Question

A P.E instructor has recently conducted a Physical Fitness test on three different groups of high school boys which are enrolled in three different classes during three different periods of class during the day. Group 1 is enrolled in the first period P.E. class, Group 2 is enrolled in the third period class, and Group 3 is enrolled in the sixth period class. The P.E. instructor would like to determine if there are significant differences between the three groups of boys in each of the three P.E. classes. The instructor would like to determine if the scores may differ based on the time of day that the P.E. participants actually took the Physical Fitness Test (ex. morning, noon, afternoon). Please address the following questions below. P.E. Test Scores Group 1 P.E. Test Scores Group 2 P.E. Test Scores Group 3 100 What type of hypothesis test will we conduct and why? What are our n values for the three groups? What are the mean scores for our three groups? What are our standard deviations for our three groups? What is the value of our test statistic? f. Are our results statistically significant or not? Explain your answer.

Answer 1

A) Anova One-way test will be conducted to determine whether there are differences between the three groups because Anova is a statistical technique that is generally used to determine the difference between 2 or more groups using an F statistic.

Null Hypothesis Ho = There is no difference between the means scores of three groups

Alternate Hypothesis Ha = There is a significant difference between the scores of three groups

B) Here, n refers to the sample size for each group.

n for P.E. Test Scores Group 1 = 5

n for P.E. Test Scores Group 2 = 5

n for P.E. Test Scores Group 2 = 5

C) Following formula can be used to calculate the mean scores:

Mean= Sum of all scores

Mean Score (Group 1) = (100+99+87+97+92)/5 = 95

Mean Score (Group 2) = (90+87+95+97+91)/5 = 92

Mean Score (Group 3) = (88+82+98+90+96)/5 = 90.8

D) Following formula can be used to calculate the Standard Deviation:

StandardDeviation= n-1 where, x= Target Score X = Mean Score n=Sample size

However, Standard deviation can also be easily calculated by putting the data in excel and using =stdev.s command

100 99 90.8 Mean Standard Deviation =st STANDARDIZE STDEV.P cal values and text in the sample) STDEV.S STDEVA fr STDEVPA o STEYX STDEV STDEVP

95 97 92 98 90 91 92 90.8 4 6.418723 96 Mean Standard Deviation 95 5.43139

Therefore,

The standard deviation of Group 1 = 5.4313

The standard deviation of Group 2 = 4

The standard deviation of Group 3 = 6.4187

E) To calculate the test statistic, we can use excel to compute the value quickly:

Just put the data into a spreadsheet and go to data --> data analysis --> Anova: Single Factor

B 5 File Home Insert Page Layout Formulas Data Review View Help À Lì From Text/CSV Co Recent Sources # Queries & Connections | 21 A2 B Te From Web Existing Connections Properties Get Refresh Sort Data- From Table/Range All - 3 Edit Links Get & Transform Data Queries & Connections H9 : foc А B C D E F G H Tell me what you want to do Clear E To Reapply Filter Text to V Advanced Columns De Sort & Filter Data Tools E Data Analysis ?, Solver What If Forecast Analysis Sheet Forecast Group - Ungroup- Subtotal Outline 5 Analyze I J K L M N O P Q R Group 1 Group 2 Group 3 100 87 Data Analysis ? x Analysis Tools 97 ок | 92 97 90 96 92 90.8 4 6.418723 Cancel Mean Standard Deviation 5.43139 Help Anova: Single Factor Anova: Two-Factor With Replication Anova: Two-Factor Without Replication Correlation Covariance Descriptive Statistics Exponential Smoothing F-Test Two-Sample for Variances Fourier Analysis Histogram

C D E F G H I J K L M N O P Q Group 1 Group 2 Group 3 Anova: Single Factor ? x 90 Input Input Range: OK 99 87 SCS5: SES9 O Columns O Rows Cancel Grouped By: Help 92 ----96 Labels in first row Alpha: 0.05 Mean Standard Deviation 5.43139 4 6.418723 Output options O Output Range: New Worksheet Ply: O New Workbook

Anova: Single Factor Count SUMMARY Groups Column 1 Column 2 Column 3 Sum 475 460 454 Average Variance 95 29.5 92 16 90.8 41.2 5 df ANOVA Source of Variation Between Groups Within Groups SS 46.8 346.8 MS 23.4 0.809689 28.9 P-value F crit 0.46789 3.885294 2 12 Total 393.6 14

We got the f statistic value = 0.80

F) We are assuming the significance level to be 0.05.

The p value from the computation is p=0.46789 which is higher than significance level (0.05). Whenever, the calculate p value is more than significance level, we accept the null hypothesis.

This means that there is no difference between the mean scores of three groups.

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