drinking and alcohol-related problems from ages 18 to 37 in a U.S. national sample. Journal of Studies on Alcohol, 61, 290-300.
download paper contact first author show abstract
Adam Milam
posted on Wednesday, October 26, 2016 - 9:40 am
I am conducting parallel process GMM with continuous indicators...new at this. Both processes have 4 time points. I am having difficulty interpreting the results. I ran separate GMM to identify the appropriate number of classes (process 1 - 4 classes; process 2 - 3 classes). When I run the parallel process the class structure seems to change (i.e. different slope and intercept); how do I interpret the different classes now with the 12 class patterns? Should the classes hold up and should there be consistency of slope and intercept for the different class patterns (should pattern 1 1 and pattern 1 2 have similar slope and intercept for process 1?). Also how do I find the conditional probabilities of class membership given membership in another class in the Mplus output.
Bengt O. Muthen
posted on Wednesday, October 26, 2016 - 2:46 pm
It all depends on how you set up the model. I would use 2 latent class variables, one for each process, and then use the dot command like (see UG):
to impose exactly the equality constraints you want.
Daniel Lee
posted on Wednesday, December 07, 2016 - 12:07 pm
Hi Dr. Muthen,
I ran a multigroup conditional LGM, and I am having some trouble wrapping my head around the fixed effects. In particular, the mean of the intercept factor (=~4.80 [on a Likert type scale was 0-5]) for one of the groups was extremely high! However, when I looked at the mean of dependent variable at the first time point for that group, the mean was a lot less. I know that the mean of the intercept is an adjusted mean at the first time point for a particular group, but the intercept of this particular group seems way to high (especially since most of the covariates are not significant).
I was wondering if my understanding of the mean of the intercept is correct (an adjusted mean for a group at the first point), and if there are other factors in a multigroup-LGM that might inflate the intercept estimate (especially if most covariates are not significant).
Bengt O. Muthen
posted on Wednesday, December 07, 2016 - 2:45 pm
Please send the output to Support along with license number - so we know exactly what you are looking at.
Emma C Burns
posted on Friday, June 02, 2017 - 12:09 am
Hi Dr. Muthen,
Could you please help clarify the following interpretation:
If the mean slope of the outcome is significantly negative, and when that is regressed on a covariate, the regression coefficient is also negative, does that mean the covariate is predicting a more or less steep slope?
We know that a positive regression for a positive slope means the steepness of the slope increases, but we are unsure how to interpret it for negative slopes.
Bengt O. Muthen
posted on Friday, June 02, 2017 - 2:06 pm
It is simple:
If x increases by a certain amount, the negative coefficient says that a negative value is added to the slope. This means that the slope decreases. So a big x value gives a larger negative slope, that is, a more steep slope downwards.
Uzay Dural
posted on Monday, June 12, 2017 - 12:30 am
Hello Drs. Muthen,
I conducted conditional multiple-group (experimental versus control groups) MLGM. The interaction between gender (men = 1) and a time invariant covariate is significantly predicting the negative slope in the experimental group. (gender x covariate --> slope = -.237).
Does this mean: as the covariate increases the slope decreases for men (compared to women) in the experimental group?
It is probably not, and I am confused. Due to limited sample size I could not conduct 4-group MLGM (female experimental, female control, male exp., male control). Instead, should I focus on the experimental group and conduct multiple group MLGM with gender groups (covariate --> slope)? Or which post-hoc analysis should I consult to?
Thank you very much in advance!
Bengt O. Muthen
posted on Monday, June 12, 2017 - 6:06 pm
Q1: Yes
Q2: With limited sample size you may want to represent the 4 different groups by 3 dummy variables.
Uzay Dural
posted on Tuesday, June 13, 2017 - 1:02 am
Thank you very much Dr. Muthen!
A follow up: is it possible to get standard errors of model estimated means to plot the interaction effects?
Bengt O. Muthen
posted on Tuesday, June 13, 2017 - 6:29 pm
Yes, you can even get plots with confidence intervals for interactions - see the Table 1.8 runs on the web page for our book examples:
Daniel Lee
posted on Wednesday, September 27, 2017 - 6:06 am
Hello, I ran an growth model for variable Z and the intercept and slope term was significant. However, upon including a predictor (X1), the slope term for Z was no longer significant, but X1 significantly predicted the slope of Z (e.g., .50). Can I interpret this result even though the slope term for variable Z is not significant in the model?
