Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals185​, x overbarequals32.4 ​hg, sequals6.1 hg. Construct a confidence interval estimate of the mean. Use a 95​% confidence level. Are these results very different from the confidence interval 31.6 hgless thanmuless than33.8 hg with only 15 sample​ values, x overbarequals32.7 ​hg, and sequals1.9 ​hg?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 32.4 hg

sample standard deviation = s = 6.1 hg

sample size = n = 185

Degrees of freedom = df = n - 1 = 185 - 1 = 184

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,184 = 1.973

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

= 1.973 * (6.1 / 185)

Margin of error = E = 0.9

The 95% confidence interval estimate of the population mean is,

   - E < < + E

32.4 - 0.9 < < 32.4 + 0.9

( 31.5 hg < < 33.3 hg )

Yes, because the confidence interval limits are not similar