Question

use the information given to compute a confidence interval for the population mean μ

x̅ = 102, *s* = 18, *n* = 81, 90% confidence
level

x̅ = 102, s = 18, n = 81, 99% confidence level

Answer #1

Solution :

Given that,

Point estimate = sample mean = = 102

sample standard deviation = s = 18

sample size = n = 81

Degrees of freedom = df = n - 1 = 81 - 1 = 80

(a)

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t_{ /2,df} =
t_{0.005,80} = 1.664

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

= 1.664 * (18 / 81)

= 3.3

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

- E < < + E

102 - 3.3 < < 102 + 3.3

98.7 < < 105.3

(b)

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t_{ /2,df} =
t_{0.05,80} = 2.639

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

= 2.639 * (18 / 81)

= 5.3

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

- E < < + E

102 - 5.3 < < 102 + 5.3

96.7 < < 107.3

(96.7 , 107.3)

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