Thank you!
Bengt O. Muthen
posted on Wednesday, September 27, 2017 - 3:31 pm
Perhaps you are looking at the intercept instead of the mean for the slope when you add the covariate.
Daniel Lee
posted on Thursday, September 28, 2017 - 5:01 am
Hi, in response to your previous post, the intercept was still significant after including the covariate (i.e., x1), but the slope was no longer significant. But x1 significantly predicted the non-significant, slope term.
I'm wondering if that means that the trajectory is flat when the covariate is at 0, but an increase to the covariate increases the rate of change in the growth factor (effect size of x1 to s was .5). I would love your input.
Bengt O. Muthen
posted on Thursday, September 28, 2017 - 10:32 am
I was talking about the intercept for the slope growth factor regressed on the covariate.
Aida Soriano
posted on Monday, October 02, 2017 - 1:41 am
Dear Dr. Muthen,
I�m working on a LGM, in which i and s of a variable influence i and s of in-role and extra-role performance. Also I included some predictors. The fit was not good. Modification indices suggested to include residual covariances between both outcomes (same time) and between one outcome (different times). If I include the ones that you see below, I got good fit. But, does it have sense? How can I justify including a residual covariance between INPE2-EXPE2, and not between INPE1-EXPE1, INPE3-EXPE3..??
ix sx|
iy sy|
iye sye|
iy ON ix X1;
sy ON sx;
iye ON ix X1;
sye ON sx;
ix ON X1 X2 X3;
iy WITH
; !they were non-significant so I fixed it to 0
EXPE4 WITH INPE4; !in-role and extra-role in the same time
INPE3 WITH INPE1; !same variable in diff times
Thank you,
Bengt O. Muthen
posted on Monday, October 02, 2017 - 5:16 pm
This question is suitable for SEMNET.
Aida Soriano
posted on Monday, October 02, 2017 - 11:01 pm
What does this mean? Should i post it in SEMNET?
I'm really interested in knowing your opinion about covariate the residuals suggested by the modification indices or not.
Thank you,
Bengt O. Muthen
posted on Tuesday, October 03, 2017 - 11:45 am
A priori, I would include correlations between outcomes at the same time point for all time points and all outcomes at each time point. This is because there are presumably many left-out time-specific covariates that influence all outcomes at a certain time point.
The outcomes may also have residual correlations across time due to left-out covariates that influence all time points.
AT Jothees
posted on Monday, April 16, 2018 - 5:07 am
Dear Linda,
I am running higher order latent growth curve analysis with multiple indicators measured at five occasions. I have managed to save the slope and intercept score for presentation of trajectories in figures. When I plot slope (e.g cognitive function) against age. I see that y value (slope) range from -3, 0, 3; and with increasing age (x value), there is a significant decline in slope.
I have a question:
a) when the slope trajectory line cross "0" on the y-axis, is this means that there is no difference in slope for reference group in x-axis?
Bengt O. Muthen
posted on Tuesday, April 17, 2018 - 4:02 pm
No - y=0 means that you have a flat trajectory at that x value.
posted on Tuesday, August 07, 2018 - 3:53 pm
Is it possible to have two groups that do not have a significantly different initial status but one group is significantly faster (slope is greater for one group and WALD test on slope shows significance) then the other yet they do not end up much different from one another (this over 4 time points)? I have this result and I'm not sure what to make of it.
Bengt O. Muthen
posted on Tuesday, August 07, 2018 - 4:50 pm
I assume you are talking about a linear growth model with fixed time scores. If so, yes I can imagine this happening with a large sample making the slope mean significant but not resulting in an outcome difference that is substantively important.
Wim Van den Broeck
posted on Tuesday, August 14, 2018 - 2:16 am
I have a dataset for which I want to test factorial measurement invariance over 8 time points (1 factor of math achievement). At each time point there are 60-80 dichotomous items, of which a few are anchor items. Is it computationally feasible to do this in MPLus? Just asking before I proceed.
Tihomir Asparouhov
posted on Tuesday, August 14, 2018 - 1:55 pm
It shouldn't be a problem. See
All three methods should be feasible: alignment, BSEM, random loadings.
