Question

A student researcher compares the heights of American students and non-American students from the student body of a certain college in order to estimate the difference in their mean heights. 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. Determine the 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. Assume that the population variances are equal and that the two populations are normally distributed.
Step 1 of 3 : Find the point estimate that should be used in constructing the confidence interval.

ANSWER: 3.3

Answer 1

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. =  $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$  ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean height of American students = 70.2 inches

$\bar X_2$ = sample mean height of non-American students = 66.9 inches

$s_1$ = sample standard deviation of American students = 2.73 inches

$s_2$ = sample standard deviation of non-American students = 3.13 inches

$n_1$ = sample of American students = 12

$n_2$ = sample of non-American students = 18

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(12-1)\times 2.73^{2} +(18-1)\times 3.13^{2} }{12+18-2} }$  = 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 ($\mu_1-\mu_2$) is ;

P(-2.467 < $t_2_8$ < 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 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 2.467) = 0.98

P( $-2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.98

P( $(\bar X_1-\bar X_2)-2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ($\mu_1-\mu_2$) < $(\bar X_1-\bar X_2)+2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.98

98% confidence interval for ($\mu_1-\mu_2$) =

[ $(\bar X_1-\bar X_2)-2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ]

= [$(70.2-66.9)-2.467 \times {s_p\sqrt{\frac{1}{12} +\frac{1}{18} } }$ , $(70.2-66.9)+2.467 \times {s_p\sqrt{\frac{1}{12} +\frac{1}{18} } }$]

= [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].