# Non-Parametric Statistics, or What to Do When the Assumptions for a Parametric Test Fail

Question 1: A medical researcher believes the number of ear infections in swimmers can be reduced if the swimmers use ear plugs. A sample of ten people was selected, and the number of ear infections for a four-month period was recorded. During the first two months, the swimmers did not use the ear plugs; during the second two months, they did. At the beginning of the second two-month period, each swimmer was examined to make sure that no infections were present. The data are shown below. At $$\alpha = 0.05$$, can the researcher conclude that using ear plugs effects the number of ear infections?

Solution: We need to test the hypotheses

\begin{align} & {{H}_{0}}:\text{ ear infections are the same with or without the ear plugs} \\ &{{H}_{A}}:\text{ swimmers get less ear infections with ear plugs} \\ \end{align}

We use a Sign-Test. We use Statdisk to get the following output:

The $$x$$ statistics is equal to 2 (the less frequent number of signs). The critical value is 1. Since $$x$$ is not less than or equal to the critical value, we fail to reject the null hypothesis. This means that we don't have enough evidence to support the claim that the number of ear infections in swimmers can be reduced if the swimmers use ear plugs.

Question 2: Research indicates that people who volunteer to participate in research studies tend to have higher intelligence than nonvolunteers. To test this phenomenon, a researcher obtains a sample of 200 high school students. The students are given a description of a psychological research study and asked whether they would volunteer to participate. The researcher also obtains an IQ score for each student and classifies the students into high, medium, and low IQ groups. Do the following data indicate a significant relationship between IQ and volunteering? Test at the .05 level of significance.

Solution: The following table shows the corresponding contingency table:

 Observed High Medium Low Total Volunteer 43 73 34 150 Not Volunteer 7 27 16 50 Total 50 100 50 200

We are interested in testing the following null and alternative hypotheses:

\begin{align}{{H}_{0}}:\,\,\, \text{Volunteer Status}\text{ and }\text {IQ}\text{ are independent} \\ {{H}_{A}}:\,\,\,\text{Volunteer Status}\text{ and }\text {IQ}\text{ are NOT independent} \\ \end{align}

From the table above we compute the table with the expected values

 Expected High Medium Low Volunteer 37.5 75 37.5 Not Volunteer 12.5 25 12.5

The way those expected frequencies are calculated is shown below:

${E}_{{1},{1}}= \frac{ {R}_{1} \times {C}_{1} }{T}= \frac{{150} \times {50}}{{200}}={37.5},\,\,\,\, {E}_{{1},{2}}= \frac{ {R}_{1} \times {C}_{2} }{T}= \frac{{150} \times {100}}{{200}}={75},\,\,\,\, {E}_{{1},{3}}= \frac{ {R}_{1} \times {C}_{3} }{T}= \frac{{150} \times {50}}{{200}}={37.5}$

$,\,\,\,\, {E}_{{2},{1}}= \frac{ {R}_{2} \times {C}_{1} }{T}= \frac{{50} \times {50}}{{200}}={12.5},\,\,\,\, {E}_{{2},{2}}= \frac{ {R}_{2} \times {C}_{2} }{T}= \frac{{50} \times {100}}{{200}}={25},\,\,\,\, {E}_{{2},{3}}= \frac{ {R}_{2} \times {C}_{3} }{T}= \frac{{50} \times {50}}{{200}}={12.5}$

Finally, we use the formula $$\frac{{{\left( O-E \right)}^{2}}}{E}$$ to get

 (fo - fe)²/fe High Medium Low Volunteer 0.8067 0.0533 0.3267 Not Volunteer 2.42 0.16 0.98

The calculations required are shown below:

$\frac{ {\left( {43}-{37.5} \right)}^{2} }{{37.5}} ={0.8067},\,\,\,\, \frac{ {\left( {73}-{75} \right)}^{2} }{{75}} ={0.0533},\,\,\,\, \frac{ {\left( {34}-{37.5} \right)}^{2} }{{37.5}} ={0.3267},\,\,\,\, \frac{ {\left( {7}-{12.5} \right)}^{2} }{{12.5}} ={2.42}$

$,\,\,\,\, \frac{ {\left( {27}-{25} \right)}^{2} }{{25}} ={0.16},\,\,\,\, \frac{ {\left( {16}-{12.5} \right)}^{2} }{{12.5}} ={0.98}$

Hence, the value of Chi-Square statistics is

${{\chi }^{2}}=\sum{\frac{{{\left( {{O}_{ij}}-{{E}_{ij}} \right)}^{2}}}{{{E}_{ij}}}}={0.8067} + {0.0533} + {0.3267} + {2.42} + {0.16} + {0.98} = 4.747$

The critical Chi-Square value for $$\alpha =0.05$$ and $$\left( 3-1 \right)\times \left( 2-1 \right)=2$$ degrees of freedom is $$\chi _{C}^{2}= {5.991}$$. Since $${{\chi }^{2}}=\sum{\frac{{{\left( {{O}_{ij}}-{{E}_{ij}} \right)}^{2}}}{{{E}_{ij}}}}= {4.747}$$ < $$\chi _{C}^{2}= {5.991}$$, then we fail to reject the null hypothesis, which means that we don't have enough evidence to reject the null hypothesis of independence.

