Question

n = 30,  = 84.6, s = 10.5, 90% confidence

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 84.6

sample standard deviation = s = 10.5

sample size = n = 30

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

At 90% confidence level

= 1-0.90% =1-0.9 =0.10

/2 =0.10/ 2= 0.05

t/2,df = t0.05,29= 1.70

t /2,df = 1.70

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

= 1.70 * (10.5 / 30)

Margin of error = E =3.257

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

- E < <  + E

84.6 -3.257 < < 84.6 +3.257

81.343< < 87.857

(81.343 ,87.857)