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Lori Jeffrey is a successful sales representative for a major publisher of college textbooks. Historically, Lori obtains a book adoption on \(25 \%\) of her sales calls. Viewing her sales calls for one month as a sample of all possible sales calls, assume that a statistical analysis of the data yields a standard error of the proportion of .0625 a. How large was the sample used in this analysis? That is, how many sales calls did Lori make during the month? b. Let \(\bar{p}\) indicate the sample proportion of book adoptions obtained during the month. Show the sampling distribution of \(\bar{p}\) c. Using the sampling distribution of \(\bar{p},\) compute the probability that Lori will obtain book adoptions on \(30 \%\) or more of her sales calls during a one-month period.

Short Answer

Expert verified
a. 48 sales calls; b. \( \bar{p} \sim N(0.25, 0.0625) \); c. 0.2119 probability.

Step by step solution

01

Identify the formula for standard error

The standard error of the proportion is given by the formula: \[ SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \] where \( \hat{p} \) is the sample proportion (0.25) and \( n \) is the sample size.
02

Calculate the sample size

Given that the standard error \( SE = 0.0625 \), plug in the values into the formula:\[ 0.0625 = \sqrt{\frac{0.25 \times (1 - 0.25)}{n}} \]Square both sides to eliminate the square root:\[ 0.00390625 = \frac{0.25 \times 0.75}{n} \]Solve for \( n \):\[ n = \frac{0.25 \times 0.75}{0.00390625} = \frac{0.1875}{0.00390625} = 48 \]Thus, Lori made 48 sales calls during the month.
03

State the sampling distribution of \( \bar{p} \)

The sampling distribution of \( \bar{p} \), under the central limit theorem, is approximately normal with mean \( \mu_{\bar{p}} = \hat{p} = 0.25 \) and standard deviation \( SE = 0.0625 \). Hence, \( \bar{p} \sim N(0.25, 0.0625) \).
04

Calculate the probability for \( \bar{p} \geq 0.30 \)

First, find the z-score for 0.30:\[ z = \frac{0.30 - 0.25}{0.0625} = \frac{0.05}{0.0625} = 0.8 \]Using the standard normal distribution table, the probability \( P(Z \leq 0.8) \approx 0.7881 \). Therefore, the probability \( P(Z \geq 0.8) = 1 - 0.7881 = 0.2119 \).

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Key Concepts

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

Sample Size Calculation
To determine the sample size effectively, understanding the concept of sample size calculation is crucial. The sample size is the number of observations or data points that you need to gather in your study to ensure you get reliable results. In many statistical analyses, the sample size is determined using formulas that consider the desired level of accuracy and the estimated variability in the data.

In Lori’s case, we used the standard error of proportion formula to find the sample size. The formula is:\[ SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \]where:
  • \( SE \) is the standard error of the proportion, which is a measure of how much the sample proportion \( \hat{p} \) can vary from the true population proportion.
  • \( \hat{p} \) is the sample proportion, given as 0.25 for Lori.
  • \( n \) is our unknown sample size.
By solving this formula with a given standard error, we calculated that Lori made 48 sales calls. This ensures that the sample provides a reliable estimate with minimum error.
Sampling Distribution
A sampling distribution gives us a picture of how a statistic like the sample mean or proportion varies from sample to sample. It's built by taking multiple samples and calculating the statistic for each one. This collection of statistics creates a distribution that gives us insights into the behavior of the sample.

For Lori's situation, we're interested in the sampling distribution of the sample proportion \( \bar{p} \). Here, the distribution tells us how the sample proportion of Lori’s book adoptions might vary if different samples of 48 sales calls were taken.
  • The mean of the sampling distribution is the population proportion \( \mu_{\bar{p}} = \hat{p} = 0.25 \).
  • The standard deviation of the sampling distribution is the standard error \( SE = 0.0625 \).
This distribution is approximately normal due to the Central Limit Theorem, giving it a bell-shaped curve. Understanding this allows us to make probability statements about future samples.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics that often comes up when dealing with sampling distributions. It states that when you have a large enough sample size, the distribution of the sample mean (or sample proportion) will approach a normal distribution, regardless of the shape of the original population distribution.

