Answer:
98% confidence interval for the true mean difference between the mean height of the American students and the mean height of the non-American students is [0.56 inches , 6.04 inches].
Step-by-step explanation:
We are given that a random sample of 12 American students had a mean height of 70.2 inches with a standard deviation of 2.73 inches.
A random sample of 18 non-American students had a mean height of 66.9 inches with a standard deviation of 3.13 inches.
Firstly, the Pivotal quantity for 98% confidence interval for the difference between the true means is given by;
P.Q. = [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] ~ [tex]t__n__1-_n__2-2[/tex]
where, [tex]\bar X_1[/tex] = sample mean height of American students = 70.2 inches
[tex]\bar X_2[/tex] = sample mean height of non-American students = 66.9 inches
[tex]s_1[/tex] = sample standard deviation of American students = 2.73 inches
[tex]s_2[/tex] = sample standard deviation of non-American students = 3.13 inches
[tex]n_1[/tex] = sample of American students = 12
[tex]n_2[/tex] = sample of non-American students = 18
Also, [tex]s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }[/tex] = [tex]\sqrt{\frac{(12-1)\times 2.73^{2} +(18-1)\times 3.13^{2} }{12+18-2} }[/tex] = 2.98
Here for constructing 98% confidence interval we have used Two-sample t test statistics.
So, 98% confidence interval for the difference between population means ([tex]\mu_1-\mu_2[/tex]) is ;
P(-2.467 < [tex]t_2_8[/tex] < 2.467) = 0.98 {As the critical value of t at 28 degree
of freedom are -2.467 & 2.467 with P = 1%}
P(-2.467 < [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] < 2.467) = 0.98
P( [tex]-2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] < [tex]{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}[/tex] < [tex]2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] ) = 0.98
P( [tex](\bar X_1-\bar X_2)-2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] < ([tex]\mu_1-\mu_2[/tex]) < [tex](\bar X_1-\bar X_2)+2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] ) = 0.98
98% confidence interval for ([tex]\mu_1-\mu_2[/tex]) =
[ [tex](\bar X_1-\bar X_2)-2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] , [tex](\bar X_1-\bar X_2)+2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] ]
= [[tex](70.2-66.9)-2.467 \times {s_p\sqrt{\frac{1}{12} +\frac{1}{18} } }[/tex] , [tex](70.2-66.9)+2.467 \times {s_p\sqrt{\frac{1}{12} +\frac{1}{18} } }[/tex]]
= [0.56 , 6.04]
Therefore, 98% confidence interval for the true mean difference between the mean height of the American students and the mean height of the non-American students is [0.56 inches , 6.04 inches].
