135k views
5 votes
A sample of 40 individuals at a shopping mall found that the mean number of visits to a restaurant per week was 2.88 with a standard deviation of 1.59. Find a 99% confidence interval for the mean num-ber of restaurant visits. Use the appropriate formula and verify your result using the Confidence Intervals workbook.

1 Answer

3 votes

Answer:

The confidence interval is between 2.23 and 3.53

Step-by-step explanation:

The confidence interval (C) = 99% = 0.99

α = 1 - C = 1 - 0.99 = 0.01

α/2 = 0.01/2 = 0.005

The z score of α/2 corresponds to the z score of 0.495 (0.5 - 0.005) which is 2.576

The margin of error (E) is given as:


E=z_{(\alpha)/(2) }*(\sigma)/(√(n) )\\\\where\ n=sample\ size,\sigma=standard\ deviation\\\\Given\ that\ \sigma=1.59,n=40,z_{(\alpha)/(2) }=2.576\ hence: \\\\E=2.576*(1.59)/(√(40) ) =0.65

The confidence interval = mean ± margin of error = 2.88 ± 0.65 = (2.23, 3.53)

The confidence interval is between 2.23 and 3.53

User Kengcc
by
8.3k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories