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Chapter 7: Problem 35

The president of Doerman Distributors, Inc., believes that \(30 \%\) of the firm's orders come from first-time customers. A simple random sample of 100 orders will be used to estimate the proportion of first-time customers. a. Assume that the president is correct and \(p=.30 .\) What is the sampling distribution of \(\bar{p}\) for this study? b. What is the probability that the sample proportion \(\bar{p}\) will be between .20 and \(.40 ?\) c. What is the probability that the sample proportion will be between .25 and \(.35 ?\)

Short Answer

Expert verified
a) Normal distribution with mean 0.30 and SD 0.0458. b) Probability is 0.9701. c) Probability is 0.7242.

Step by step solution

01

Understanding the Sampling Distribution

According to the problem, the proportion of first-time customers is assumed to be \( p = 0.30 \). The sampling distribution of the sample proportion \( \bar{p} \) can be described using a normal distribution when the sample size is large enough, typically for \( n \geq 30 \). Since we have a sample size of 100, we meet this condition. Therefore, the sampling distribution of \( \bar{p} \) is approximately normal with mean \( \mu_{\bar{p}} = p = 0.30 \) and standard deviation \( \sigma_{\bar{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.30 \times 0.70}{100}} = 0.0458 \).
02

Calculating Probability for .20 < \( \bar{p} \leq \) .40

To find the probability that \( \bar{p} \) is between 0.20 and 0.40, we standardize these values using the calculated standard deviation. The corresponding z-scores are: \[ Z_{0.20} = \frac{0.20 - 0.30}{0.0458} = -2.18 \] \[ Z_{0.40} = \frac{0.40 - 0.30}{0.0458} = 2.18 \] Using the standard normal distribution table, we find the probability from \(-2.18\) to \(2.18\) which is \(P(-2.18 < Z < 2.18) \approx 0.9850 - 0.0149 = 0.9701 \).
03

Calculating Probability for .25 < \( \bar{p} \leq \) .35

Next, we calculate the probability for the sample proportion being between 0.25 and 0.35 using the same standard deviation. The z-scores are: \[ Z_{0.25} = \frac{0.25 - 0.30}{0.0458} = -1.09 \] \[ Z_{0.35} = \frac{0.35 - 0.30}{0.0458} = 1.09 \] The probability between these z-scores is \(P(-1.09 < Z < 1.09) \approx 0.8621 - 0.1379 = 0.7242 \).

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Key Concepts

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

Normal Distribution
A normal distribution is a continuous probability distribution that is symmetrical around its mean. It forms a bell-shaped curve when plotted on a graph. This type of distribution is incredibly important in statistics because it appears in various real-world situations.

Some key characteristics of a normal distribution include:
  • The mean, median, and mode of the distribution are all equal.
  • It is perfectly symmetrical about the mean.
  • The area under the curve represents the whole probability, which is 1.
  • The distribution is completely described by its mean and standard deviation.
In terms of the sampling distribution of proportions, if a sample size is large enough (usually when it's 30 or more), the distribution of the sample proportions tends to follow a normal distribution. This is part of the Central Limit Theorem, which is fundamental in statistics and allows us to make inferences about population parameters based on sample statistics.
Sample Proportion
The sample proportion, often denoted as \( \bar{p} \), is an estimate of a population proportion derived from a sample. In this context, the sample proportion represents the fraction of first-time customers within the sampled orders.

To calculate the sample proportion, use the formula:
  • \( \bar{p} = \frac{x}{n} \)
where \( x \) is the number of successes in the sample (e.g., first-time customers), and \( n \) is the total number of observations in the sample.

