Question

Here are summary statistics for randomly selected weights of newborn​ girls: n = 187​, x = 33.9 ​hg, s = 6.1 hg. Construct a confidence interval estimate of the mean. Use a 99​% confidence level. Are these results very different from the confidence interval 32.4 hg<U< 35.0 hg with only 16 sample​ values, x = 33.7 ​hg, and s=1.8 ​hg?

A) What is the confidence interval for the population mean

B) Are the results between the two confidence intervals very​ different?

A.​Yes, because the confidence interval limits are not similar.

B.No, because the confidence interval limits are similar.

C.​No, because each confidence interval contains the mean of the other confidence interval.

D. ​Yes, because one confidence interval does not contain the mean of the other confidence interval.

Answer 1

Solution :

Given that,

A) Point estimate = sample mean = = 33.9

sample standard deviation = s = 6.1

sample size = n = 187

Degrees of freedom = df = n - 1 = 187- 1 = 186

t _/2,df = 2.6

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

= 2.6 * (6.1 / 187)

Margin of error = E = 1.161

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

- E < < + E

33.9 - 1.161 < < 33.9 +1.161

32.7 < < 35.1

(32.7,35.1)

B)  A.​Yes, because the confidence interval limits are not similar.