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 number of restaurant visits. Use the appropriate formula and verify your result using the confidence intervals workbook.

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Problem 17, P. 241 Sample Standard Deviation, 5 1.59 Sample size (n) 40 Degrees Freedom (df) = n-1 39 Hint: Use the formula below & Table A. 2 P. 636 - 638 Sample mean, X 2.88 Find 95% Confidence Interval for Mean xit Upper limit a (6.3) (Specify formulas for upper & lower limits in cells C8 & C9) Lower limit (They should match the values in B26 & B25 below.) Why is the 95% Confidence Interval narrower?: Find 99% Confidence Interval for Mean Upper limit (Specify formulas for upper & lower limits in cells Ci1 & C12) Lower limit (They should match the values in F26 & F25 below.) Confidence Interval for Population Mean, Standard Deviation Unknown TEMPLATE Confidence Interval for Population Mean, Standard Deviation Unknown TEMPLATE Note: confidence level = .95 Note: confidence level = .99 Alpha 0.05 Alpha 0.01 Sample standa 1.59 Sample standard dev 1.59 Sample size 40 Sample size 40 Sample average 2.88 Sample average 2.88 Confidence Int 95% Confidence Interval 99% t-value 2.0227 t-value 2.7079 Error 0.51 Error 0.68 Lower 2.37 Lower 2.20 Upper 3.39 Upper 3.56