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Chapter 7: Problem 47

Purpose of sampling distribution You'd like to estimate the proportion of all students in your school who are fluent in more than one language. You poll a random sample of 50 students and get a sample proportion of 0.12. Explain why the standard deviation of the sampling distribution of the sample proportion gives you useful information to help gauge how close this sample proportion is to the unknown population proportion.

Short Answer

Expert verified
The standard deviation indicates how much sample proportions might vary, helping gauge closeness to the true population proportion.

Step by step solution

01

Identify Known Values

First, identify the known values from the problem. We are told that the sample size is 50 and the sample proportion is 0.12. Let the sample proportion be denoted as \(\hat{p} = 0.12\) and sample size \(n = 50\).
02

Understand Sampling Distribution

The sampling distribution of a sample proportion represents the distribution of the sample proportion values that one would expect to see if they took many random samples of the same size from the population.
03

Define Standard Deviation of Sampling Distribution

The standard deviation of the sampling distribution is given by the formula \( \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \), where \(p\) is the population proportion and \(n\) is the sample size. Since the population proportion \(p\) is unknown, we use the sample proportion \(\hat{p}\) to estimate it, therefore, \( \sigma_{\hat{p}} \approx \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \).
04

Calculate Estimated Standard Deviation

Substitute the known values into the formula to calculate the estimated standard deviation: \( \sigma_{\hat{p}} \approx \sqrt{\frac{0.12(1-0.12)}{50}} \). Compute the value to get \( \sigma_{\hat{p}} \approx \sqrt{\frac{0.12 \times 0.88}{50}} \approx \sqrt{\frac{0.1056}{50}} \approx \sqrt{0.002112} \approx 0.046 \).
05

Interpret Obtained Standard Deviation

The standard deviation of the sampling distribution of the sample proportion (approximately 0.046) provides an estimate of how much sample proportions are expected to fluctuate around the true population proportion for repeated samples of this size.
06

Use of Standard Deviation for Estimation

The standard deviation informs us about the variability of sample proportions. A smaller standard deviation would suggest that the sample proportion is more likely to be close to the true population proportion, indicating that the sample proportion might be a good estimate of the population proportion.

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Key Concepts

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

Understanding Sample Proportion
To begin with, let's explore what a sample proportion is. When you're dealing with large groups or populations, it may not be feasible to gather information from every single member. Instead, a smaller, more manageable group known as a sample is often used. The sample proportion, denoted as \( \hat{p} \), is simply the fraction or percentage of the sample that has a particular attribute of interest. In the context of our exercise, the sample proportion of 0.12 indicates that 12% of the 50 surveyed students are fluent in more than one language. Calculating the sample proportion is straightforward: divide the number of individuals in the sample with the attribute by the total number within the sample. This gives us a snapshot or estimate of what might be occurring at the population level. While this estimate isn't perfect, due to random sampling error, it is a crucial starting point for further analysis.
Delving into Standard Deviation
The concept of standard deviation is key when understanding the reliability of the sample proportion. The standard deviation of the sampling distribution of the sample proportion provides insights into how much variation one might expect if multiple samples were taken. Mathematically, we use the formula \( \sigma_{\hat{p}} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( n \) is the sample size. The standard deviation tells us how much the sample proportions will bounce around the actual population proportion if you repeated the sampling process multiple times. A smaller standard deviation means that the sample proportions are clustered tightly around the population proportion, which suggests a higher level of reliability in the sample as an estimate of the population. In practical terms, a smaller standard deviation is reassuring, indicating that the sample-proportion estimate is likely to be close to the true population proportion.
Grasping Population Proportion
Population proportion is a fundamental concept that can be envisioned as the proportion of the entire population having a certain attribute, similar to the sample proportion but for the entire population. The population proportion is denoted by \( p \). However, unlike the sample proportion, it is often unknown, and researchers endeavor to estimate it through sample data.Understanding the population proportion is crucial because it represents the real scenarios among the population. The objective of using sample data is to make reliable inferences about this population proportion. While the sample proportion provides a point estimate of the population proportion, the idea is to create a bridge between the sample and the population. Ensuring the sample accurately reflects the population helps to predict the population proportion with more confidence. This understanding assists in various fields from social sciences to marketing, where knowing how a population behaves or attributes is invaluable.

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Most popular questions from this chapter

Restaurant profit? Jan's All You Can Eat Restaurant charges \(\$ 8.95\) per customer to eat at the restaurant. Restaurant management finds that its expense per customer, based on how much the customer eats and the expense of labor, has a distribution that is skewed to the right with a mean of \(\$ 8.20\) and a standard deviation of \(\$ 3\). a. If the 100 customers on a particular day have the characteristics of a random sample from their customer base, find the mean and standard deviation of the sampling distribution of the restaurant's sample mean expense per customer. b. Find the probability that the restaurant makes a profit that day, with the sample mean expense being less than \$8.95. (Hint: Apply the central limit theorem to the sampling distribution in part a.)

