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A marketing research firm wishes to estimate the percent of homeowners who are dissatisfied with their present homeowner's insurance policy. A simple random sample of 400 homeowners led to 80 who were dissatisfied with their homeowner's insurance policy.
a. Compute the 95% confidence interval for the percent of homeowners who are dissatisfied with their homeowner's insurance policy.
b. Interpret this confidence interval.
c. How large a sample size will need to be selected if we wish to have a 95% confidence interval that is accurate to within 1%.
Pollsters try to determine whether or not a person is a "likely voter" before they count their opinion in a poll. If we assume 40% of the registered voters will actually vote, in a random sample of 100 registered voters we can be 95% confident tha..
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How much evidence is there that the mean composite satisfaction rating exceeds 42?
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The number of passengers on the Carnival Sensation during one-week cruises in the Caribbean follows the normal distribution.
He would like to determine whether there are more units produced on the afternoon shift than on the day shift.
A researcher administers a treatment to a sample of participants selected from a population with µ = 80. If the researcher obtains a sample mean of M = 88, which combination of factors is most likely to result in rejecting the null hypothesis?
Assume that a simple random sample has been selected from a normally distributed population. Find the test statistic, P-value, critical value(s), and state the final conclusion.
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