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A certain chromosome defect occurs in only 1 in 200 adult Caucasian males. A random sample of 100 adult Caucasian males will be selected. The proportion of men in this sample who have the defect, \(\hat{p},\) will be calculated. a. What are the mean and standard deviation of the sampling distribution of \(\hat{p}\) ? b. Is the sampling distribution of \(\hat{p}\) approximately normal? Explain. c. What is the smallest value of \(n\) for which the sampling distribution of \(\hat{p}\) is approximately normal?

Short Answer

Expert verified
a. Mean is 0.005 and the standard deviation is 0.022. b. No, the sample distribution of \(\hat{p}\) is not approximately normal as \(np = 0.5 < 5\) and \(n(1 - p) = 99.5 > 5\). c. Minimum sample size, \(n\), for \(\hat{p}\) to be approximately normal is 1000.

Step by step solution

01

Calculate Mean and Standard Deviation of \(\hat{p}\)

In this step, we utilize the formulas for mean and standard deviation of a sample proportion. The mean (\(\mu_{\hat{p}}\)) is equal to the success proportion in the population (\(p\)), which is \(\frac{1}{200}\) or 0.005 in this case. The standard deviation (\(\sigma_{\hat{p}}\)) is calculated by the formula, \(\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}}\), where \(n\) is the sample size of 100 men.
02

Check the Normality of the Sampling Distribution

The normality of the sampling distribution can be justified by applying the rule of thumb which states that, for the sampling distribution of a proportion to be approximately normal, both \(np\) and \(n(1 - p)\) should be greater than or equal to 5. Given \(p = 0.005\) and \(n = 100\), we check whether both the quantities are greater than 5 or not.
03

Find the Minimum Sample Size for Normality

We must find the smallest value of \(n\) which makes the sampling distribution approximately normal. This value can be calculated by ensuring that both \(np\) and \(n(1 - p)\) are greater than or equal to 5.

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Key Concepts

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

Sample Proportion
In statistics, a sample proportion is a helpful tool when trying to understand how often an event occurs in a sample compared to the entire population. When you're dealing with a sample proportion, you essentially answer the question: "How common is this feature within my sample?" To calculate the sample proportion (\(\hat{p}\)), you use the formula:
  • Number of successes in the sample divided by the total sample size.
So, if you tested 100 people for a trait and found that 5 of them exhibit it, your sample proportion would be:
  • \(\hat{p} = \frac{5}{100} = 0.05\)
This value represents the portion of your sample that displays the feature of interest. As with any measurement, the sample proportion may differ from the actual proportion in the population. It's crucial to understand this core concept because it's the foundation for making inferences about a larger population.
Normality
The normality of the sampling distribution is vital because it influences how we interpret data and make predictions. A sampling distribution of a sample proportion is roughly normal if certain conditions are met, like seeing how often a particular trait appears in many similar-sized samples.

Rule of Thumb for Normality

To decide if the sampling distribution is normal, we use a simple rule of thumb: both \(np\) and \(n(1 - p)\) should be at least 5.
  • "\(np\)" checks the expected number of successes.
  • "\(n(1 - p)\)" checks the expected number of failures.
For example, if you have a small success rate, like detecting a rare trait, you need a large sample to ensure normality. Once normality is established, you can use the properties of the normal distribution to make statistical inferences about the population.
Mean and Standard Deviation
The mean and standard deviation are key elements in understanding the sampling distribution of a sample proportion. The mean of the sampling distribution for the proportion (\(\mu_{\hat{p}}\)) actually equals the true proportion of the population (\(p\)).

Mean

Given a population proportion (\(p\)), your mean of the sampling distribution is straightforwardly given by \(\mu_{\hat{p}} = p\). Essentially, it reflects where the center of your data points lies, over repeated sampling.

Standard Deviation

The standard deviation (\(\sigma_{\hat{p}}\)) is slightly more complex. It indicates how much variation exists in the sampling distribution and is calculated using:\[\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}}\]This formula allows you to measure the spread of your sample proportions from the mean. A smaller standard deviation indicates less spread, meaning your samples tend to be closer to the population proportion.Understanding these concepts will enable you to effectively critique data because you'll know both its typical behavior (mean) and variability (standard deviation).

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Most popular questions from this chapter

A random sample is to be selected from a population that has a proportion of successes \(p=0.65\). Determine the mean and standard deviation of the sampling distribution of \(\hat{p}\) for each of the following sample sizes: a. \(n=10\) d. \(n=50\) b. \(n=20\) e. \(n=100\) c. \(n=30\) f. \(n=200\)

The report "California's Education Skills Gap: Modest Improvements Could Yield Big Gains" (Public Policy Institute of California, April \(16,2008,\) www.ppic.org) states that nationwide, \(61 \%\) of high school graduates go on to attend a two-year or four-year college the year after graduation. The proportion of high school graduates in California who go on to college was estimated to be \(0.55 .\) Suppose that this estimate was based on a random sample of 1,500 California high school graduates. Is it reasonable to conclude that the proportion of California high school graduates who attend college the year after graduation is different from the national figure? (Hint: Use what you know about the sampling distribution of \(\hat{p}\). You might also refer to Example \(8.5 .)\)

In a study of pet owners, it was reported that \(24 \%\) celebrate their pet's birthday (Pet Statistics, Bissell Homecare, Inc., 2010 ). Suppose that this estimate was from a random sample of 200 pet owners. Is it reasonable to conclude that the proportion of all pet owners who celebrate their pet's birthday is less than \(0.25 ?\) Use what you know about the sampling distribution of \(\hat{p}\) to support your answer.

A random sample of 100 employees of a large company included 37 who had worked for the company for more than one year. For this sample, \(\hat{p}=\frac{37}{100}=0.37\). If a different random sample of 100 employees were selected, would you expect that \(\hat{p}\) for that sample would also be \(0.37 ?\) Explain why or why not.

A random sample of 50 registered voters in a particular city included 32 who favored using city funds for the construction of a new recreational facility. For this sample, \(\hat{p}=\frac{32}{50}=\) 0.64 . If a second random sample of 50 registered voters was selected, would it surprise you if \(\hat{p}\) for that sample was not equal to 0.64 ? Why or why not?

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