Comparing pizza brands \(\quad\) The owners of Aunt Erma's Restaurant plan an advertising campaign with the claim that more people prefer the taste of their pizza (which we'll denote by A) than the current leading fast-food chain selling pizza (which we'll denote by \(\mathrm{D}\) ). To support their claim, they plan to randomly sample three people in Boston. Each person is asked to taste a slice of pizza \(A\) and a slice of pizza \(D\). Subjects are blindfolded so they cannot see the pizza when they taste it, and the order of giving them the two slices is randomized. They are then asked which pizza tastes better. Use a symbol with three letters to represent the responses for each possible sample. For instance, ADD represents a sample in which the first subject sampled preferred pizza \(A\) and the second and third subjects preferred pizza \(\mathrm{D}\) a. Identify the eight possible samples of size \(3,\) and for each sample report the proportion that preferred pizza \(A\). b. In the entire Boston population, suppose that exactly half would prefer pizza \(\mathrm{A}\) and half would prefer pizza D. Explain why the sampling distribution of the sample proportion who prefer Aunt Erma's pizza, when \(n=3,\) is \begin{tabular}{cc} \hline Sample Proportion & Probability \\ \hline 0 & \(1 / 8\) \\ \(1 / 3\) & \(3 / 8\) \\ \(2 / 3\) & \(3 / 8\) \\ 1 & \(1 / 8\) \\ \hline \end{tabular} c. In part b, we can also find the probabilities for each possible sample proportion value using the binomial distribution. Use the binomial with \(n=3\) and \(p=0.50\) to show that the probability of a sample proportion of \(1 / 3\) equals \(3 / 8 .\) (Hint: This equals the probability that \(x=1\) person out of \(n=3\) prefer pizza A. It's especially helpful to use the binomial formula when \(p\) differs from \(0.50,\) since then the eight possible samples listed in part a would not be equally likely.)

Using control charts to assess quality In many industrial production processes, measurements are made periodically on critical characteristics to ensure that the process is operating properly. Observations vary from item to item being produced, perhaps reflecting variability in material used in the process and/or variability in the way a person operates machinery used in the process. There is usually a target mean for the observations, which represents the long-run mean of the observations when the process is operating properly. There is also a target standard deviation for how observations should vary around that mean if the process is operating properly. A control chart is a method for plotting data collected over time to monitor whether the process is operating within the limits of expected variation. A control chart that plots sample means over time is called an \(\overline{\boldsymbol{x}}\) -chart. As shown in the following, the horizontal axis is the time scale and the vertical axis shows possible sample mean values. The horizontal line in the middle of the chart shows the target for the true mean. The upper and lower lines are called the upper control limit and lower control limit, denoted by UCL and \(\mathbf{L C L .}\) These are usually drawn 3 standard deviations above and below the target value. The region between the LCL and UCL contains the values that theory predicts for the sample mean when the process is in control. When a sample mean falls above the UCL or below the LCL, it indicates that something may have gone wrong in the production process. a. Walter Shewhart invented this method in 1924 at Bell Labs. He suggested using 3 standard deviations in setting the UCL and LCL to achieve a balance between having the chart fail to diagnose a problem and having it indicate a problem when none actually existed. If the process is working properly ("in statistical control") and if \(n\) is large enough that \(\bar{x}\) has approximately a normal distribution, what is the probability that it indicates a problem when none exists? (That is, what's the probability a sample mean will be at least 3 standard deviations from the target, when that target is the true mean?) b. What would the probability of falsely indicating a problem be if we used 2 standard deviations instead for the UCL and LCL? c. When about nine sample means in a row fall on the same side of the target for the mean in a control chart, this is an indication of a potential problem, such as a shift up or a shift down in the true mean relative to the target value. If the process is actually in control and has a normal distribution around that mean, what is the probability that the next nine sample means in a row would (i) all fall above the mean and (ii) all fall above or all fall below the mean? (Hint: Use the binomial distribution, treating the successive observations as independent.)

Multiple choice: CLT The central limit theorem implies a. All variables have approximately bell-shaped data distributions if a random sample contains at least about 30 observations. b. Population distributions are normal whenever the population size is large. c. For sufficiently large random samples, the sampling distribution of \(\bar{x}\) is approximately normal, regardless of the shape of the population distribution. d. The sampling distribution of the sample mean looks more like the population distribution as the sample size increases.

Sample = population Let \(X=\) GPA for students in your school. a. What would the sampling distribution of the sample mean look like if you sampled every student in the school, so the sample size equals the population size? (Hint: The sample mean then equals the population mean.) b. How does the sampling distribution compare to the population distribution if we take a sample of size \(n=1 ?\)
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