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Analyzing Waistline Data: Normality Test and T-Test, Study notes of History

Abo Akademi University (AAU)History

Instructions on how to use spss software to analyze waistline data, focusing on obtaining summary measures, checking normality, and performing both parametric and non-parametric tests. The steps to obtain descriptive statistics, shapiro-wilk test results, and mann-whitney test results. Additionally, it illustrates how to perform an independent samples t-test, assuming and not assuming equal variances.

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2021/2022

Uploaded on 07/04/2022

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Reading SPSS Output for SPSS Assignment #2

Variable: Waistline

1. Obtaining summary measures: Click on “Analyze” → “Descriptive Statistics” → “Explore”. Move the variable “waist” into the “dependent

list” (putting “gender” in the “Factor list” will give you summary measures for males and females separately). To get the qq-plots and the

Shapiro-Wilk test, make sure you click on “plots” then check the box for “Normality plots with tests”. Below is the SPSS output that you will

get:

Descriptives

Gender Statistic Std. Error

Mean 85.0325 2.43517

Lower Bound 78.4383

99% Confidence Interval for

Mean Upper Bound 91.6267

5% Trimmed Mean 83.7472

Median 81.9500

Variance 237.202

Std. Deviation 15.40136

Minimum 66.70

Maximum 126.50

Range 59.80

Interquartile Range 22.05

Skewness .962 .374

Female

Kurtosis .611 .733

Mean 91.2850 1.55930

Lower Bound 87.0626

99% Confidence Interval for

Mean Upper Bound 95.5074

5% Trimmed Mean 91.2333

Median 91.2000

Variance 97.256

Std. Deviation 9.86185

Minimum 75.20

Maximum 108.70

Range 33.50

Interquartile Range 18.78

Skewness .037 .374

Waist

Male

Kurtosis -1.058 .733

I don’t want you to copy and paste this whole table. Just pick out the correct values to put in your tables.

2. Checking Normality. Together with the above table, you will also get results of the Shapiro-Wilk test to determine if it is reasonable to

assume that both data sets come from normal population. The result for the “waist” variable is given below

Tests of Normality

Kolmogorov-Smirnova Shapiro-Wilk

Gender Statistic df Sig. Statistic df Sig.

Female .148 40 .027 .905 40 .003Waist

Male .131 40 .084 .952 40 .090

Discover Study notes of History Abo Akademi University (AAU)

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Download Analyzing Waistline Data: Normality Test and T-Test and more Study notes History in PDF only on Docsity!

Reading SPSS Output for SPSS Assignment

Variable: Waistline

1. Obtaining summary measures: Click on “Analyze” → “Descriptive Statistics” → “Explore”. Move the variable “ waist ” into the “dependent

list” (putting “ gender ” in the “Factor list” will give you summary measures for males and females separately). To get the qq-plots and the

Shapiro-Wilk test, make sure you click on “plots” then check the box for “Normality plots with tests”. Below is the SPSS output that you will

get:

Descriptives Gender Statistic Std. Error Mean 85.0325 2. 99% Confidence Interval for Lower Bound 78. Mean (^) Upper Bound 91. 5% Trimmed Mean 83. Median 81. Variance 237. Std. Deviation 15. Minimum 66. Maximum 126. Range 59. Interquartile Range 22. Skewness .962. Female Kurtosis .611. Mean 91.2850 1. 99% Confidence Interval for Lower Bound 87. Mean (^) Upper Bound 95. 5% Trimmed Mean 91. Median 91. Variance 97. Std. Deviation 9. Minimum 75. Maximum 108. Range 33. Interquartile Range 18. Skewness .037. Waist Male Kurtosis -1.058.

I don’t want you to copy and paste this whole table. Just pick out the correct values to put in your tables.

2. Checking Normality. Together with the above table, you will also get results of the Shapiro-Wilk test to determine if it is reasonable to

assume that both data sets come from normal population. The result for the “ waist” variable is given below

Tests of Normality Kolmogorov-Smirnov a^ Shapiro-Wilk Gender (^) Statistic df Sig. Statistic df Sig. Waist Female .148 40 .027 .905 40. Male .131 40 .084 .952 40.

Note that the p-value for the Shapiro-Wilk test are 0.003 and 0.090 (in the last column under “Sig.”). This implies that the female data set is

not normal because the p-value was smaller than alpha=.05. You can also see a curve pattern in the corresponding qq-plots (see left figure below),

suggesting that the female data is not normal.

