1. A pharmaceutical company is testing a new cold medicine to determine if the drug has side affects. To test the drug, 7 patients are given the drug and 8 patients are given a placebo (sugar pill). The change in pulse rate (beats per minute) after taking the pill was as follows:
|
Given drug:
|
3
|
5
|
2
|
8
|
1
|
4
|
5
|
|
|
Given placebo:
|
2
|
3
|
1
|
1
|
2
|
0
|
2
|
4
|
Test to determine if the drug raises patients' pulse rate more than the placebo using α = 0.05
2. Corn plants grow to a mean height of 6.44 feet with a standard deviation of 0.12 feet. A new genetically modified corn plant is being developed. A researcher grew a group of 37 genetically modified corn plants and found that their mean height was 6.48 feet. Test to determine if the genetically modified plants are taller using α = 0.01.
3. A recent poll of 2350 U.S. voters found that 1034 of them currently approve of Obama's performance as president. Make a 99% confidence interval for the proportion of all U.S. voters that approve of Obama's performance. Describe the meaning of the interval you found.
4. From the population of all Americans with high cholesterol, which has a mean cholesterol level of 265 mg/dL with a standard deviation of 42, Isis Pharmaceuticals chose 38 patients for treatment with its new antisense drug. After treatment, the patients in the study had a mean cholesterol level of 249. Test to determine if the antisense drug lowered the cholesterol using α = 0.01 and the prob-value approach.
5. The IRS claims that it takes 87 minutes to prepare a tax form. You believe it takes longer than 87 minutes. You ask 19 tax filers and find the following preparation times:
61, 74, 78, 79, 87, 89, 96, 97, 98, 98, 99, 100, 100, 105, 108, 110, 117, 120, 121
Determine if the preparation time is significantly longer than 87 minutes. Use α = .025.
6. Given below are the birth weights of babies born to mothers who took special vitamin supplements while pregnant:
3.13 4.57 3.93 4.33 3.39 3.68 4.68 3.52
3.02 4.29 2.47 4.13 4.47 3.22 3.43 2.54
a. Make a 99% confidence interval for the mean weight of babies whose mothers take vitamin supplements.
b. Do a hypothesis test to determine if these babies weight is more than the mean weight for the population of all babies which is 3.39 kg using α = .005
c. In a short paragraph, describe the relationship between your answer to part (a) and your answer to part (b).
7. A student wanted to study the ages of couples applying for marriage licenses in his county. He studied a sample of 94 marriage licenses and found that in 67 cases the husband was older than the wife. Do the sample data provide evidence that the husband is usually older than the wife among couples applying for marriage licenses in that county? Explain briefly and justify your answer using the methods studied in this course.Use α= 0.05 as the level of significance.
8. The national standard for salinity (salt content) in drinking water is 500 parts per million. A government inspector tests the drinking water at 10 locations around a certain city and finds the following levels of salinity (in parts per million of course):
|
532
|
560
|
506
|
452
|
425
|
531
|
582
|
376
|
529
|
629
|
(a) Make a 95% confidence interval for the mean salinity level in this city.
(b) Is this city's salinity level above the national standard? (Use α = 0.025 and the result from part (a))
9. A researcher studied whether pregnant women who consumed more than 800 mg of caffeine per day had babies with a lower birth weight (in lbs). The results are in the table below:
|
Caffeinated women's babies weights
|
7.5
|
6.7
|
5.2
|
6.8
|
7.4
|
7.7
|
5.6
|
6.9
|
|
Non-caffeinated women's babies weights
|
7.8
|
7.6
|
8.1
|
7.5
|
7.1
|
7.2
|
7.9
|
|
Test to determine if the use of caffeine lowered birth weights using α = 0.05.
10. You intend to poll students in order to determine the proportion of all students who will vote in an upcoming election. What sample size do you need to choose to ensure that the margin of error is less than 3% assuming a 99% level of confidence? Hint: "To ensure" means that, even in the worst case scenario which is when the variation is largest, your confidence interval would still have a 99% chance of containing the true proportion of all students.