Question 3: Listed below are the numbers of years that U.S. presidents, popes since 1690 and British monarchs lived after they were inaugurated, elected, or coronated. As of the writing, the last president is Gerald Ford and the last pope is John Paul II. The times are based on data from Computer interactive Data Analysis, by Lunn and McNeil, John Wiley & Son. Use a 0.05 significance level to test the claim that the 2 samples of longevity data from popes and monarchs are from populations with the same median.

Presidents

10 29 26 28 15 23 17 25 0 20 4 1 24 16 12 4 10 17 16 0 7 24 12 4

18 21 11 2 9 36 12 28 3 16 9 25 23 32

Popes

2 9 21 3 6 10 18 11 6 25 23 6 2 15 32 25 11 8 17 19 5 15 0 26

Monarchs 17 6 13 12 13 33 59 10 7 63 9 25 36 15

Solution: We need to use a Wilcoxon test in order to assess the claim that the 2 samples are from populations with the same median. The following results are obtained:

 Wilcoxon - Mann/Whitney Test n sum of ranks 24 416 Popes 14 325 Monarchs 38 741 total 468.00 expected value 33.00 standard deviation -1.56 z, corrected for ties .1186 p-value (two-tailed) No. Label Data Rank 1 Popes 2 2.5 2 Popes 9 12.5 3 Popes 21 28 4 Popes 3 4 5 Popes 6 7.5 6 Popes 10 14.5 7 Popes 18 26 8 Popes 11 16.5 9 Popes 6 7.5 10 Popes 25 31 11 Popes 23 29 12 Popes 6 7.5 13 Popes 2 2.5 14 Popes 15 22 15 Popes 32 34 16 Popes 25 31 17 Popes 11 16.5 18 Popes 8 11 19 Popes 17 24.5 20 Popes 19 27 21 Popes 5 5 22 Popes 15 22 23 Popes 0 1 24 Popes 26 33 25 Monarchs 17 24.5 26 Monarchs 6 7.5 27 Monarchs 13 19.5 28 Monarchs 12 18 29 Monarchs 13 19.5 30 Monarchs 33 35 31 Monarchs 59 37 32 Monarchs 10 14.5 33 Monarchs 7 10 34 Monarchs 63 38 35 Monarchs 9 12.5 36 Monarchs 25 31 37 Monarchs 36 36 38 Monarchs 15 22

Since we are comparing two independent groups (Popes & Monarchs), we can use Wilcoxon Rank Sum Test.

The null hypothesis tested is

H0: The two samples are from populations with the same median.

The alternative hypothesis is

H1: The two samples are from populations with the different median.

Significance level = 0.05

Test statistic: The observed values from the pooled sample results are ranked from smallest to largest. After the rankings are obtained, the samples are separated, and the sum of the rankings is calculated for each.

The test statistic used is

$Z=\frac{{{T}_{A}}-\frac{{{n}_{2}}\left( {{n}_{1}}+{{n}_{2}}+1 \right)}{2}}{\sqrt{\frac{{{n}_{1}}{{n}_{2}}\left( {{n}_{1}}+{{n}_{2}}+1 \right)}{12}}}$

,

where TA is the sum of ranks of the smaller sample. Here n1 = 24, n2 = 14, TA = 416.

Therefore,

$Z=\frac{416-\frac{24\left( 14+24+1 \right)}{2}}{\sqrt{\frac{14*24\left( 14+24+1 \right)}{12}}}=-1.57$

Rejection criteria: Reject the null hypothesis, if the absolute value of test statistic is greater than the critical value at the 0.05 significance level.

Lower critical value = -1.96

Upper critical value = 1.96

Conclusion: Fails to reject the null hypothesis, since the absolute value of test statistic is less than the critical value. The sample does not provide enough evidence to reject the claim that the two samples are from populations with the same median.

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The text passage contains three distinct statistical hypothesis testing questions, each with a detailed solution provided.
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Given a text passage, determine:
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First, describe the content structure you found in 1-2 sentences.

