Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals225​, x overbarequals29.1 ​hg, sequals7.7 hg. Construct a confidence interval estimate of the mean. Use a 99​% confidence level. Are these results very different from the confidence interval 26.6 hgless thanmuless than32.2 hg with only 13 sample​ values, x overbarequals29.4 ​hg, and sequals3.3 ​hg?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 29.1

sample standard deviation = s = 7.7

sample size = n = 225

Degrees of freedom = df = n - 1 = 225 - 1 = 224

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t_0.005,224 = 2.598

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

= 2.598 * (7.7 / 225)

Margin of error = E = 1.3

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

- E < < + E

29.1 - 1.3 < < 29.1 + 1.3

27.8 < < 30.4