Question

n=30, x=94.0, s=16.5, 90% confidence

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 94.0

sample standard deviation = s = 16.5

sample size = n = 30

Degrees of freedom = df = n - 1 = 30 - 1 = 29

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,29 = 1.699

Margin of error = E = t_/2,df * (s /n)

= 1.699 * ( 16.5/ 30)

= 0.93

The 90% confidence interval estimate of the population mean is,

- E < < + E

94.0 - 0.93 < < 94.0 + 0.93

93.07 < < 94.93

(93.07 , 94.93)