Question

Ample random sample of population mean 13 is drawn from a population that may dared the sample means found to bex 63 and the same mandard deviation is found to be s 18. Construct a condence interval about the The 90condence intervalis Round to two decimal places as needed.)

Answer 1

Solution :

Given that,

$\bar x$ = 63

s = 18

n = 13

Degrees of freedom = df = n - 1 = 13 - 1 = 12

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_0.025,12 =2.179

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

= 2.179 * (18 / $\sqrt$ 13)

= 10.88

Margin of error = 10.88

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

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

63 - 10.88 < $\mu$ < 63 + 10.88

52.12 < $\mu$ < 73.88

(52.12, 73.88 )