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Chapter 6: Problem 11

Use the population of \(\\{34,36,41,51\\}\) of the amounts of caffeine \((m g / 12 \text { oz })\) in Coca-Cola Zero, Diet Pepsi, Dr Pepper, and Mellow Yello Zero. Assume that random samples of size \(n=2\) are selected with replacement. Sampling Distribution of the Sample Mean a. After identifying the 16 different possible samples, find the mean of each sample, then construct a table representing the sampling distribution of the sample mean. In the table, combine values of the sample mean that are the same. (Hint: See Table \(6-3\) in Example 2 on page \(258 .\) ) b. Compare the mean of the population \(\\{34,36,41,51\\}\) to the mean of the sampling distribution of the sample mean. c. Do the sample means target the value of the population mean? In general, do sample means make good estimators of population means? Why or why not?

Short Answer

Expert verified
Sample means generally target the population mean and make good estimators.

Step by step solution

01

- Identify All Possible Samples

List all possible samples of size n=2 selected with replacement from the population \( \{34, 36, 41, 51\} \). Each sample consists of two values.
02

- Calculate the Sample Means

Calculate the mean of each of the 16 possible samples. Use the formula \( \text{mean} = \frac{x_1 + x_2}{2} \) for each sample, where \( x_1 \) and \( x_2 \) are the two values in the sample.
03

- Construct the Sampling Distribution

Create a table listing each unique sample mean and its frequency of occurrence. Combine values of the sample mean that are the same.
04

- Calculate Population Mean

Find the mean of the original population \( \{34, 36, 41, 51\} \) using the formula \( \frac{\sum x}{N} \), where \( N \) is the number of data points in the population.
05

- Compare Means

Compare the mean of the sampling distribution of the sample mean to the mean of the population.
06

- Assess Accuracy of Sample Means

Determine if the sample means target the population mean and evaluate whether sample means are good estimators of population means.

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Key Concepts

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

Population Mean
The population mean is the average of all values in a given population. In our exercise, the population consists of caffeine amounts of \(\{34, 36, 41, 51\}\) mg/12 oz. To find the population mean, we use the formula: \(\frac{\sum x}{N}\).
Adding up all the caffeine amounts, we get 34 + 36 + 41 + 51 = 162. Dividing this by the number of values (4) in the population, we find the population mean: \(\frac{162}{4} = 40.5\).
The population mean gives us a single value that summarizes the data, representing the center or 'balance point' of the population distribution.
Sample Mean
The sample mean is computed to summarize a smaller, representative subset (sample) of the population. In this exercise, samples of size \(n=2\) are selected with replacement from the population \(\{34, 36, 41, 51\}\).
To find the sample mean of any two values, we use the formula: \(\text{mean} = \frac{x_1 + x_2}{2}\). For instance, for the sample \(\{34,36\}\), the sample mean is \(\frac{34 + 36}{2} = 35\).
Repeating this for all 16 possible samples gives us various sample means, which are then summarized into a sampling distribution.
Statistical Estimation
Statistical estimation involves using sample data to infer the population parameters, like the population mean.
One common method is point estimation, where the sample mean serves as an estimate of the population mean. In our exercise, by averaging the sample means of all possible samples, we obtain the 'mean of the sampling distribution of the sample mean', which we can compare to the actual population mean to evaluate accuracy.
Good estimators like the sample mean are expected to be unbiased, meaning their expected value is equal to the population parameter they estimate.
Sampling Methods
Sampling methods dictate how we select samples from a population. In this exercise, we used random sampling with replacement.
Random sampling ensures that every member of the population has an equal chance of being selected, reducing bias.
Sampling with replacement allows the same value to be chosen more than once across different samples, generating more variability in sample means.
Different sampling methods like systematic, stratified, or cluster sampling might be used based on the study objectives, population characteristics, and the required precision of estimates. Understanding the appropriate method ensures more reliable and generalizable results.

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Most popular questions from this chapter

The probability of a baby being born a boy is \(0.512 .\) Consider the problem of finding the probability of exactly 7 boys in 11 births. Solve that problem using (1) normal approximation to the binomial using Table \(A-2 ;\) (2) normal approximation to the binomial using technology instead of Table \(A-2 ;\) (3) using technology with the binomial distribution instead of using a normal approximation. Compare the results. Given that the requirements for using the normal approximation are just barely met, are the approximations off by very much?

Use the data in the table below for sitting adult males and females (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats. (Hint: Draw a graph in each case.) $$\begin{array}{|l|l|l|l|} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \mathrm{in} . & 1.1 \mathrm{in} . & \text { Normal } \\ \hline \text { Females } & 22.7 \mathrm{in} . & 1.0 \mathrm{in} . & \text { Normal } \\ \hline \end{array}$$ Significance Instead of using 0.05 for identifying significant values, use the criteria that a value \(x\) is significantly high if \(P(x \text { or greater) } \leq 0.025\) and a value is significantly low if \(P(x \text { or less }) \leq 0.025 .\) Find the female back-to-knee length, separating significant values from those that are not significant. Using these criteria, is a female back-to-knee length of 20 in. significantly low?

Do the following: If the requirements of \(n p \geq 5\) and \(n q \geq 5\) are both satisfied, estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution; if \(n p < 5\) or n \(q < 5,\) then state that the normal approximation should not be used. With \(n=8\) births and \(p=0.512\) for a boy, find \(P\) (exactly 5 boys).

In a study of 420,095 cell phone users in Denmark, it was found that 135 developed cancer of the brain or nervous system. For those not using cell phones, there is a 0.000340 probability of a person developing cancer of the brain or nervous system. We therefore expect about 143 cases of such cancers in a group of 420,095 randomly selected people. a. Find the probability of 135 or fewer cases of such cancers in a group of 420,095 people. b. What do these results suggest about media reports that suggest cell phones cause cancer of the brain or nervous system?

College Presidents There are about 4200 college presidents in the United States, and they have annual incomes with a distribution that is skewed instead of being normal. Many different samples of 40 college presidents are randomly selected, and the mean annual income is computed for each sample. a. What is the approximate shape of the distribution of the sample means (uniform, normal, skewed, other)? b. What value do the sample means target? That is, what is the mean of all such sample means?
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