A random sample of 25 kids 15 boys and 10 girls showed an av

A random sample of 25 kids; 15 boys and 10 girls; showed an average boy height of 50 inches, and an average girl height of 40 inches. The boy heights have a standard deviation of 5 inches and the girls heights have a standard deviation of 4 inches. The kid

Solution

(a) Let mu1 be the mean for boys

Let mu2 be the mean for girls

Ho: mu1=mu2 (i.e. null hypothesis)

Ha: mu1<mu2 (i.e. alternative hypothesis)

The test statistic is

t=(xbar1-xbar2)/sqrt(s1^2/n1+s2^2/n2)

=(50-40)/sqrt(5^2/15+4^2/10)

=5.53

The degree of freedom =n1+n2-2=15+10-2=23

It is a left-tailed test.

Given a=0.05, the critical value is t(0.05, df=23) =-1.71 (from student t table)

The rejeciton region is if t<-1.71, we reject Ho.

Since t=5.53 is larger than -1.71, we do not reject Ho.
So we can not conclude that boys are shorter than girls

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(b) Given a=1-0.99=0.01, t(0.005, df=23) =2.81 (from student t table)

So the lower bound is

(xbar1-xbar2)-t*sqrt(s1^2/n1+s2^2/n2)

=(50-40)-2.81*sqrt(5^2/15+4^2/10)

=4.921228

So the upper bound is

(xbar1-xbar2)+t*sqrt(s1^2/n1+s2^2/n2)

=(50-40)+2.81*sqrt(5^2/15+4^2/10)

=15.07877