This theorem is powerful because it means we can use normal probability calculations, even if we start with data that isn't normally distributed. In Lori's problem, thanks to the CLT, we can say that the sampling distribution of the sample proportion \( \bar{p} \) is normal. With our sample size of 48 sales calls, it helps us to estimate probabilities using the normal distribution, making our statistical inferences not only possible but more accurate and reliable.

The insight that the CLT provides allows us to leverage all the tools and tables of normal distribution, easing the process of statistical inference.
Standard Error of Proportion
The standard error of the proportion is a critical measurement in statistical analysis. It tells you how much the sample proportion is expected to fluctuate around the actual population proportion. The smaller this error, the more precise your sample estimate is, and the more confidence you have in your results.

The formula for the standard error of a sample proportion is:\[ SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \]
  • \( \hat{p} \) is the sample proportion.
  • \( n \) is the sample size.
In the context of Lori’s exercise, the standard error was given as 0.0625. This value was used to back-calculate the necessary sample size. The key takeaway of the standard error is that it is crucial for making conclusions about a population based on a sample. It helps in determining the spread and uncertainty of the sample proportion, aiding in making statistically sound predictions and conclusions.

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Most popular questions from this chapter

BusinessWeek surveyed MBA alumni 10 years after graduation \((\text {Business Week} \text { , September } 22,\) 2003 ). One finding was that alumni spend an average of \(\$ 115.50\) per week eating out socially. You have been asked to conduct a follow-up study by taking a sample of 40 of these MBA alumni. Assume the population standard deviation is \(\$ 35\) a. Show the sampling distribution of \(\bar{x}\), the sample mean weekly expenditure for the \(40 \mathrm{MBA}\) alumni. b. What is the probability that the sample mean will be within \(\$ 10\) of the population mean? c. Suppose you find a sample mean of \(\$ 100 .\) What is the probability of finding a sample mean of \(\$ 100\) or less? Would you consider this sample to be an unusually low spending group of alumni? Why or why not?

A population has a mean of 200 and a standard deviation of \(50 .\) A simple random sample of size 100 will be taken and the sample mean \(\bar{x}\) will be used to estimate the population mean a. What is the expected value of \(\bar{x} ?\) b. What is the standard deviation of \(\bar{x} ?\) c. Show the sampling distribution of \(\bar{x}\). d. What does the sampling distribution of \(\bar{x}\) show?

The American Association of Individual Investors (AAII) polls its subscribers on a weekly basis to determine the number who are bullish, bearish, or neutral on the short-term prospects for the stock market. Their findings for the week ending March \(2,2006,\) are consistent with the following sample results (AAII website, March 7, 2006). Bullish \(409 \quad\) Neutral \(299 \quad\) Bearish \(291\) Develop a point estimate of the following population parameters. a. The proportion of all AAII subscribers who are bullish on the stock market. b. The proportion of all AAII subscribers who are neutral on the stock market. c. The proportion of all AAII subscribers who are bearish on the stock market.

Assume a finite population has 350 elements. Using the last three digits of each of the following five-digit random numbers (e.g., \(601,022,448, \ldots),\) determine the first four elements that will be selected for the simple random sample. \\[ 98601 \quad 73022 \quad 83448 \quad 02147 \quad 34229 \quad 27553 \quad 84147 \quad 93289 \quad 14209 \\]

After deducting grants based on need, the average cost to attend the University of Southern California (USC) is \(\$ 27,175\) (U.S. News \& World Report, America 's Best Colleges, \(2009 \text { ed. }) .\) Assume the population standard deviation is \(\$ 7400 .\) Suppose that a random sample of 60 USC students will be taken from this population. a. What is the value of the standard error of the mean? b. What is the probability that the sample mean will be more than \(\$ 27,175 ?\) c. What is the probability that the sample mean will be within \(\$ 1000\) of the population mean? d. How would the probability in part (c) change if the sample size were increased to \(100 ?\)

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