Question

#3. A researcher obtained a sample mean 780 and the sample variance 49 from 121 samples. (a) Calculate the 95% confidence interval for the population mean. (b) Calculate the 99% confidence interval for the population mean.

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 780

sample standard deviation = s = 7

sample size = n = 121

Degrees of freedom = df = n - 1 = 120

a)

At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t_{$\alpha$ /2,df} = t_0.025,120 = 1.980

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

= 1.980 * ( 7/ $\sqrt$ 121)

= 0.636

The 95% confidence interval estimate of the population mean is,

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

780 - 0.636 < $\mu$ < 180 + 0.636

779.364 < $\mu$ < 780.636

(779.364 , 780.636)

b)

At 99% confidence level the z is ,

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

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

t_{$\alpha$ /2,df} = t_0.005,120 = 2.617

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

= 2.617 * (7 / $\sqrt$ 121)

= 1.665

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

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

780 - 1.665 < $\mu$ < 780 + 1.665

778.335 < $\mu$ < 781.665

(778.335 , 781.665)