Question

The manager of The Cheesecake Factory in Boston reports that on six randomly selected weekdays, the number of customers served was 165, 135, 95, 145, 125, and 265. She believes that the number of customers served on weekdays follows a normal distribution. Construct the 90% confidence interval for the average number of customers served on weekdays. (You may find it useful to reference the t table. Round intermediate calculations to at least 4 decimal places, "sample mean" and "sample standard deviation" to 2 decimal places. Round "t" value to 3 decimal places and final answers to 2 decimal places.)

Confidence interval: to

Answer 1

From given sample,

= X / n = 155

S = sqrt [ X² - n ² / n-1 ] = 58.65

t critical value at 0.10 significance level with 5 df = 2.015

90% confidence interval for is

- t * S / sqrt(n ) < < + t * S / sqrt(n )

155 - 2.015 * 58.65 / sqrt (6) < < 155 + 2.015 * 58.65 / sqrt (6)

106.75 < < 203.25

90% C I is 106.75 to 203.25