Question

27 s have a mearn 9. A group of 25, randomly selected college student score 30.0 with a standard deviation of 5.0 on a placement test. What is 90 % confidence interval for the mean score, μ, of all the students taking the test? 9/a

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 30.0

sample standard deviation = s = 5.0

sample size = n = 27

Degrees of freedom = df = n - 1 = 27 - 1 = 26

At 99% confidence level the t is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

t_{$\alpha$ /2,df} = t_0.0005,26 = 2.779

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

= 2.779 * (5.0 / $\sqrt$ 27)

= 2.7

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

$\bar x$ - E < $\mu$ < $\bar x$ + E

30.0 - 2.7 < $\mu$ < 30.0 + 2.7

27.3 < $\mu$ < 32.7

The 99% confidence interval for the mean score is, ( 27.3 , 32.7 )