Question

Suppose that in a sample of 50 college students in Illinois, the mean credit card debt was $346. Suppose that we also have reason to believe (from previous studies) that the population standard deviation of credit card debts for this group is $108. Use this information to calculate a 95% confidence interval for the mean credit card debt of all college students in Illinois.

Answer 1

A 95% confidence interval for the mean credit card debt of all college students in Illinois is [$316.06, $375.94] .

Explanation:

We are given that in a sample of 50 college students in Illinois, the mean credit card debt was $346. Suppose that we also have reason to believe that the population standard deviation of credit card debts for this group is $108.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. =
`(\bar X-\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\bar X` = sample mean credit card debt = $346

`\sigma` = population standard deviation = $108

n = sample of college students = 50

`\mu` = population mean credit card debt

Here for constructing a 95% confidence interval we have used a One-sample z-test statistics because we know about population standard deviation.

So, a 95% confidence interval for the population mean,
`\mu` is;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96} P(-1.96 <
`(\bar X-\mu)/((\sigma)/(√(n) ) )` < 1.96) = 0.95

P(
`-1.96 * {(\sigma)/(√(n) ) }` <
`{\bar X-\mu}} }` <
`1.96 * {(\sigma)/(√(n) ) }` ) = 0.95

P(
`\bar X-1.96 * {(\sigma)/(√(n) ) }` <
`\mu` <
`\bar X+1.96 * {(\sigma)/(√(n) ) }` ) = 0.95

95% confidence interval for
`\mu` = [
`\bar X-1.96 * {(\sigma)/(√(n) ) }` ,
`\bar X+1.96 * {(\sigma)/(√(n) ) }` ]

= [
`\$346-1.96 * {(\$108)/(√(50) ) }` ,
`\$346+1.96 * {(\$108)/(√(50) ) }` ]

= [$316.06, $375.94]

Therefore, a 95% confidence interval for the mean credit card debt of all college students in Illinois is [$316.06, $375.94] .