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# Non-Parametric Statistics, or What to Do When the Assumptions for a Parametric Test Fail

Question 1: A medical researcher believes the number of ear infections in swimmers can be reduced if the swimmers use ear plugs. A sample of ten people was selected, and the number of ear infections for a four-month period was recorded. During the first two months, the swimmers did not use the ear plugs; during the second two months, they did. At the beginning of the second two-month period, each swimmer was examined to make sure that no infections were present. The data are shown below. At $$\alpha = 0.05$$, can the researcher conclude that using ear plugs effects the number of ear infections?

Solution: We need to test the hypotheses

\begin{align} & {{H}_{0}}:\text{ ear infections are the same with or without the ear plugs} \\ &{{H}_{A}}:\text{ swimmers get less ear infections with ear plugs} \\ \end{align}

We use a Sign-Test. We use Statdisk to get the following output:

The $$x$$ statistics is equal to 2 (the less frequent number of signs). The critical value is 1. Since $$x$$ is not less than or equal to the critical value, we fail to reject the null hypothesis. This means that we don't have enough evidence to support the claim that the number of ear infections in swimmers can be reduced if the swimmers use ear plugs.

Question 2: Research indicates that people who volunteer to participate in research studies tend to have higher intelligence than nonvolunteers. To test this phenomenon, a researcher obtains a sample of 200 high school students. The students are given a description of a psychological research study and asked whether they would volunteer to participate. The researcher also obtains an IQ score for each student and classifies the students into high, medium, and low IQ groups. Do the following data indicate a significant relationship between IQ and volunteering? Test at the .05 level of significance.

Solution: The following table shows the corresponding contingency table:

 Observed High Medium Low Total Volunteer 43 73 34 150 Not Volunteer 7 27 16 50 Total 50 100 50 200

We are interested in testing the following null and alternative hypotheses:

\begin{align}{{H}_{0}}:\,\,\, \text{Volunteer Status}\text{ and }\text {IQ}\text{ are independent} \\ {{H}_{A}}:\,\,\,\text{Volunteer Status}\text{ and }\text {IQ}\text{ are NOT independent} \\ \end{align}

From the table above we compute the table with the expected values

 Expected High Medium Low Volunteer 37.5 75 37.5 Not Volunteer 12.5 25 12.5

The way those expected frequencies are calculated is shown below:

${E}_{{1},{1}}= \frac{ {R}_{1} \times {C}_{1} }{T}= \frac{{150} \times {50}}{{200}}={37.5},\,\,\,\, {E}_{{1},{2}}= \frac{ {R}_{1} \times {C}_{2} }{T}= \frac{{150} \times {100}}{{200}}={75},\,\,\,\, {E}_{{1},{3}}= \frac{ {R}_{1} \times {C}_{3} }{T}= \frac{{150} \times {50}}{{200}}={37.5}$

$,\,\,\,\, {E}_{{2},{1}}= \frac{ {R}_{2} \times {C}_{1} }{T}= \frac{{50} \times {50}}{{200}}={12.5},\,\,\,\, {E}_{{2},{2}}= \frac{ {R}_{2} \times {C}_{2} }{T}= \frac{{50} \times {100}}{{200}}={25},\,\,\,\, {E}_{{2},{3}}= \frac{ {R}_{2} \times {C}_{3} }{T}= \frac{{50} \times {50}}{{200}}={12.5}$

Finally, we use the formula $$\frac{{{\left( O-E \right)}^{2}}}{E}$$ to get

 (fo - fe)²/fe High Medium Low Volunteer 0.8067 0.0533 0.3267 Not Volunteer 2.42 0.16 0.98

The calculations required are shown below:

$\frac{ {\left( {43}-{37.5} \right)}^{2} }{{37.5}} ={0.8067},\,\,\,\, \frac{ {\left( {73}-{75} \right)}^{2} }{{75}} ={0.0533},\,\,\,\, \frac{ {\left( {34}-{37.5} \right)}^{2} }{{37.5}} ={0.3267},\,\,\,\, \frac{ {\left( {7}-{12.5} \right)}^{2} }{{12.5}} ={2.42}$

$,\,\,\,\, \frac{ {\left( {27}-{25} \right)}^{2} }{{25}} ={0.16},\,\,\,\, \frac{ {\left( {16}-{12.5} \right)}^{2} }{{12.5}} ={0.98}$

Hence, the value of Chi-Square statistics is

${{\chi }^{2}}=\sum{\frac{{{\left( {{O}_{ij}}-{{E}_{ij}} \right)}^{2}}}{{{E}_{ij}}}}={0.8067} + {0.0533} + {0.3267} + {2.42} + {0.16} + {0.98} = 4.747$

The critical Chi-Square value for $$\alpha =0.05$$ and $$\left( 3-1 \right)\times \left( 2-1 \right)=2$$ degrees of freedom is $$\chi _{C}^{2}= {5.991}$$. Since $${{\chi }^{2}}=\sum{\frac{{{\left( {{O}_{ij}}-{{E}_{ij}} \right)}^{2}}}{{{E}_{ij}}}}= {4.747}$$ < $$\chi _{C}^{2}= {5.991}$$, then we fail to reject the null hypothesis, which means that we don't have enough evidence to reject the null hypothesis of independence.

