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Chapter 6 and 8 review. By Henry Mesa Portland Community College. Here are a few problems to try out. Use the arrow keys on your keyboard to move forward or backwards on the slide. Enjoy the animation. 1. What is the purpose of creating a confidence interval?.
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Chapter 6 and 8 review By Henry Mesa Portland Community College
Here are a few problems to try out. Use the arrow keys on your keyboard to move forward or backwards on the slide. Enjoy the animation.
1. What is the purpose of creating a confidence interval? General answer: To estimate a parameter. With respect to section 6.1: estimate the population mean, . With respect to section 8.1: to estimate the population proportion parameter, p-hat.
If I create a 99% confidence interval to estimate μ, what is the probability that the interval contains the parameter μ? In the long run, 99% of the time the interval created will contain the parameter.
3. I want to estimate μ with a 90% confidence interval so that my margin of error is at most 4 units. I know that σ = 20 units. How big should be my sample size? n = 68
4. I create a 95% confidence interval to estimate μ. If I then construct a 99% confidence interval using the same data, will my interval be larger or smaller than the 95% confidence interval? The 99% confidence interval will be wider compared to the 95% confidence interval.
The news editor at KGW news wants to estimate the number of Portland residents who would support same sex civil unions. During the 5:00 pm broadcast of the news, a story is done about the legalities of same sex civil unions, and the end of the story segment, viewers are asked to call in and vote on the question “Would you support same sex civil unions?” Two phone numbers are provided, 1-888- 789- 4000 for “Yes”, 1-888-789-4001 for “No”. At the10:00pm broadcast the news reported that 55% of Portland residents are in support of same sex civil unions, with a margin of error of plus/minus 4% points. Is this result trustworthy? Explain.
No, the confidence interval created can not be used to make an inference about the sentiment of the entire population. The reason is that the method used to gather the data is not a simple random sample, but rather a volunteer response sample (chapter 3 page 249)
The student union wants to estimate the mean age of the student body at their school. A preliminary simple random sample of 40 students is gathered. The students will use the standard deviation from the sample for σ, σ = 4.6 years. The sample mean is 27.4 years. Create a 95% confidence interval. (26.0, 28.8)
A person wants to estimate the probability that a die tossed completely at random comes up six. The person tosses a die 64 times, and the number of times a six appears is 14. Create a 95% confidence interval. (0.1175, 0.3201)
A test for breast cancer is 90% effective in detecting the cancer in women who do have breast cancer. • Suppose that 30 women who have breast cancer go in for testing. What is the expected number test that would detect breast cancer? • What is the probability having all the 30 test having a positive result (all the cancers are detected)? • c. What is the probability of having 25 or less of the tests detect the cancer out of the 30?
A test for breast cancer is 90% effective in detecting the cancer in women who do have breast cancer. • Suppose that 30 women who have breast cancer go in for testing. What is the expected number test that would detect breast cancer? • = np • = 30(0.9) = 27 We would expect 27 tests to come back with positive results. Remember this is a long run average.
A test for breast cancer is 90% effective in detecting the cancer in women who do have breast cancer. • What is the probability having all the 30 test having a positive result (all the cancers are detected)? P(All 30 tests are positive) = = 0.4239
A test for breast cancer is 90% effective in detecting the cancer in women who do have breast cancer. • c. What is the probability of having 25 or less of the tests detect the cancer out of the 30? Let X count the number of test that are positive out of the 30. P(X ≤ 25) = Binomdist(25, 30, 0.9, true) = 0.1755