The table below contains the amount that a sample of nine customers spent for lu

Question

The table below contains the amount that a sample of nine customers spent for lunch ($) at a fast-food restaurant. 4.59 5.47 5.81 6.56 7.21 7.59 8.12 8.64 9.64 a. Construct a 99% confidence interval estimate for the population mean amount spent for lunch ($) at the fast-food restaurant, assuming a normal distribution. b. Interpret the interval constructed in (a). Click here to view page 1 of the table of critical values of t. LOADING... Click here to view page 2 of the table of critical values of t. LOADING... a. The 99% confidence interval estimate is from $ nothing to $ nothing. (Round to two decimal places as needed.) b. Which statement below best interprets this interval? A. The mean amounts in dollars spent for lunch at the fast-food restaurant of 99% of all samples of the same size are contained in the interval. B. We have 99% confidence that the mean amount in dollars spent for lunch at the fast-food restaurant for the sample is contained in the interval. C. We have 99% confidence that the population mean amount in dollars spent for lunch at the fast-food restaurant is contained in the interval.

Explanation / Answer

n = 9 ,

mean= 7.07 Standard deviation = 1.62097

t-distribution with n-1 = 8 df

t0.005 = 3.355

hence 99 % confidence interval is (7.07 - 3.355*1.62097/sqrt(9),7.07 + 3.355 *1.62097/sqrt(9))

=(5.2572,8.8827847)

b) option C) is correct

a range of values so defined that there is a specified probability that the value of a parameter lies within it.

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