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Assume a 2015 Gallup Poll asked a national random sample of 482 adult women to state their current weight. Assume the mean weight in the sample was

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Answer #1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 164

Population standard deviation = $\sigma$ = 36

Sample size = n = 482

Z_$\alpha$/2 = 1.645

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

= 1.645 * (36 / $\sqrt$ 482)

= 2.70

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

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

164 - 2.70 < $\mu$ < 160 + 2.70

161.3 < $\mu$ < 162.70

Lower value = 161.3

Upper value = 162.70

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Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 164

Population standard deviation = $\sigma$ = 36

Sample size = n = 482

Z_$\alpha$/2 = 1.645

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

= 1.645 * (36 / $\sqrt$ 482)

= 2.70

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

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

164 - 2.70 < $\mu$ < 160 + 2.70

161.3 < $\mu$ < 162.70

Lower value = 161.3

Upper value = 162.70