Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Excel File Edit View Insert Format Tools Data Window Help 81%.> Thu 9:31 PM a@E

ID: 3337506 • Letter: E

Question

Excel File Edit View Insert Format Tools Data Window Help 81%.> Thu 9:31 PM a@E Q Search Sheet Insert Page Layout Formulas Data Review View +Share A cut AutoSum A Fill Calibri (Body)-11,A-Ar Wrap Text Copy Format A -Merge & Center' | $, % , % 4% Conditional Format Cell Formatting as Table Sty es EEE Insert Delete Format Sort & Filter H28 A consultant for a large university studied the number of hours per week freshmen watch TV versus the number of hours seniors do. The results of this study follow. Is there enough evidence to show the mean number of hours per week freshmen watch TV is different from the mean number of hours seniors do at =0.005? 16 I 19 21 78740 39749 25 For the hrpothesis stated above in terms of Seniors Freshmen: 28 What isare the critica value(s)? Question Question 2What is the decision? Question 3What ts the p value? Fill in only gne of the following statements. If the Z table is approprate, If the t table is appropriate, pvalue Click to Grade Your Work Daily Problem Ready ® 100% 26

Explanation / Answer

H0: 1 - 2 = 0 i.e. (1 = 2)

H1: 1 - 2 0 i.e. (1 2)

            

Assuming population variances are equal, we would have to calculate pooled-variance t-Test. Freshman would be considered as population 1 and Seniors would be considered as population 2

Sp^2= (n1-1)S1^2+(n2-1)S2^2/(n1-1)+(n2-1)

         = (8-1)*7.8740^2+(4-1)*3.9749^2/7+3

         = 48.14

tSTAT=(X1-X2)-(µ1-µ2)/Sp^2(1/n1+1/n2)

       =(18.4-11.3)-0/48.14(1/8+1/4)

       =7.1/4.2488

       =1.67

tCRIT is +/- 3.169 and hence cannot reject the null hypothesis

p-value one sided would be .062938 and hence on both the sides it would be .125875