Mba510 Week 5 Problem Set
GRADE 9/9
Lind Chapter 9; Exercise 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 population mean is unknown. The best estimate of this value is the sample mean of 60 pounds.
b. Explain why we need to use the t distribution. What assumption do you need to make?
According to Lind, et al. (2005), when population standard deviation is unknown, and the sample is smaller than 30, a t distribution should be used. We need to assume that the sample is from a normal population (pp. 291-293).
c. For a 90 percent confidence interval, what is the value of t?
For a 90 percent confidence interval, and df = 15, t= 1.753
d. Develop the 90 percent confidence interval for the population mean.
Xbar = 60; s = 20; n = 16
Xbar ±t(s/√n) = 60 ± 1.753 (20/√16) = 60± 1.753 (5) = 60± 8.765 = [51.24, 68.77]
e. Would it be reasonable to conclude that the population mean is 63 pounds?
Yes because 63 pounds is among the confidence intervals. Lind Chapter 9; Exercise 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?
Xbar = 20.4; s = 0.5; E= 0.2; 95% confidence level
At 95% confidence level, z = 1.96 n = (zs/E)² = [(1.96)(0.5)/0.2]² = 24.01
24.01 boxes must the processor sample to be 95 percent confident that the same mean does not differ from the population mean by more than 0.2 pounds. Lind Chapter 10; Exercise 6
The MacBurger
Lind Chapter 9; Exercise 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 population mean is unknown. The best estimate of this value is the sample mean of 60 pounds.
b. Explain why we need to use the t distribution. What assumption do you need to make?
According to Lind, et al. (2005), when population standard deviation is unknown, and the sample is smaller than 30, a t distribution should be used. We need to assume that the sample is from a normal population (pp. 291-293).
c. For a 90 percent confidence interval, what is the value of t?
For a 90 percent confidence interval, and df = 15, t= 1.753
d. Develop the 90 percent confidence interval for the population mean.
Xbar = 60; s = 20; n = 16
Xbar ±t(s/√n) = 60 ± 1.753 (20/√16) = 60± 1.753 (5) = 60± 8.765 = [51.24, 68.77]
e. Would it be reasonable to conclude that the population mean is 63 pounds?
Yes because 63 pounds is among the confidence intervals. Lind Chapter 9; Exercise 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?
Xbar = 20.4; s = 0.5; E= 0.2; 95% confidence level
At 95% confidence level, z = 1.96 n = (zs/E)² = [(1.96)(0.5)/0.2]² = 24.01
24.01 boxes must the processor sample to be 95 percent confident that the same mean does not differ from the population mean by more than 0.2 pounds. Lind Chapter 10; Exercise 6
The MacBurger