Question

I did a survey on how times a week people take a shower, this is a random sample

The confidence level is 90%

I asked 10 girls from my family in group one. 6,5,4,5,3,6,8,7,5,9.

I asked 10 guys from my family in group two. 7,4,8,5,7,5,6,5,5,4

Group 1

Mean:

Sample Size: 10

Standard Deviation:

Group 2

Mean:

Sample Size: 10

Standard Deviation:

(___________ , _____________)

DF:

Answer 1

there is no difference of number of shower taken by girls and guys.

here you want to see weather the girls and guys take different shower so here

here we use t-test assuming equality of variance of both girls and boys and

here we use t-test with

null hypothesis H0:mean1=mean2 and alternate hypothesis H1:mean1≠mean2

statistic t=|(mean1-mean2)|/((s_p*(1/n1 +1/n2)^1/2)=0.2798 with df is n=n1+n2-2 =18

and s_p²=((n1-1)s1²+(n2-1)s2²)/n=2.556

since the two-tailed p-value is more than typical level of significnce alpha=0.1, so we fail to reject H0 or accept H0 and conclude that there is no difference of number of shower taken by girls and guys.

	sample	mean	s	s2	n	(n-1)s2
	girls	5.8000	1.8135	3.2889	10	29.6000
	boys	5.6000	1.3499	1.8222	10	16.4000
	difference=	0.2000	sum=	5.1111	20	46.0000
	sp2=	2.5556
	sp=	1.5986
	SE=	0.7149
	t=	0.2798
two tailed	p-value=	0.7829
critical	t(0.1)	1.7341

%confidence level=100-%level of significance