/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 Comparing pizza brands \(\quad\)... [FREE SOLUTION] | 91Ó°ÊÓ

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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.)

Short Answer

Expert verified
There are 8 samples: AAA, AAD, ADA, ADD, DAA, DAD, DDA, DDD with varying proportions. The binomial distribution confirms the probability of 1/3 as 3/8.

Step by step solution

01

Identify Possible Samples

List all possible combinations for three individuals when each can choose either pizza A or pizza D. The combinations are: AAA, AAD, ADA, ADD, DAA, DAD, DDA, DDD.
02

Calculate Proportion for Each Sample

For each combination, calculate the proportion of individuals who preferred pizza A. For example, for AAA, the proportion is 3/3 = 1; for AAD, it's 1/3 = 2/3; etc. The result is as follows: AAA (1), AAD (2/3), ADA (2/3), ADD (1/3), DAA (2/3), DAD (1/3), DDA (1/3), DDD (0).
03

Explain Sampling Distribution

Since each person has a 50% probability of preferring pizza A or D, each of the outcomes in part a has an equal probability of occurring. With n=3, the binomial probability for each configuration must be considered. Thus, the computed proportions and probabilities match the provided table: 0 (1/8), 1/3 (3/8), 2/3 (3/8), 1 (1/8).
04

Use Binomial Distribution

Apply the binomial formula P(X=k) = (n choose k) * p^k * (1-p)^(n-k), where n=3, k=1, and p=0.50. For k=1, the probability is P(X=1) = 3C1 * (0.5)^1 * (0.5)^2 = 3 * 0.5 * 0.25 = 3/8, confirming the probability of the 1/3 proportion.

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Key Concepts

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

Sampling Methods
Sampling methods are crucial in statistical analysis as they determine how representative the collected data will be for the entire population. In the exercise, a simple random sample method is used. A simple random sample means each individual within the population has an equal chance of being selected. This ensures that the sample is unbiased and that the results are valid for drawing conclusions about the entire population.
When sampling, it's essential to consider the sample size. In the given scenario, the sample size is three, which is relatively small. However, small samples can still provide useful insights, especially when the sampling process is well-structured and controlled, as demonstrated by the randomization and blinding implemented in the pizza tasting setup.
Randomization in the order of tasting and blinding participants are additional steps that strengthen the randomness and objectivity of the process. By randomizing the tasting order, potential biases due to any tasting sequence are minimized. By blinding the participants, the influence of visual cues on their decisions is eliminated. This approach ensures that the responses are solely based on taste preferences.
Binomial Distribution
The binomial distribution is a probability distribution that summarizes the likelihood that a value will take on one of two independent states. In this case, each participant in the sampling process can either prefer pizza A or pizza D. We use the binomial distribution to model this scenario because it deals with the number of successes in a set number of trials, where each trial has only two possible outcomes.
To apply the binomial distribution, we identify the number of trials ( ext{n}=3 ext), which is the sample size, and the probability of success ( ext{p}=0.50 ext), which is the probability that a person will prefer pizza A over pizza D.
The formula for a binomial probability is given by:\[P(X=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}\] where \ \(k\) is the number of successes (people preferring pizza A). Using this formula, we can calculate the probability for each possible profile of preferences (0, 1, 2, or 3 people choosing A), ensuring we understand how likely each scenario is within a fair sampling context.
Proportion Calculation
Proportion calculation is a simple yet powerful statistical tool. It helps summarize and represent the diversity within a sample. In this exercise, we're interested in calculating the proportion of participants in each sample who prefer pizza A.
To find the proportion, divide the number of people who choose pizza A by the total sample size. For example, in a sample where two out of three individuals prefer pizza A, the proportion would be \( \frac{2}{3}\). It gives a quantitative measure of the preference for pizza A in that sample.
The calculated proportions from each sample configuration help in constructing the sampling distribution of the sample proportion. This distribution shows the frequency of different possible proportions from our samples, providing insight into how preference for pizza A is distributed among the population. Knowing these proportions from multiple samples allows us to use statistical tools like the binomial distribution to calculate specific probabilities, enhancing our interpretation and conclusions. Understanding proportions and their applications in sampling can greatly enhance decision-making processes and our grasp of population trends.

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Most popular questions from this chapter

Bank machine withdrawals An executive in an Australian savings bank decides to estimate the mean amount of money withdrawn in bank machine transactions. From past experience, she believes that \(\$ 50\) (Australian) is a reasonable guess for the standard deviation of the distribution of withdrawals. She would like her sample mean to be within \(\$ 10\) of the population mean. Estimate the probability that this happens if she randomly samples 100 withdrawals. (Hint: Find the standard deviation of the sample mean. How many standard deviations does \(\$ 10\) equal?)

iPhone apps For the population of individuals who own an iPhone, suppose \(p=0.25\) is the proportion that has a given app. For a particular iPhone owner, let \(x=1\) if they have the app and \(x=0\) otherwise. For a random sample of 50 people who have an iPhone: a. State the population distribution (that is, the probability distribution of \(X\) for each observation). b. Find the mean of the sampling distribution of the sample proportion who have the app among the 50 people. c. Find the standard deviation of the sampling distribution of the sample proportion who have the app among the 50 people. d. Explain what the standard deviation in part \(\mathrm{c}\) describes.

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.)

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.

Random variability in baseball A baseball player in the major leagues who plays regularly will have about 500 atbats (that is, about 500 times he can be the hitter in a game) during a season. Suppose a player has a 0.300 probability of getting a hit in an at-bat. His batting average at the end of the season is the number of hits divided by the number of at-bats. This batting average is a sample proportion, so it has a sampling distribution describing where it is likely to fall. a. Describe the shape, mean, and standard deviation of the sampling distribution of the player's batting average after a season of 500 at-bats. b. Explain why a batting average of 0.320 or of 0.280 would not be especially unusual for this player's yearend batting average. (That is, you should not conclude that someone with a batting average of 0.320 one year is necessarily a better hitter than a player with a batting average of \(0.280 .)\)

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