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

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].

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{(1)/(n_1) +(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{((n_1-1)s_1^(2) +(n_2-1)s_2^(2) )/(n_1+n_2-2) }` =
`\sqrt{((12-1)* 2.73^(2) +(18-1)* 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{(1)/(n_1) +(1)/(n_2) } }` < 2.467) = 0.98

P(
`-2.467 * {s_p\sqrt{(1)/(n_1) +(1)/(n_2) } }` <
`{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}` <
`2.467 * {s_p\sqrt{(1)/(n_1) +(1)/(n_2) } }` ) = 0.98

P(
`(\bar X_1-\bar X_2)-2.467 * {s_p\sqrt{(1)/(n_1) +(1)/(n_2) } }` < (
`\mu_1-\mu_2`) <
`(\bar X_1-\bar X_2)+2.467 * {s_p\sqrt{(1)/(n_1) +(1)/(n_2) } }` ) = 0.98

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

[
`(\bar X_1-\bar X_2)-2.467 * {s_p\sqrt{(1)/(n_1) +(1)/(n_2) } }` ,
`(\bar X_1-\bar X_2)+2.467 * {s_p\sqrt{(1)/(n_1) +(1)/(n_2) } }` ]

= [
`(70.2-66.9)-2.467 * {s_p\sqrt{(1)/(12) +(1)/(18) } }` ,
`(70.2-66.9)+2.467 * {s_p\sqrt{(1)/(12) +(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].