1. A recent poll of 1500 U.S. voters found that 855 of them currently approve of Obama’s performance as president. (http://www.gallup.com/poll/113980/Gallup-Daily-Obama-Job-Approval.aspx) 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. (Extra credit if you can explain how it is possible that 57% approve of Obama and yet the Democrats lost the election . . . Just kidding about the extra credit . . .)
2. New car owners are currently keeping their cars for an average (mean) of 6.5 years with a standard deviation of 2.2 years. (http://www.cnbc.com/2015/07/28/americans-holding-onto-their-cars-longer-than-ever.html). You have sampled 37 owners of new Mercedes Benz and found a mean length of ownership of 7.1 years. Test to determine if Mercedes’ owners hold their car longer than the average for all brands using α = 0.01 and the prob-value approach.

3. 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:

Test to determine if the drug raises patients’ pulse rate more than the placebo using α = 0.025

1. Corn plants grow to a mean height of 6.44 feet with a standard deviation of 0.31 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 6.58 feet. Test to determine if the genetically modified plants are taller using α = 0.01.

5. The IRS claims that it takes 87.3 minutes to prepare a tax form. You believe it takes longer than 87.3 minutes. You ask 21 tax filers and find the following preparation times:

60, 63, 65, 70, 76, 79, 84, 85, 88, 90, 90, 98, 98, 99, 100, 101, 110, 111, 115, 123, 137

Determine if the preparation time is significantly longer than 87.3 minutes. Use α = .025.

6. Given below are the birth weights of babies born to mothers who took special vitamin supplements while pregnant:

3.23 4.37 3.93 4.33 3.39 3.68 4.68 3.52

3.02 4.29 2.47 4.13 4.47 3.22 3.33 2.64

a. Make a 95% 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 = .025

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.

1. 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):
2. 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:

• (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))

Test to determine if the use of caffeine lowered birth weights using α = 0.05.

EXTRA CREDIT: 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.

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