Homework Question 10 (of 17) 10.00 points The manager of The Cheesecake Factory

Question

Homework Question 10 (of 17) 10.00 points The manager of The Cheesecake Factory in Boston reports that on six randomly selected weekdays, the number of customers served was 145, 190, 170, 225, 230, and 260. She believes that the number of customers served on weekdays follows a normal distribution. Construct the 95% confidence interval for the average number of customers served on weekdays. Use Table 2. (Round intermediate calculations to 4 decimal places, sample mean and "sample standard deviation" to 2 decimal places. Round "r value to 3 decimal places and final answers to 2 decimal places.) Hints References eBook & Resources

Explanation / Answer

Question 10

Solution:

Here, we have to find the confidence interval for the population mean.

We are given

Confidence level = 95%

Sample size = n = 6

Degrees of freedom = n - 1 = 6 – 1 = 5

Critical value t = 2.571

From the given sample, we have

Sample standard deviation = SD = 42.62

Sample mean = Xbar = 203.33

The confidence interval formula is given as below:

Confidence interval = Xbar -/+ t*SD/sqrt(n)

Confidence interval = 203.33 -/+ 2.571*42.62/sqrt(6)

Confidence interval = 203.33 -/+ 44.7341

Lower limit = 203.33 – 44.7341 = 158.60

Upper limit = 203.33 + 44.7341 = 248.06

Confidence interval = (158.6, 248.06)

Question 16

We are given

Sample size = n = 16

Sample mean = Xbar = 132000

Sample standard deviation = SD = 28000

Confidence level = 95%

Degrees of freedom = n – 1 = 16 – 1 = 15

Critical t value = 2.571

The confidence interval formula is given as below:

Confidence interval = Xbar -/+ t*SD/sqrt(n)

Confidence interval = 132000 -/+ 2.571*28000/sqrt(16)

Confidence interval = 132000 -/+ 44.7341

Lower limit = 132000 – 44.7341 = 131955.27

Upper limit = 132000 + 44.7341 = 132044.73

Confidence interval = (131955.27, 132044.73)

Question 17

(All answers are given by using the t-table.)

Part a

Alpha = 0.01, df = 23

Critical t value = 2.500

Part b

Alpha = 0.05, df = 23

Critical t value = 1.714

Part c

Alpha = 0.01, df = 8

Critical t value = 2.896

Part e

Alpha = 0.05, df = 8

Critical t value = 1.860

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