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.

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.

Step-by-step 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

1^3 + 2^3 + 3^3 +4^3 = (1+2+3+4)^2

step-by-step explanation:

i am only able to answer the first part. i am not sure about the second part. sorry

it is a secret for it not to be compromised sorry

step-by-step explanation: