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

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

what is the intpretation of this interval?

Answer 1

(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)=s_p*(1/n1 +1/n2)^1/2)=0.51 and

s_p²=((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