Hypothesis Testing, Confidence Intervals

Mathematics 1670 Assignment 3 – Chapters 9-12

Hypothesis testing, confidence intervals,

Name: ______ID______

Due Date November 18

1) A researcher wanted to compare the pulse rates of identical twins to see whether there was any difference. Eight sets of twins were selected. The rates are given in the table as number of beats per minutes. At the = 0.01, is there a significant difference in the average pulse rates of twins? (5 1 for each step)

TWIN A 87 92 78 83 90 84 93 88 TWIN B 83 95 79 83 86 93 80 86

ma= m b

ma m b  a = 0.01

1 CONCLUSION: Accept H0

p-value = 0.651 <0.01 Accept Ho

2. Health Care Knowledge Systems reported that an insured woman spends on average 2.3 days in the hospital for a routine childbirth, while an uninsured woman spends on average 1.9 days. Assume two samples of 16 women each were used and the standard deviations are both equal to 0.6 days. At the 0.01, test the claim that the means are equal. Classical p-value

2 1.H0 : m 1= m 2

H1 : m 1 m 2 2. a = 0.01

CONCLUSION: Accept H0 p-value = 0.069 < 0.01 Accept H0

3 4 3. The manufacturer of a patent medicine claimed that it was 90% effective in relieving an allergy for a period of 8 hours. In a sample of 200 people who had the allergy, the medicine provided relief for 160 people. Determine whether the manufacturer’s claim is correct. Use = 0.05. Classical p-value

H0:  ≠  = 0.05

CONCLUSION: Accept H1

5 3. A survey of 1000 students nationwide showed a mean ACT score of 21.4. A survey of 500 Ohio scores showed a mean of 20.8. If the standard deviation in each case of 3, can we conclude that Ohio is below the national average? Use = 0.05. (5) (1 for each step)

1.H0 : m 1 m 2

H1 : m 1> m 2 2. a = 0.05

CONCLUSION accept H1. p-value = 0.00013< 0.05

6 4. A dietian read in a survey that at least 55% of adults do not eat breakfast at least 3 days a week. To verify this, she selected a random sample of 80 adults and asked them how many days a week they skipped breakfast. A total of 50% responded that they skipped breakfast at least 3 days a week. At the=0.10 test the claim. (5)

1.H0 :p 1 = 0.55

H1 :p 1 0.55 2. a = 0.10

Two tailed

7 CONCLUSION accept H0. p-value = 0.369>0.10

8 5. A travel agent claims that the average of the number of rooms in hotels in a large city is 500. At = 0.01 is the claim realistic? The data for a sample of seven hotels are shown. 713 300 292 311 598 401 618

Classical p-value

1.H0 :m = 500

H1 :m 500 2. a = 0.01

CONCLUSION accept H0. p-value = 0.589>0.01

9 10 6. In a sample of 200 surgeons, 15% though the government should control health care. In a sample of 200 general practitioners, 21% felt the same way. At =0.01, is there a difference in the proportions? (5)

1.H0 :p 1= p 2

H1 :p 1 p 2 2. a = 0.01

CONCLUSION accept H0. p-value = 0.118>0.01

11 12 7. The President and CEO of Cliff Hanger International Airlines is concerned about high cholesterol levels of the pilots. In an attempt to improve the situation a sample of seven pilots is selected to take part in a special program, in which each pilot is given a special diet by the company physician. After six months each pilot’s cholesterol level is checked again. At the 0.01 significance level can we conclude that the program was effective in reducing cholesterol levels?

Pilot Before After 1 255 210 2 230 225 3 290 215 4 242 215 5 300 240 6 250 235 7 215 190

a. State the null and alternate hypotheses.

H0: _d 0 _H1: _d>0

13 b. Shade the critical area.

c. Compute the value of the test statistic.

d. Compute the p-value. P=0.0045

e. What is your decision regarding the null hypothesis?

Conclusion accept H1 the programme was effective!

14 8. The number of unhealthy days based on the Air Quality Index for a random sample of metropolitan areas is shown. Construct a 98% confidence interval based on the data. 61 12 6 40 27 38 93 5 13 40