Question

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Answer 1

Solution

Given that,

$\bar x$ = 82.82

$\sigma$ = 14.97

n = 74

At 90% confidence level the z is ,

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.10

$\alpha$ / 2 = 0.10 / 2 = 0.05

Z$\alpha$/2 = Z0.05 = 1.645

Margin of error = E = Z$\alpha$/2* ($\sigma$$\sqrt$/n)

= 1.645 * (14.97 / $\sqrt$ 74 )

= 2.86

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

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

82.82 - 2.86 < $\mu$ < 82.82 +2.86

79.96 < $\mu$ < 85.68

(79.96 , 85.68)