Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals180​, x overbarequals30.8 ​hg, sequals7.5 hg. Construct a confidence interval estimate of the mean. Use a 95​% confidence level. Are these results very different from the confidence interval 29.7 hgless thanmuless than32.9 hg with only 17 sample​ values, x overbarequals31.3 ​hg, and sequals3.1 ​hg?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 30.8

sample standard deviation = s = 7.5

sample size = n = 180

Degrees of freedom = df = n - 1 = 179

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,179 = 1.973

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

= 1.973 * (7.5 / 180)

= 1.103

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

- E < < + E

30.8 - 1.1 < < 30.8 + 1.1

29.7 < < 31.9

These results are different from the confidence interval,

29.7 < < 32.9