Question

A hamburger chain sells large hamburgers. When we take a sample of 40 hamburgers and weigh​ them, we find that the mean is 0.52 pounds and the standard deviation is 0.3 pound.

Technology Output N 40 Mean 0.52   STDEV. 0.3000 One-Sample T SE Mean 0.0474 95% CI ​(0.4241, 0.61590)

a. A technology input menu for calculating a confidence interval requires a sample​ size, a sample​ mean, and a sample standard deviation. State how you would fill in these numbers.

Sample Size: Answer ____ Sample Mean: Answer ____   Standard Deviation: Answer _____ (Type an integer or a decimal do not round)

b. Choose the correct interpretation of the confidence interval below and fill in the answer boxes to complete your choice. ​(Round to four decimal places as needed. Use ascending​ order.) Using the accompanying technology​ output, report the confidence interval in a carefully worded sentence.

A.We are​ 95% confident that the population mean is between Answer ___ and ___.

B.There is a​ 95% probability that the recorded sample mean is between Answer ___ and ___.

C.There is a​ 95% probability that the population mean is between Answer ___ and ___.

D.We are​ 95% confident that the recorded sample mean is between Answer ___ and ___.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 0.52

sample standard deviation = s = 0.3

sample size = n = 40

Degrees of freedom = df = n - 1 = 40 - 1 = 39

SE = s / n = 0.3 / 40 = 0.0474

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,39 = 2.023

Margin of error = E = t/2,df * SE

= 2.023 * 0.0474

Margin of error = E = 0.0959

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

  ± E  

= 0.52  ± 0.0959

= ( 0.4241,0.6159 )

A.We are​ 95% confident that the population mean is between 0.4241 and 0.6159