Question 3: Listed below are the numbers of years that U.S. presidents, popes since 1690 and British monarchs lived after they were inaugurated, elected, or coronated. As of the writing, the last president is Gerald Ford and the last pope is John Paul II. The times are based on data from Computer interactive Data Analysis, by Lunn and McNeil, John Wiley & Son. Use a 0.05 significance level to test the claim that the 2 samples of longevity data from popes and monarchs are from populations with the same median.

Presidents

10 29 26 28 15 23 17 25 0 20 4 1 24 16 12 4 10 17 16 0 7 24 12 4

18 21 11 2 9 36 12 28 3 16 9 25 23 32

Popes

2 9 21 3 6 10 18 11 6 25 23 6 2 15 32 25 11 8 17 19 5 15 0 26

Monarchs 17 6 13 12 13 33 59 10 7 63 9 25 36 15

Solution: We need to use a Wilcoxon test in order to assess the claim that the 2 samples are from populations with the same median. The following results are obtained:

 Wilcoxon - Mann/Whitney Test n sum of ranks 24 416 Popes 14 325 Monarchs 38 741 total 468.00 expected value 33.00 standard deviation -1.56 z, corrected for ties .1186 p-value (two-tailed) No. Label Data Rank 1 Popes 2 2.5 2 Popes 9 12.5 3 Popes 21 28 4 Popes 3 4 5 Popes 6 7.5 6 Popes 10 14.5 7 Popes 18 26 8 Popes 11 16.5 9 Popes 6 7.5 10 Popes 25 31 11 Popes 23 29 12 Popes 6 7.5 13 Popes 2 2.5 14 Popes 15 22 15 Popes 32 34 16 Popes 25 31 17 Popes 11 16.5 18 Popes 8 11 19 Popes 17 24.5 20 Popes 19 27 21 Popes 5 5 22 Popes 15 22 23 Popes 0 1 24 Popes 26 33 25 Monarchs 17 24.5 26 Monarchs 6 7.5 27 Monarchs 13 19.5 28 Monarchs 12 18 29 Monarchs 13 19.5 30 Monarchs 33 35 31 Monarchs 59 37 32 Monarchs 10 14.5 33 Monarchs 7 10 34 Monarchs 63 38 35 Monarchs 9 12.5 36 Monarchs 25 31 37 Monarchs 36 36 38 Monarchs 15 22

Since we are comparing two independent groups (Popes & Monarchs), we can use Wilcoxon Rank Sum Test.

The null hypothesis tested is

H0: The two samples are from populations with the same median.

The alternative hypothesis is

H1: The two samples are from populations with the different median.

Significance level = 0.05

Test statistic: The observed values from the pooled sample results are ranked from smallest to largest. After the rankings are obtained, the samples are separated, and the sum of the rankings is calculated for each.

The test statistic used is

$Z=\frac{{{T}_{A}}-\frac{{{n}_{2}}\left( {{n}_{1}}+{{n}_{2}}+1 \right)}{2}}{\sqrt{\frac{{{n}_{1}}{{n}_{2}}\left( {{n}_{1}}+{{n}_{2}}+1 \right)}{12}}}$

,

where TA is the sum of ranks of the smaller sample. Here n1 = 24, n2 = 14, TA = 416.

Therefore,

$Z=\frac{416-\frac{24\left( 14+24+1 \right)}{2}}{\sqrt{\frac{14*24\left( 14+24+1 \right)}{12}}}=-1.57$

Rejection criteria: Reject the null hypothesis, if the absolute value of test statistic is greater than the critical value at the 0.05 significance level.

Lower critical value = -1.96

Upper critical value = 1.96

Conclusion: Fails to reject the null hypothesis, since the absolute value of test statistic is less than the critical value. The sample does not provide enough evidence to reject the claim that the two samples are from populations with the same median.

This tutorial is brought to you courtesy of MyGeekyTutor.com
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The text passage contains three distinct statistical hypothesis testing questions, each with a detailed solution provided.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS