Question

2. A simple random sample of 15 college students showed a mean credit score of 655 and standard deviation 20. Construct the 99% confidence interval estimate of the mean credit score for all college students. a. State the critical value. b. Compute the margin of error. c. State the confidence interval.

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 655

sample standard deviation = s = 20

sample size = n = 15

Degrees of freedom = df = n - 1 = 15 -1 = 14

a) At 99% confidence level

$\alpha$ = 1 - 99%

$\alpha$ =1 - 0.99 =0.01

$\alpha$/2 = 0.005

t$\alpha$/2,df = z 0.005 , 14 = 2.977

b) Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 2.977 * ( 20 / $\sqrt$ 15)

Margin of error = E =15.37

c) The 99% confidence interval estimate of the population mean is,

$\bar x$  ± E

655 ± 15.37

639.63 ,670.37