For large sample sizes, the distribution of the sample proportion \( \bar{p} \) can be approximated by a normal distribution. The mean of this sampling distribution is equal to the population proportion, \( p \), and the standard deviation is given by the formula:
  • \( \sigma_{\bar{p}} = \sqrt{\frac{p(1-p)}{n}} \)
This allows us to make probabilistic statements about the range within which the true population proportion might lie.
Z-score
A z-score quantifies the number of standard deviations a data point is from the mean of a data set. This transformation into z-scores allows us to understand where a specific sample proportion falls in relation to the overall sampling distribution.

Calculating a z-score for a sample proportion involves the formula:
  • \( Z = \frac{\bar{p} - p}{\sigma_{\bar{p}}} \)
where \( \bar{p} \) is the sample proportion, \( p \) is the assumed population proportion, and \( \sigma_{\bar{p}} \) is the standard deviation of the sample proportion.

Z-scores are used to determine the probability of observing a sample proportion within a certain range. By transforming sample proportions to z-scores, we can easily consult standard normal distribution tables to find these probabilities. This method is particularly useful in studies where we assess the likelihood of observing outcomes by calculating probabilities between two z-scores.

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Most popular questions from this chapter

A researcher reports survey results by stating that the standard error of the mean is \(20 .\) The population standard deviation is 500 a. How large was the sample used in this survey? b. What is the probability that the point estimate was within 卤25 of the population mean?

Three firms carry inventories that differ in size. Firm A's inventory contains 2000 items, firm \(\mathrm{B}\) 's inventory contains 5000 items, and firm \(\mathrm{C}\) 's inventory contains 10,000 items. The population standard deviation for the cost of the items in each firm's inventory is \(\sigma=144\) A statistical consultant recommends that each firm take a sample of 50 items from its inventory to provide statistically valid estimates of the average cost per item. Managers of the small firm state that because it has the smallest population, it should be able to make the estimate from a much smaller sample than that required by the larger firms. However, the consultant states that to obtain the same standard error and thus the same precision in the sample results, all firms should use the same sample size regardless of population size. a. Using the finite population correction factor, compute the standard error for each of the three firms given a sample of size 50 . b. What is the probability that for each firm the sample mean \(\bar{x}\) will be within 卤25 of the population mean \(\mu ?\)

The Cincinnati Enquirer reported that, in the United States, \(66 \%\) of adults and \(87 \%\) of youths ages 12 to 17 use the Internet (The Cincinnati Enquirer, February 7,2006 ). Use the reported numbers as the population proportions and assume that samples of 300 adults and 300 youths will be used to learn about attitudes toward Internet security. a. Show the sampling distribution of \(\bar{p}\) where \(\bar{p}\) is the sample proportion of adults using the Internet. b. What is the probability that the sample proportion of adults using the Internet will be within 卤.04 of the population proportion? c. What is the probability that the sample proportion of youths using the Internet will be within 卤.04 of the population proportion? d. Is the probability different in parts (b) and (c)? If so, why? e. Answer part (b) for a sample of size \(600 .\) Is the probability smaller? Why?

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. \\[ \begin{array}{lllllll} 98601 & 73022 & 83448 & 02147 & 34229 & 27553 & 84147 & 93289 & 14209 \end{array} \\]

Americans have become increasingly concerned about the rising cost of Medicare. In \(1990,\) the average annual Medicare spending per enrollee was \(\$ 3267 ;\) in \(2003,\) the average annual Medicare spending per enrollee was \(\$ 6883\) (Money, Fall 2003 ). Suppose you hired a consulting firm to take a sample of fifty 2003 Medicare enrollees to further investigate the nature of expenditures. Assume the population standard deviation for 2003 was \(\$ 2000\). a. Show the sampling distribution of the mean amount of Medicare spending for a sample of fifty 2003 enrollees. b. What is the probability the sample mean will be within \(\pm \$ 300\) of the population mean? c. What is the probability the sample mean will be greater than \(\$ 7500 ?\) If the consulting firm tells you the sample mean for the Medicare enrollees they interviewed was \(\$ 7500,\) would you question whether they followed correct simple random sampling procedures? Why or why not?
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