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Chapter 7: Problem 20

Marketing: Bargain Hunters In a marketing survey, a random sample of 1001 supermarket shoppers revealed that 273 always stock up on an item when they find that item at a real bargain price. See reference in Problem \(19 .\) (a) Let \(p\) represent the proportion of all supermarket shoppers who always stock up on an item when they find a real bargain. Find a point estimate for \(p\). (b) Find a \(95 \%\) confidence interval for \(p\). Give a brief explanation of the meaning of the interval. (c) Interpretation As a news writer, how would you report the survey results on the percentage of supermarket shoppers who stock up on real-bargain items? What is the margin of error based on a \(95 \%\) confidence interval?

Short Answer

Expert verified
The point estimate for \( p \) is 0.2727, and the 95% confidence interval is (0.2458, 0.2996). The margin of error is 2.69%.

Step by step solution

01

Understand the Problem

We are given that, out of 1001 shoppers, 273 always stock up on a real bargain. We need to estimate the overall proportion \( p \) of all shoppers who would behave in the same way using this sample.
02

Calculate the Point Estimate for p

The point estimate for the proportion \( p \) is the sample proportion, \( \hat{p} \). This is calculated as the number of shoppers who stock up divided by the total number of shoppers surveyed: \( \hat{p} = \frac{273}{1001} \approx 0.2727 \).
03

Set Up for Confidence Interval

To find a 95% confidence interval for \( p \), we use the formula for a confidence interval for a proportion: \( \hat{p} \pm z \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( z \) is the z-score corresponding to a 95% confidence level, typically 1.96.
04

Compute the Standard Error

Calculate the standard error (SE) using the formula: \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.2727(1-0.2727)}{1001}} \approx 0.0137 \).
05

Calculate the Confidence Interval

Using the standard error and the z-score, the 95% confidence interval for \( p \) is: \( 0.2727 \pm 1.96 \times 0.0137 \). This gives a range from approximately 0.2458 to 0.2996.
06

Explain the Confidence Interval

The confidence interval (0.2458, 0.2996) suggests that we are 95% confident that the true proportion of all shoppers who stock up on a real bargain is between 24.58% and 29.96%.
07

Report Results and Margin of Error

As a news writer, you can report: "In a survey, about 25% to 30% of supermarket shoppers always stock up on bargains." The margin of error for this confidence interval is approximately \( 1.96 \times 0.0137 = 0.0269 \), or about 2.69%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Point Estimate
When working with data, especially in situations like surveys, it's important to estimate the true parameter for a whole population using sample data. In this case, we're interested in the proportion of supermarket shoppers who consistently stock up when they find a good bargain. The **point estimate** is a crucial concept here.The point estimate is essentially the estimate of this proportion based on our sample. Here, it is computed using the formula:\[ \hat{p} = \frac{273}{1001} \approx 0.2727 \]
This means that approximately 27.27% of the surveyed shoppers always stock up when there's a bargain. It's crucial to understand that this point estimate gives us a single best guess at what the true proportion (\( p \)) in the entire population would be if we could survey everyone.
Proportion
Proportion plays a significant role in statistics, particularly in survey data analysis. It's the part of a whole expressed in relation to the entire group. In our context, we are interested in the proportion of shoppers who stock up on bargains, represented by \( p \).To calculate the proportion in our sample, you take the number of cases of interest (shoppers who stock up, in this case) and divide it by the total number of cases surveyed. In mathematical terms:\[ \hat{p} = \frac{273}{1001} \approx 0.2727 \]
This - Tells us that around 27.27% of the sampled shoppers have the behavior of stocking up.- Assists in making predictions and decisions like resource allocation or marketing strategies for the shoppers.By understanding and calculating proportions, businesses can better cater to customer needs and enhance market tactics.
Margin of Error
The **margin of error** is a core statistical concept that indicates the range within which we can expect the true population parameter to lie with a certain level of confidence, usually expressed as a percentage.In our case, the margin of error is calculated from the confidence interval for a proportion, using the formula:\[ ME = z \times \text{SE} \]
Where - \( z \) is the z-score corresponding to the desired confidence level (1.96 for 95%).- SE is the standard error calculated earlier.From the provided solution:\[ ME = 1.96 \times 0.0137 \approx 0.0269 \]
This roughly translates to a 2.69% margin of error, meaning the true proportion of all supermarket shoppers stocking up on bargains lies within 2.69% of the point estimate in either direction. This gives our confidence interval of approximately (24.58%, 29.96%).Understanding margin of error helps us gauge the reliability of the survey results. It's essential for interpreting survey data and reporting accurate information in fields like marketing, economics, and policy-making.

