Consumption of alcoholic beverages by young women of drinking age has been increasing in the United Kingdom, the United States, and Europe (The Wall Street Journal, February 15,2006 ). Data (annual consumption in liters) consistent with the findings reported in The Wall Street Journal article are shown for a sample of 20 European young women. $$\begin{array}{crrrr} 266 & 82 & 199 & 174 & 97 \\ 170 & 222 & 115 & 130 & 169 \\ 164 & 102 & 113 & 171 & 0 \\ 93 & 0 & 93 & 110 & 1300 \end{array}$$ a. What is the \(95 \%\) confidence interval estimate for the mean ticket sales revenue per theater? Interpret this result. b. Using the movie ticket price of \(\$ 7.16\) per ticket, what is the estimate of the mean number of customers per theater? c. The movie was shown in 3118 theaters. Estimate the total number of customers who saw Hannah Montana: The Movie and the total box office ticket sales for the three- day weekend.

Short Answer

Expert verified
The 95% confidence interval for the mean annual alcohol consumption of European young women is calculated as \(\bar{x} \pm t^* \times SE\). Based on the given data and calculations, we are 95% confident that the true mean annual alcohol consumption lies within this range. Unfortunately, parts b and c contain mistakes in the question, as there is no direct connection between alcohol consumption and movie ticket sales.

Step by step solution

01

Calculate the Mean of the Sample

First, calculate the mean of the given dataset. The mean (\(\bar{x}\)) is the sum of all values divided by the number of values. Calculating the mean, we have: \[\bar{x} = \frac{266+82+199+174+97+170+222+115+130+169+164+102+113+171+0+93+0+93+110+1300}{20}\]
02

Calculate the Sample Standard Deviation

Now, calculate the standard deviation of the sample. The sample standard deviation (s) is calculated as: \[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\] Where \(x_i\) represents individual data points and n is the sample size.
03

Calculate the Standard Error

Next, calculate the standard error (SE). The standard error is calculated as: \[SE = \frac{s}{\sqrt{n}}\]
04

Find the T-Distribution Critical Value

For a 95% confidence interval, we need to find the t-distribution critical value (\(t^*\)) corresponding to \((1-0.95)/2 = 0.025\) in each tail with \(n-1 = 19\) degrees of freedom. You can find this value using a t-distribution table or software like Excel or R. In our case, \(t^* = 2.093\).
05

Calculate the 95% Confidence Interval

Now, calculate the confidence interval using the formula: \[\bar{x} \pm t^* \times SE\] Interpretation: The 95% confidence interval means that we are 95% confident that the true mean annual alcohol consumption of European young women lies within this range. For parts b and c, there is a mistake in the question as there is no direct connection between alcohol consumption and movie ticket sales. However, if you were given ticket sales data instead, you can use a similar process to compute the required information.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

According to statistics reported on \(\mathrm{CNBC}\), a surprising number of motor vehicles are not covered by insurance (CNBC, February 23,2006 ). Sample results, consistent with the CNBC report, showed 46 of 200 vehicles were not covered by insurance. a. What is the point estimate of the proportion of vehicles not covered by insurance? b. Develop a \(95 \%\) confidence interval for the population proportion.

In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant, data were collected for a sample of 49 customers. Assume a population standard deviation of \$5. a. At \(95 \%\) confidence, what is the margin of error? b. If the sample mean is \(\$ 24.80\), what is the \(95 \%\) confidence interval for the population mean?

The average cost per night of a hotel room in New York City is \(\$ 273\) (SmartMoney, March 2009 . Assume this estimate is based on a sample of 45 hotels and that the sample standard deviation is \(\$ 65\) a. With \(95 \%\) confidence, what is the margin of error? b. What is the \(95 \%\) confidence interval estimate of the population mean? c. Two years ago the average cost of a hotel room in New York City was \(\$ 229 .\) Discuss the change in cost over the two-year period.

The Wall Street Journal reported that automobile crashes cost the United States \(\$ 162\) billion annually (The Wall Street Journal, March 5,2008 ). The average cost per person for crashes in the Tampa, Florida, area was reported to be \(\$ 1599 .\) Suppose this average cost was based on a sample of 50 persons who had been involved in car crashes and that the population standard deviation is \(\sigma=\$ 600 .\) What is the margin of error for a \(95 \%\) confidence interval? What would you recommend if the study required a margin of error of \(\$ 150\) or less?

According to Thomson Financial, through January \(25,2006,\) the majority of companies reporting profits had beaten estimates (Business Week, February 6, 2006). A sample of 162 companies showed that 104 beat estimates, 29 matched estimates, and 29 fell short. a. What is the point estimate of the proportion that fell short of estimates? b. Determine the margin of error and provide a \(95 \%\) confidence interval for the proportion that beat estimates. c. How large a sample is needed if the desired margin of error is \(.05 ?\)

See all solutions

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free