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Problem Set IV

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Week 5

Problem Set IV Complete and submit the following problem set:

• Lind Chapter 9: Exercises 12, 28

• Lind Chapter 10: Exercises 6, 18

• Lind Chapter 12: Exercise 30

Lind Chapter 9:

12. The American Sugar Producers Association wants to estimate the mean yearly sugar consumption. A sample of 16 people reveals the mean yearly consumption to be 60 pounds with a standard deviation of 20 pounds.

A. What is the value of the population mean? What is the best estimate of this value?

The value of the population mean is unknown. However, we have the sample mean to guesstimate it at about 60 pounds.

B. Explain why we need to use the t distribution. What assumption do you need to make?

Given that the standard deviation of the population is unknown, we need to use the t distribution. We need to assume that the population follows normal distribution.

C. For a 90 percent confidence interval, what is the value of t?

For a 90 percent confidence interval, the value of t is 2.131

D. Develop the 90 percent confidence interval for the population mean.

(49.345, 70.655)

E. Would it be reasonable to conclude that the population mean is 63 pounds?

Yes, because 63 is in the above 90% confidence interval

28. A processor of carrots cuts the green top off each carrot, washes the carrots, and inserts six to a package. Twenty packages are inserted in a box for shipment. To test the weight of the boxes, a few were checked. The mean weight was 20.4 pounds, the standard deviation 0.5 pounds. How many boxes must the processor sample to be 95 percent confident that the sample mean does not differ from the population mean by more than 0.2 pounds?

There should be a minimum of 17 boxes.

Lind Chapter 10:

6. The MacBurger restaurant chain claims that the waiting time of customers for service is normally distributed, with a mean of 3 minutes and a standard deviation of 1 minute. The quality-assurance department found in a sample of 50 customers at the Warren Road

MacBurger that the mean waiting time was 2.75 minutes. At the .05 significance level, can we conclude that the mean waiting time is less than 3 minutes?

Ho: The mean waiting time is 3 minutes

Ha: The mean waiting time is less than 3 minutes.

Z = ( 2.75 - 3) / 1/ √50

Z = -1.77

At 0.05 level the table value is -1.645

Given that the calculated value lies in the rejection region, reject Ho.

The mean waiting time is less than 3 minutes.

18. The management of White Industries is considering a new method of assembling its golf cart. The present method requires 42.3 minutes, on the average, to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, and the standard deviation of the sample was 2.7 minutes. Using the .10 level

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