Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals202​, x overbarequals28.3 ​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 27.8 hgless thanmuless than29.6 hg with only 17 sample​ values, x overbarequals28.7 ​hg, and sequals1.8 ​hg? What is the confidence interval for the population mean mu​? nothing hgless thanmuless than nothing hg ​(Round to one decimal place as​ needed.) Are the results between the two confidence intervals very​ different? A. ​Yes, because the confidence interval limits are not similar. B. ​Yes, because one confidence interval does not contain the mean of the other confidence interval. C. ​No, because each confidence interval contains the mean of the other confidence interval. D. ​No, because the confidence interval limits are similar.

Answer 1

The confidence interval is 27.5 hg less than mu less than 29.1 hg

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

Explanation:

Confidence interval is given as mean +/- margin of error (E)

mean = 28.3 hg

sd = 6.1 hg

n = 202

degree of freedom = n-1 = 202-1 = 201

confidence level (C) = 95% = 0.95

significance level = 1 - C = 1 - 0.95 = 0.05 = 5%

critical value corresponding to 201 degrees of freedom and 5% significance level is 1.97196

E = t×sd/√n = 1.97196×6.1/√202 = 0.8 hg

Lower limit = mean - E = 28.3 0.8 = 27.5 hg

Upper limit = mean + E = 28.3 + 0.8 = 29.1 hg

95% confidence interval is (27.5, 29.1)

When mean is 28.3, sd = 6.1 and n = 202, the confidence limits are 27.5 and 29.1 which is different from 27.8 and 29.6 which are the confidence limits when mean is 28.7, sd = 1.8 and n = 17