Jennifer Barton-Crosby
posted on Monday, November 05, 2018 - 8:50 am
I�d like to check whether or not my interpretation of a dual process growth model in which the growth factors of process A predict the growth factors of process B is correct. Both processes follow a quadratic curve with an initial decrease followed by an increase. Essentially, I would like to establish if negative relationships between growth factors correspond with steeper increases and/or decreases, and if positive relationships correspond with slower/gentler increases and/or decreases.
For example:
When the intercept of process A predicts the intercept of process B, the relationship is positive � does this mean that high initial levels of process A predict high initial levels of process B?
When the intercept of process A predicts the linear slope of process B and the relationship is positive, does this mean that high initial levels of process A predict a slower decline in process B?
When there is a positive relationship between the linear slopes of both processes, does this mean that a slow decrease on process A predicts a slow decrease on process B?
Thank you.
Bengt O. Muthen
posted on Monday, November 05, 2018 - 3:49 pm
When you say "predict", you need either time ordering between the 2 processes - A happening before B - or strong theory to back that.
Q1-Q4: Yes
posted on Wednesday, July 24, 2019 - 1:43 am
Dear Drs. Muthen,
I have run a conditional 3-class GMM-CI model.
When inspecting the logistic coefficients vs the odds ratio of the three different predictors, the related p values change.
For example:
Why does this happen?
Which one should I use in a report then? I was actually using odds ratio.
Yours Sincerely,
Bengt O. Muthen
posted on Wednesday, July 24, 2019 - 5:34 pm
Q1: see our FAQs on Odds ratios
Q2: Use CIs for ORs.
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When performing a restoration, a rotisserie is often helpful, allowing one a comfortable working position while providing access to places that would otherwise be nigh-on impossible to get to without an expensive lift. We came across this ingenious solution while up at Trevor's a couple of months back — Trev's restoring a Mini that used to be a crisps delivery van in Dublin. It's in relatively good shape for a 40-year old car that was never seen as more than an appliance, but the suspension needed to be rebuilt and there was a healthy amount of floorpan corrosion.
For a light, unibody car like a Mini, Trev's rough-but-elegant solution is perfect. After gutting the interior and yanking the front and rear subframes, he bolted up these simple rockers made of tubing and angle iron to the body, which affords him 0, 45 and 90-degree working positions. It's really stable, and it can easily and safely be moved from position to position by one man. Note, while this same technique could be applied to other, larger vehicles, such as Camaros or Mopar B-Bodies, we'd recommend heavier construction, and at least a couple of guys to rotate the car. Something as light as a stripped-out Mini body doesn't have a whole lot in the way of inertia, making it highly unlikely to over-rotate when flipping the vehicle. But for a shadetree resto project like Trev's, we can't think of a more KISS-simple solution.
And You Will Know Us by the Trail of Parts: Using Car Ramps [Internal]


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Dr Hiten Vyas is the founder of Stuttering Hub Limited, a UK-based life coaching company. He is an NLP Master Practitioner, trained in the United States and uses NLP in his coaching practice. He offers face-to-face coaching locally in Leicestershire UK, and internationally through telephone coaching.
Early on, almost every aspect of his life was crippled because of stammering. He then entered the self-development field in 2003, and he has successfully created positive and long lasting changes in his life.
It is his passion, to help people overcome confidence and anxiety problems and achieve the lives they want.
In his spare time he enjoys improvisational comedy, public speaking, yoga, meditation and travelling. He is also a major UK blogger in the area of personal development and has been doing so since 2008.
He has a PhD in Biomedical Information Systems.
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REDWOOD CITY — A former San Mateo County real estate broker faces dozens of fraud charges after fleecing three clients, including an accomplice's relative and a friend, out of more than $803,000, a prosecutor said Monday.
Andrew Abed Khoury, 42, who worked out of Daly City but moved to Arizona City, Ariz., pleaded not guilty Friday in San Mateo County Superior Court to 36 counts of real estate fraud. Between May 2006 and May 2012 he used the ruse of real estate investment to scam three people, said District Attorney Steve Wagstaffe. Cindy Teahan, who is a co-defendant in the case, said one of the victims was her mother-in-law.
Teahan, 47, of Pacifica, faces six felonies, including three counts of falsifying documents, in connection with the fraud. Prosecutors say she worked at several escrow companies and knew Khoury after doing a lot of business with him.
"I didn't know a thing about it. I didn't forge any documents," Teahan said.
