Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 21

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.

Start your free trial

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

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 65 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample standard deviation was \(5.2 .\) Provide \(90 \%\) and \(95 \%\) confidence intervals for the population mean number of weekly customer contacts for the sales personnel.

Playbill magazine reported that the mean annual household income of its readers is \(\$ 119,155(\text {Playbill}, \text { January } 2006) .\) Assume this estimate of the mean annual household income is based on a sample of 80 households, and, based on past studies, the population standard deviation is known to be \(\sigma=\$ 30,000\) a. Develop a \(90 \%\) confidence interval estimate of the population mean. b. Develop a \(95 \%\) confidence interval estimate of the population mean. c. Develop a \(99 \%\) confidence interval estimate of the population mean. d. Discuss what happens to the width of the confidence interval as the confidence level is increased. Does this result seem reasonable? Explain.

For a \(t\) distribution with 16 degrees of freedom, find the area, or probability, in each region. a. To the right of 2.120 b. To the left of 1.337 c. To the left of -1.746 d. To the right of 2.583 e. Between -2.120 and 2.120 f. Between -1.746 and 1.746

If a perfectly competitive firm sells 100 units of output at a market price of \(\$ 100\) per unit, its marginal revenue per unit is a. \(\$ 1\) b. \(\$ 100\) c. more than \(\$ 1,\) but less than \(\$ 100\). d. less than \(\$ 100\).

Nielsen Media Research conducted a study of household television viewing times during the 8 P.M. to 11 P.M. time period. The data contained in the file named Nielsen are consistent with the findings reported (The World Almanac, 2003). Based upon past studies, the population standard deviation is assumed known with \(\sigma=3.5\) hours. Develop a \(95 \%\) confidence interval estimate of the mean television viewing time per week during the 8 P.M. to 11 P.M. time period.
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