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A sample of 84 golfers showed that their average score on a particular golf course was 89.05 with a standard deviation of 3.47.
Answer each of the following (show all work and state the final answer to at least two decimal places.):
(A) Find the 98% confidence interval of the mean score for all 84 golfers.
(B) Find the 98% confidence interval of the mean score for all golfers if this is a sample of 115 golfers instead of a sample of 84.
(C) Which confidence interval is smaller and why?
Using the value in the given data table finding the missing values
We suspect that automobile insurance premiums (in dollars) may be steadily decreasing with the driver's driving experience (in years), so we choose a random sample of drivers who have similar automobile insurance coverage and collect data about th..
At.05 level of significance determine if correlation between rental cost and apartment size is significant.
Use a test of the difference between proportions when answering these questions.
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Explain how the Central Limit Theorem can help you convince your boss that while you can't get rid of sampling error the results from your statistical work (that is based on sampling) can still be useful.
Compute the coefficient of determination for bone loss and women's age and describe what this statistic means.
As it is a 2 _ 2 table, try also two-tailed two-sample z test for ð1 = ð2 and verify that z2 is same as your chi-square statistic. Which test do you favour? Describe why?
At the start of the study the mean of the group was 24 and the end of the study the mean of the group was 30. The standard deviation of the difference scores was 12.0 and we had 16 subjects.
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You're considering investing in a portfolio of the common stocks of four publicly-traded companies with betas as follows:
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