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Most popular questions from this chapter

Jerry tested 30 laptop computers owned by classmates enrolled in a large computer science class and discovered that 22 were infected with keystroke- tracking spyware. Is it appropriate for Jerry to use his data to estimate the proportion of all laptops infected with such spyware? Explain.

Basic Computation: Confidence Interval for \(\mu_{1}-\mu_{2}\) Consider two independent distributions that are mound-shaped. A random sample of size \(n_{1}=36\) from the first distribution showed \(\bar{x}_{1}=15\), and a random sample of size \(n_{2}=40\) from the second distribution showed \(\bar{x}_{2}=14\) (a) Check Requirements If \(\sigma_{1}\) and \(\sigma_{2}\) are known, what distribution does \(\bar{x}_{1}-\bar{x}_{2}\) follow? Explain. (b) Given \(\sigma_{1}=3\) and \(\sigma_{2}=4\), find a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (c) Check Requirements Suppose \(\sigma_{1}\) and \(\sigma_{2}\) are both unknown, but from the random samples, you know \(s_{1}=3\) and \(s_{2}=4\). What distribution approximates the \(\bar{x}_{1}-\bar{x}_{2}\) distribution? What are the degrees of freedom? Explain. (d) With \(s_{1}=3\) and \(s_{2}=4\), find a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (e) If you have an appropriate calculator or computer software, find a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using degrees of freedom based on Satterthwaite's approximation. (f) Interpretation Based on the confidence intervals you computed, can you be \(95 \%\) confident that \(\mu_{1}\) is larger than \(\mu_{2} ?\) Explain.

Basic Computation: Sample Size What is the minimal sample size needed for a \(99 \%\) confidence interval to have a maximal margin of error of \(0.06\) (a) if a preliminary estimate for \(p\) is \(0.8\) ? (b) if there is no preliminary estimate for \(p ?\)

Ballooning: Air Temperature How hot is the air in the top (crown) of a hot air balloon? Information from Ballooning: The Complete Guide to Riding the Winds by Wirth and Young (Random House) claims that the air in the crown should be an average of \(100^{\circ} \mathrm{C}\) for a balloon to be in a state of equilibrium. However, the temperature does not need to be exactly \(100^{\circ} \mathrm{C}\). What is a reasonable and safe range of temperatures? This range may vary with the size and \((\mathrm{dec}-\) orative) shape of the balloon. All balloons have a temperature gauge in the crown. Suppose that 56 readings (for a balloon in equilibrium) gave a mean temperature of \(\bar{x}=97^{\circ} \mathrm{C}\). For this balloon, \(\sigma \approx 17^{\circ} \mathrm{C}\). (a) Compute a \(95 \%\) confidence interval for the average temperature at which this balloon will be in a steady-state equilibrium. (b) Interpretation If the average temperature in the crown of the balloon goes above the high end of your confidence interval, do you expect that the balloon will go up or down? Explain.

Profits: Banks Jobs and productivity! How do banks rate? One way to answer this question is to examine annual profits per employee. Forbes Top Companies, edited by J. T. Davis (John Wiley \& Sons), gave the following data about annual profits per employee (in units of one thousand dollars per employee) for representative companies in financial services. Companies such as Wells Fargo, First Bank System, and Key Banks were included. Assume \(\sigma \approx 10.2\) thousand dollars.$$ \begin{array}{lllllllllll} 42.9 & 43.8 & 48.2 & 60.6 & 54.9 & 55.1 & 52.9 & 54.9 & 42.5 & 33.0 & 33.6 \\ 36.9 & 27.0 & 47.1 & 33.8 & 28.1 & 28.5 & 29.1 & 36.5 & 36.1 & 26.9 & 27.8 \\ 28.8 & 29.3 & 31.5 & 31.7 & 31.1 & 38.0 & 32.0 & 31.7 & 32.9 & 23.1 & 54.9 \\ 43.8 & 36.9 & 31.9 & 25.5 & 23.2 & 29.8 & 22.3 & 26.5 & 26.7 & & \end{array} $$ (a) Use a calculator or appropriate computer software to verify that, for the preceding data, \(\bar{x} \approx 36.0\). (b) Let us say that the preceding data are representative of the entire sector of (successful) financial services corporations. Find a \(75 \%\) confidence interval for \(\mu\), the average annual profit per employee for all successful banks. (c) Interpretation Let us say that you are the manager of a local bank with a large number of employees. Suppose the annual profits per employee are less than 30 thousand dollars per employee. Do you think this might be somewhat low compared with other successful financial institutions? Explain by referring to the confidence interval you computed in part (b). (d) Interpretation Suppose the annual profits are more than 40 thousand dollars per employee. As manager of the bank, would you feel somewhat better? Explain by referring to the confidence interval you computed in part (b). (e) Repeat parts (b), (c), and (d) for a \(90 \%\) confidence level.
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