One- and Two-Sample Tests of Hypothesis, Variance, and Chi-Squared Analysis Problem Sets

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Chapter 10
31. A new weight-watching company, Weight Reducers International, advertises that those who join will lose, on the average, 10 pounds the first two weeks with a standard deviation of 2.8 pounds. A random sample of 50 people who joined the new weight reduction program revealed the mean loss to be 9 pounds. At the .05 level of significance, can we conclude that those joining Weight Reducers on average will lose less than 10 pounds? Determine the p-value.

Answer:

H0: = 10 pounds
H1: < 10 pounds

Reject the null hypothesis if Z < -1.65

9.0 – 10.00
Z= ---------------- = -2.53
2.8/sqrt(50)

The null hypothesis is rejected, the average weight loss is less than 10 pounds.

p-value = .5000 - .4943 = .0057

32. Dole Pineapple, Inc. is concerned that the 16-ounce can of sliced pineapple is being overfilled. Assume the standard deviation of the process is .03 ounces. The quality control department took a random sample of 50 cans and found that the arithmetic mean weight was 16.05 ounces. At the 5 percent level of significance, can we conclude that the mean weight is greater than 16 ounces? Determine the p-value.

Answer:

H0: = = 16 ounces
H1: = > 16 pounds

Reject the null hypothesis if Z > 1.65

16.05 – 16.00
Z= ---------------- = 11.79
.03/sqrt(50)

The null hypothesis is rejected, since 11.79 is greater than 1.65. The 16-ounce cans can be overfilled.

p-value ~ 0.

38. A recent article in The Wall Street Journal reported that the 30-year mortgage rate is now less than 6 percent. A sample of eight small banks in the Midwest revealed the following 30-year rates (in percent):

At the .01 significance level, can we conclude that the 30-year mortgage rate for small banks is less than 6 percent? Estimate the p-value.

Answer:

H0: = 6 percent
H1: < 6 percent

Reject the null hypothesis if t < -2.99

5.638 – 6.0
t= ---------------- = -1.61
.6346/sqrt(8)

T is not < -2.99 so the null hypothesis is accepted. The 30-year mortgage rate is not less than 6 percent.

p-value = .669

Chapter 11
27. A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. The information is summarized below.

At the .01 significance level, is there a difference in the mean number of times men and women order take-out dinners in a month? What is the p-value?

Answer:

H0: = There is no difference in the mean for take out dinner times for men and women. H1: = There is a difference in the mean for take out dinner times for men and women.

Reject the null hypothesis if z > 2.33

24.51 – 22.69 1.82
Z= ------------------------------- = ------- = 1.871
Sqrt(4.482/35 + 3.862/40) .973

Z < 2.33, thus the null hypothesis is accepted. There is no difference in the mean for take out dinner times for men and women.

P-value = (.5 - .4693) = .031

46. Grand Strand Family Medical Center is specifically set up to treat minor medical emergencies for visitors to the Myrtle Beach area. There are two facilities, one in the Little River Area and the other in Murrells Inlet. The Quality Assurance Department wishes to compare the mean waiting time for patients at the two locations. Samples of the waiting times, reported in minutes, follow:

Assume the population standard deviations are not the same. At the .05 significance level, is there a difference in the mean waiting time?

Answer:

H0 = There is no difference in the mean for waiting time on Little River vs. Murrells Inlet location.
H1 = There is difference in the mean for waiting time on Little River location vs. Murrells Inlet location.

With 20 degree of freedom and .05 significance level, the critical t values are -2.086 and 2.086.

Two-sample T for Little River vs Murrells Inlet

N Mean StDev SE Mean...
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