Question

If X = 70, S = 9, and n= 36, and assuming that the population is normally distributed, construct a 99% confidence interval estimate of the population mean, u. Click here to view page 1 of the table of critical values for the t distribution. Click here to view page 2 of the table of critical values for the t distribution. (Round to two decimal places as needed.)

Answer 1

Solution :

Given that,

sample mean = $\bar x$ = 70

sample standard deviation = s = 9

sample size = n = 36

Degrees of freedom = df = n - 1 = 36 - 1 = 35

At 99% confidence level

$\alpha$ = 1 - 99%

$\alpha$ =1 - 0.99 =0.01

$\alpha$/2 = 0.005

t$\alpha$/2,df = t_0.005,35 = 2.7238

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

= 2.7238 * ( 9/ $\sqrt$ 36)

Margin of error = E = 4.0857

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

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

70 - 4.0857 < $\mu$ < 70 + 4.0857

(74.09 < $\mu$ < 65.91)