Problem 1
(You should not use MegaStat to answer this question)
Recent study stated that if a person chewed gum, the average number of sticks of gum he or she chewed daily was 8. To test the claim, a researcher selected a random sample of 36 gum chewers and found the mean number of sticks of gum chewed per day was 9. The standard deviation of the population is 1.
- At a = 0.05, is the number of sticks of gum a person chews per day actually greater than 8? Calculate the value of the test statistic and state the hypotheses, and conclusion.
- Find a 95% confidence interval for the average number of sticks of gum he or she chewed daily
- Explain what does “a 95% confidence” means.
- What would your conclusion be if the p-value of the test is equal to 0.0?
- State and check the assumptions of the test you used in (a)
Problem 2
(You can use MegaStat to answer this question)
A random sample of the average debt (in dollars) at graduation from 24 of the top 100 public colleges and universities is listed below.
16,012 15,784 16,597 18,105 12,665 14,734
17,225 16,953 15,309 15,297 14,437 14,835
13,607 13,374 19,410 18,385 22,312 16,656
- Is there sufficient evidence at the 10% significance level to conclude that the population mean debt at graduation is less than $18,000?
- What is the required conditions for the statistical test in part (a)?
Problem 3
(You should not use MegaStat to answer this question)
For a certain year a study reports that the percentage of college students using credit cards was 83%. A college dean of student services feels that this is too high for her university, so she randomly selects 50 students and finds that 40 of them use credit cards.
- At a = 0.10, is she correct about her university? Calculate the value of the test statistic and state the alternatives, and conclusion.
- What would your conclusion be if the p-value of the test is equal to 0.0?
- Estimate the population proportion with 90% confidence.
- Check the assumptions of the test you used in (a)
Problem 4
(You should not use MegaStat to answer this question)
Health Care Knowledge Systems reported that an insured woman and an uninsured woman spend on average same number of days in the hospital for a routine childbirth, Assume two samples of 35 women each were used in both samples. The summary statistics are as follows:
|
Sample size
|
Mean
|
Standard deviation
|
Insured woman
|
35
|
2.3
|
0.6
|
Uninsured woman
|
35
|
1.9
|
0.5
|
F-test for equality of variance
|
|
|
|
|
0.36
|
variance: Insured woman
|
|
0.25
|
variance: uninsured woman
|
|
1.44
|
F
|
|
|
|
.2926
|
p-value
|
|
|
|
|
|
|
|
- Test the claim that the means are equal. Calculate the value of the test statistic and state the alternatives, and conclusion. Use a = 0.01,
- Find the 99% confidence interval for the differences of the means.
- At a = 0.01, test the claim that the means are equal. Use the confidence interval method to do the test.
Problem 5
(You can use MegaStat to answer this question)
Listed below is the moisture content (by percent) for random samples of different fruits and vegetables.
- At 5% significance level, can it be concluded that fruits differ from vegetables in average moisture content? Calculate the p-value and state the alternatives, and conclusion.
- Chick the assumptions of the above test.
Problem 6
(You can use MegaStat to answer this question)
As an aid for improving students’ study habits, nine students were randomly selected to attend a seminar on the importance of education in life. The table shows the number of hours each student studied per week before and after the seminar.
Before:
9 12 6 15 3 18 10 13 7
After:
9 17 9 20 2 21 15 22 6
- At 10% significance level, did attending the seminar increase the number of hours the students studied per week? Calculate the value of the test statistic and state the alternatives, and conclusion.