3. Non-parameteric Test: Because the female data is not normal, we cannot use the ordinary 2-sample independent t-test. Instead, we are going

to use a non-parametric test, called the Mann-Whitney test. You can do this by choosing, “ Analyze ”→ “ Nonparametric Test ” → “

Independent Samples ”. Choose the variable you wish to test and use Gender for “ grouping variable ”. If you don’t see Gender in the list of

variables, then create a new variable that will indicate the grouping of the values (you can do this by simply going to an empty column and

type 1 for males and 2 for females). After you moved your grouping variable into the box “Grouping variable”, click on “define” to tell SPSS

who are in group 1 and who are in group 2. Simply type the code that you used for group 1 and group 2. If you used 1 and 2, then just type 1

in the first box and 2 in the second box. Then hit “continue”, then click “ok”. You will then get something like the table below:

Test Statistics a Waist Mann-Whitney U 531. Wilcoxon W 1351. Z -2. Asymp. Sig. (2-tailed). a. Grouping Variable: VAR

The p-value that you want to look for is the value in the last row, labeled as “Asymp. Sig. (2-tailed)”. In this particular example, the p-value is 0.010.

Because this p-value is smaller than alpha=0.05, we reject the null hypothesis and conclude that the mean waistline for males is not equal to the mean

waistline for females.

4. T-test: Let’s suppose we that we can assume that both data sets are reasonably normal, just so that I can illustrate how to perform the t-test.

You can access the t-test procedure by choosing “ Analyze ”→ “ Compare Means ” → “ Independent Samples T Test ”. Use Gender for

“ grouping variable ” (Just like you did earlier), then click on “Define Groups”, and type “M” for group 1 and “F” for group 2., then hit

“continue” and then click “ok”. You will then get the table below

Independent Samples Test Levene's Test for Equality of Variances t-test for Equality of Means 95% Confidence Interval of the Difference F Sig. t df Sig. (2-tailed) Mean Difference Std. Error Difference (^) Lower Upper Waist Equal variances assumed 7.430 .008 2.162 78 .034 6.25250 2.89162 .49573 12. Equal variances not assumed 2.162 66.378 .034 6.25250 2.89162 .47982 12.

Analyzing Waistline Data: Normality Test and T-Test, Study notes of History

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Partial preview of the text

Download Analyzing Waistline Data: Normality Test and T-Test and more Study notes History in PDF only on Docsity!

Reading SPSS Output for SPSS Assignment

Variable: Waistline

1. Obtaining summary measures: Click on “Analyze” → “Descriptive Statistics” → “Explore”. Move the variable “ waist ” into the “dependent

list” (putting “ gender ” in the “Factor list” will give you summary measures for males and females separately). To get the qq-plots and the

Shapiro-Wilk test, make sure you click on “plots” then check the box for “Normality plots with tests”. Below is the SPSS output that you will

get:

I don’t want you to copy and paste this whole table. Just pick out the correct values to put in your tables.

2. Checking Normality. Together with the above table, you will also get results of the Shapiro-Wilk test to determine if it is reasonable to

assume that both data sets come from normal population. The result for the “ waist” variable is given below

Note that the p-value for the Shapiro-Wilk test are 0.003 and 0.090 (in the last column under “Sig.”). This implies that the female data set is

not normal because the p-value was smaller than alpha=.05. You can also see a curve pattern in the corresponding qq-plots (see left figure below),

suggesting that the female data is not normal.

3. Non-parameteric Test: Because the female data is not normal, we cannot use the ordinary 2-sample independent t-test. Instead, we are going

to use a non-parametric test, called the Mann-Whitney test. You can do this by choosing, “ Analyze ”→ “ Nonparametric Test ” → “

Independent Samples ”. Choose the variable you wish to test and use Gender for “ grouping variable ”. If you don’t see Gender in the list of

variables, then create a new variable that will indicate the grouping of the values (you can do this by simply going to an empty column and

type 1 for males and 2 for females). After you moved your grouping variable into the box “Grouping variable”, click on “define” to tell SPSS

who are in group 1 and who are in group 2. Simply type the code that you used for group 1 and group 2. If you used 1 and 2, then just type 1

in the first box and 2 in the second box. Then hit “continue”, then click “ok”. You will then get something like the table below:

The p-value that you want to look for is the value in the last row, labeled as “Asymp. Sig. (2-tailed)”. In this particular example, the p-value is 0.010.

Because this p-value is smaller than alpha=0.05, we reject the null hypothesis and conclude that the mean waistline for males is not equal to the mean

waistline for females.

4. T-test: Let’s suppose we that we can assume that both data sets are reasonably normal, just so that I can illustrate how to perform the t-test.

You can access the t-test procedure by choosing “ Analyze ”→ “ Compare Means ” → “ Independent Samples T Test ”. Use Gender for

“ grouping variable ” (Just like you did earlier), then click on “Define Groups”, and type “M” for group 1 and “F” for group 2., then hit

“continue” and then click “ok”. You will then get the table below

When using the t-test, you need to decide if you can assume equal variances. From the table above, we see that the p-value for the Levene’s test for

equality of variance is 0.008 (under “sig”). Since this value is less than alpha=0.05, this implies that the variances cannot be assumed to be equal.

Therefore, you should use the t-test result given in the second row “Equal variances not assumed”. The corresponding t_obs is 2.162, df=66.378, and

the p-value is 0.034. Since this p-value is smaller than alpha=0.05, we reject the null hypothesis.