Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals173​, x overbarequals31.6 ​hg, sequals7.6 hg. Construct a confidence interval estimate of the mean. Use a 99​% confidence level. Are these results very different from the confidence interval 29.1 hgless thanmuless than33.7 hg with only 18 sample​ values, x overbarequals31.4 ​hg, and sequals3.4 ​hg?

Answer 1

Solution :

degrees of freedom = n - 1 = 173 - 1 = 172

t/2,df = t0.005,172 = 2.605

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

= 2.605 * ( 7.6 / 173)

Margin of error = E = 1.5

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

- E < < + E

31.6 - 1.5 < < 31.6 + 1.5

30.1 hg < < 33.1 hg

Yes, because the confidence interval limits are not similar