Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals223​, x overbarequals27.9 ​hg, sequals6.3 hg. Construct a confidence interval estimate of the mean. Use a 90​% confidence level. Are these results very different from the confidence interval 26.4 hgless thanmuless than28.4 hg with only 16 sample​ values, x overbarequals27.4 ​hg, and sequals2.2 ​hg?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 27.9 hg

sample standard deviation = s = 6.3 hg

sample size = n = 223

Degrees of freedom = df = n - 1 = 223 - 1 = 222

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t0.05,222 = 1.652

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

= 1.652 * ( 6.3 / 223)

Margin of error = E = 0.7

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

- E < < + E

27.9 - 0.7 < < 27.9 + 0.7

( 27.2 hg < < 28.6 hg )

Yes, because the confidence interval limits are not similar