Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals159​, x overbarequals32.4 ​hg, sequals6.8 hg. Construct a confidence interval estimate of the mean. Use a 98​% confidence level. Are these results very different from the confidence interval 31.2 hgless thanmuless than34.6 hg with only 16 sample​ values, x overbarequals32.9 ​hg, and sequals2.6 ​hg?

Answer 1

Solution :

degrees of freedom = n - 1 = 159 - 1 = 158

t/2,df = t0.01,158 = 2.350

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

= 2.350 * ( 6.8 / 159)

Margin of error = E = 1.3

The 98% confidence interval estimate of the population mean is,

- E < < + E

32.4 - 1.3 < < 32.4 + 1.3

( 31.1 hg < < 33.7 hg )

Yes, because the confidence interval limits are not similar