The purported scams vary between the three victims. In one case, Khoury convinced a 59-year-old female friend to refinance her two houses and then give the proceeds to him for a "real estate investment," Wagstaffe said. She lost $511,215.
In another, he's accused of talking Teahan's 83-year-old mother-in-law into handing over $202,000 she made from selling her house. Wagstaffe said the woman lived in a seniors home and was in poor health — a description Teahan refuted.
During a final instance, Khoury is accused of convincing a 74-year-old man to give him $100,000 to invest. He told the man the money was secured by a "deed of trust" in a secure property, but it wasn't, Wagstaffe said. Instead, the man's cash purportedly went to pay down Khoury's credit card bills.
Khoury was being held at San Mateo County Jail in lieu of $1 million bail. He's due back in court June 12, said Wagstaffe. Defense attorney Majeed Samara declined to comment on the case.
Teahan is out of custody on $100,000 bail, but it wasn't immediately clear when she was due in court.
Contact Joshua Melvin at 650-348-4335. Follow him at Twitter.com/melvinreport.


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Antwerp is located about 150 km from Brussels and was home to the master of the Baroque style, Paul Peter Rubens. The city has the largest Gothic church in the Low Countries, where you can admire two of the most beautiful masterpieces of the baroque painter. Antwerp is the second largest town of Belgium, the largest port of Belgium and the second largest port of Europe. You will be pleasantly surprised by the history of the city and by the fusion of architectural styles from various periods of time.
• Departure from Brussels by train with our guide, who will tell you about the history, the architecture, the legends and much more for a visit of maximum 3 hours,
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• Free time of 3 hours,
• Return to Brussels by train with our guide. If you would like to stay in Antwerp for a longer time, our guide will give you all the necessary information to return to Brussels.
Price: 34€
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Starting time: 9:30 am
Meeting point:
• Grand-Place in front of the "Tourist Information Office". Find our guides! They hold a white umbrella
Duration: 9 hours
Language: English
• Professional guide, expert in the history and the legends of Antwerp
• Round trip by train from Brussels (Brussels-Antwerp-Brussels)
Not included:
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Size of the group: to ensure the best service possible, our tours are limited to 25 participants per group. For larger groups, please contact us.
Bravo Discovery offers you two booking options:
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We are a group: we recommend our "Day trip to Antwerp" private guided tour starting from 220,00€. Send us a message to receive a personalised offer.
The Rubens House "Rubenshuis": the museum is open daily from 10 a.m. until 5 p.m. Price: 8€
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• The Central Train Station, known as the 'iron cathedral' and considered as one of the most beautiful railway stations in the world
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Yuste Monastery
Contact details
Monasterio de San Jerónimo de Yuste. s/n
10430 Cuacos de Yuste, Cáceres (Extremadura)
Tel.:+34 927172197
- Autonomous region: Extremadura
- Province/Island: Cáceres
Spanish emperor Charles was a great lover of gardens and nature, as can be clearly seen from his retreat to Yuste in 1554. He received news of the monastery's setting from Brussels: "… in a cool, deserted spot with water from many fountains, with endless fruit of all types, winter and summer… it is the most peaceful, mild and enjoyable spot that exists in Spain" – and he decided to build there, "a suitable house in which to be able to live with only the most vital personal servants." The emperor wished to be in contact with nature in this place to aid meditation and prayer. He would "take time to create a garden at the top, where there is a terrace, which he wants to cover over, placing a fountain in the middle surrounded by a garden with many orange trees and flowers, and he wishes to do the same in the lower part". Thus, he would create a pond with tiles that, along with the one on the lower terrace, would make the palace resemble the house where he was born in Ghent.
Practical information
- From Oct 01 to Mar 31
From Tuesday to Sunday
Public holidaysFrom 10:00 AM to 6:00 PM
- Closing days: Monday
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From Tuesday to Sunday
Public holidaysFrom 10:00 AM to 8:00 PM
- Closing days: Monday
- General: €7
- Reduced: €4
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Sony Ericsson C901 GreenHeart US Pictures
Sony Ericsson C901 GreenHeart is one of the first eco-friendly products of the manufacturer. Sony Ericsson has used reduced packaging materials, recycled plastics, optimized display light and waterborne paint in order to reduce the lifetime CO2 footprint of the device by 15%. C901 GreenHeart is a quad-band GSM with 3G support and sports a 2.2" QVGA display, 5-megapixel camera with Xenon flash.