the owner of a fitness center is interested in estimating the difference in mean

Question

the owner of a fitness center is interested in estimating the difference in mean years that female members have been with the club compared with male members. he wishes to develop a 99% confidence interval estimate. The data are showen in the accompanying table. Assuming that the same data are approximately normal and that the two populations have equal variances, develop and interpret the confidence interval estimate. dicuss the results

2

2

3.5

7

show formulas in excel

what is the intpretation of this interval?

gender 1=male 2=female 1 2 2 2 1 2 2 2 1 2 1 2 1 2 2 1 2 1 2 2 1 1 2 2

2

2 2 2 1 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 1 2 1 2 1

2

years with the club 3.5 1 2.5 2 3.5 4 5.5 1 2 1.5 3 6.5 1.5 3 1 3.5 4.5 1.5 1 0 3 3.5 4.5 6

3.5

5.5 2 1.5 3 3.5 1.5 5.5 2 3 2 2 2.5 0 2.5 2 5 1 2.5 1.5 1 4.5 2.5 1 4.5

7

show formulas in excel

Explanation / Answer

(1-alpha)*100% confidence interval for difference of population mean=

=difference of sample mean±t(0.05/2, n)*SE(of difference of sample mean)

95% confidence interval =0.53±2.02*0.51=0.53±1.03=(-0.50, 1.56)

since confidence interval contains 0, so difference of two sample means is not significant at 5% level of significance

following information has been generate using ms-excel

SE(difference of sample mean)=sp*(1/n1 +1/n2)1/2)=0.51 and

sp2=((n1-1)s12+(n2-1)s22)/n=2.81 and with df is n=n1+n2-2=16+34-2=48

sample mean s s2 n (n-1)s2 male 2.46875 1.175709006 1.382291667 16 20.734375 female 3 1.858640755 3.454545455 34 114 difference= 0.53125 134.734375 sp2= 2.81 sp= 1.68 SE= 0.51

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