Question

                                

                               After:             302       295       289       291         261        298        281  

                               Before:          275       289       287       258         248        291        278

n =

df =

xbar_d =

s_d =

1. State the null hypothesis and the alternative hypothesis for testing the difference. Make sure you clearly indicate the parameters for “after the class” or “before the class”.

Ho:

Ha:

2. Calculate the test statistic. Specify which test statistic you are finding (“z” or “t”).

3. Find the p-value using the appropriate table. Draw the curve showing p-value, critical value(s)(α=.01), rejection region.

4. At a significance level of 0.01, would you reject or retain the null hypothesis? Circle your response and explain why you chose this response, using the p-value you found in part c.

            Reject Ho / Do NOT Reject Ho               Why?

sketch your work for C

Answer 1

SOLUTION-

WE USE MINITAB-16 TO DO THE CALCULATIONS:

A.) LET   BE THE AVERAGE FOR THE 'AFTER'-CLASS AND   BE THE AVERAGE FOR THE 'BEFORE'-CLASS.

TO TESTING THE DIFFERENCE OFMEAN FOR BOTH THE CLASS, WE FRAME THE FOLLOWING HYPOTHESIS,

B.) THE CLASSES ARE DEPENDENT, WE PERFORM A PAIRED-T TEST TO TEST THE HYPOTHESIS-

STEP- ENTER THE DATA IN COLUMNS> STAT> BASIC STATISTICS> PAIRED-T> SELECT THE SAMPLES> UNDER 'OPTIONS', SET THE CONFIDENCE LEVEL AS 99.0 AND ALTERNATE AS 'NOT EQUAL'> OK.

THE VALUE OF THE TEST STATISTIC IS, T= 2.81

C.) THE P-VALUE OBSERVED IS 0.031

D.) AT , THE OBSERVED P-VALUE IS MORE THAN THE LEVEL OF SIGNIFICANCE. SO WE ACCEPT THE NULL HYPOTHESIS AND DO NOT REJECT

HENCE, WE CONCLUDE THAT THERE IS NO SIGNIFICANT DIFFERENCE BETWEEN BOTH THE CLASSES.

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