Name

  1. Automobile accidents in the U.S. cost Americans more than $160 billion annually. The average cost per person for crashes in a retirement community in Florida was reported to be $1800. This average cost was based on a sample of 50 persons who had been involved in car accidents. The population standard deviation is \(\sigma\) = 800.
  1. Calculate the margin of error for a 95% confidence interval around the population mean.

\[me = z\frac{\sigma}{\sqrt{n}}\]

\[me = 221.745\]

  1. Report the upper and lower bounds associated with this confidence interval and interpret the result.

\[CI = (1578.255,2021.745)\]

  1. What would you recommend if the study required a margin of error of $150 or less?

Increase the sample size. Increasing the sample size provides more information and a narrower confidence interval.

  1. A survey of 12 Fast Food restaurants provided a sample mean customer satisfaction index of 75. Assume the population standard deviation for the index is \(\sigma\) = 10.
  1. Construct a 95% confidence interval around the population mean and report the upper and lower bounds.

\[CI = (69.342,80.658)\]

  1. How would you interpret the confidence interval for this case?

We are 95% confident that the true population mean lies inside this interval. There is a very good chance that the true population mean lies between 80.7 and 69.3.

  1. Use the sale variable in the J-Ville-House data set to answer the following questions.
  1. What is the sample mean sales price for a house in Jacksonville from this sample?

\[Avg\_Sales = 85659.38\]

  1. Construct a 95% confidence interval for the population mean sales price for houses in Jacksonville.

\[CI = 81002.035, 90316.726\]

  1. Interpret the confidence interval in the specific context of the question.

We are 95% confident that the true population mean lies inside this interval. There is a high likelihood that the true population mean sales price lies between $81,002 and $90,317.

  1. A random sample of 200 registered voters yielded a sample proportion of 0.42 in favor of a tax cut proposal. Use this information to answer the following question.
  1. Construct a 99% confidence interval around the population proportion. Interpret this value in the context of the question. Also, Interpret the margin of error.

\[CI = (0.33,0.51) \]

  1. Given the confidence interval, how would you characterize the support for the tax cuts? That is, do you have enough information to suggest people support or don’t support the tax cuts? Explain your answer.

Since the confidence interval includes 0.50 we don’t have enough information to draw a conclusion on support of the tax. Given the confidence interval, it’s possible that the population proportion could be as high as 0.51, which would indicate support for the tax, or as low as 0.33, which would indicate no support for the tax.

  1. Annual starting salaries for college graduates with degrees in Finance are generally expected to be between $30,000 and $50,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired.
  1. What is the planning value for the population standard deviation? 5000
  2. How large a sample should be taken if the desired margin of error is $500? 385
  3. How large a sample should be taken if the desired margin of error is $200? 2401
  4. How large a sample should be taken if the desired margin of error is $100? 9604
  5. Would you recommend trying to obtain the $100 margin of error? Explain.

A margin of error of $100 is not recommended. The number of observations required to get this narrow margin of error is not worth it.