Question

Here are summary statistics for randomly selected weights of newborn​ girls: n=238​, x=26.6 ​hg, s=6.5 hg. Construct a confidence interval estimate of the mean. Use a 99​% confidence level. Are these results very different from the confidence interval 25.1 hg< mu <28.3 hg with only 17 sample​ values, x = 26.7 ​hg, and s=2.2 ​hg?

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 26.6 hg

sample standard deviation = s = 6.5 hg

sample size = n = 238

Degrees of freedom = df = n - 1 = 238 - 1 = 237

At 99% confidence level

$\alpha$ = 1 - 99%

$\alpha$ =1 - 0.99 =0.01

$\alpha$/2 = 0.005

t$\alpha$/2,df = t_{0.005, 237} = 2.597

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

= 2.597 * (6.5 / $\sqrt$ 238)

Margin of error = E = 1.1

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

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

26.6 - 1.1 < $\mu$ < 26.6 + 1.1

(25.5 < $\mu$ < 27.7)