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A particular brand of dishwasher soap is sold in three sizes: \(25 \mathrm{oz}, 40 \mathrm{oz}\), and \(65 \mathrm{oz}\). Twenty percent of all purchasers select a \(25-0 z\) box, \(50 \%\) select a \(40-0 z\) box, and the remaining \(30 \%\) choose a \(65-\mathrm{oz}\) box. Let \(X_{1}\) and \(X_{2}\) denote the package sizes selected by two independently selected purchasers. a. Determine the sampling distribution of \(\bar{X}\), calculate \(E(\bar{X})\), and compare to \(\mu\). b. Determine the sampling distribution of the sample variance \(S^{2}\), calculate \(E\left(S^{2}\right)\), and compare to \(\sigma^{2}\).

Short Answer

Expert verified
The expected value of \(\bar{X}\) equals \(\mu\) at 44.5, and the expected sample variance equals \(\sigma^2\) at 64.125.

Step by step solution

01

Determine the Probability Distribution

Calculate the probability distribution for the different package sizes. The probabilities are given as follows:- Probability of selecting a 25 oz box, \( P(X=25) = 0.20 \)- Probability of selecting a 40 oz box, \( P(X=40) = 0.50 \)- Probability of selecting a 65 oz box, \( P(X=65) = 0.30 \)
02

Calculate Expected Value \(E(X)\)

The expected value \(E(X)\) can be calculated using the formula \( E(X) = \sum x_i p_i \). Therefore,\[E(X) = 25 \cdot 0.20 + 40 \cdot 0.50 + 65 \cdot 0.30\]\[E(X) = 5 + 20 + 19.5 = 44.5\]
03

Calculate \(E(\bar{X})\)

For the sample mean \(\bar{X}\) of size 2, the expected value \(E(\bar{X})\) is the same as \(E(X)\) due to the linearity of expectation. Thus, \(E(\bar{X}) = E(X) = 44.5 \).
04

Variance \(\sigma^2\) for Distribution of \(X\)

Calculate the variance \(\sigma^2\) of the distribution of \(X\). This requires:\[\sigma^2 = \sum (x_i - E(X))^2 p_i \]\[= (25 - 44.5)^2 \cdot 0.20 + (40 - 44.5)^2 \cdot 0.50 + (65 - 44.5)^2 \cdot 0.30\]\[= 64.125\]
05

Sampling Distribution of \(S^2\) and \(E(S^2)\)

For a sample size of 2, the unbiased estimator of the variance is given by \(S^2 = \frac{1}{n-1} \sum (x_i - \bar{x})^2\), where \(n=2\).In a two-sample scenario, \(E(S^2) = \sigma^2\) because the sample variance is an unbiased estimator of the population variance. Thus, \(E(S^2) = 64.125\).
06

Comparison\(\bar{X}\) and \(\mu\), \(E(S^2)\) and \(\sigma^2\)

Both \(E(\bar{X}) = 44.5\) and \(\mu = 44.5\); thus they are the same.Similarly, \(E(S^2) = 64.125\) equates the population variance \(\sigma^2 = 64.125\).

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Key Concepts

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

Probability Distribution
When dealing with probability distributions, it's essential to understand how different outcomes are associated with probabilities. For this exercise on dishwasher soap box sizes, we define a discrete probability distribution that relates the package sizes to their likelihood of selection.
  • For a 25 oz box, the probability is 0.20.
  • A 40 oz box has a probability of 0.50.
  • The 65 oz box is selected with a probability of 0.30.
These probabilities must sum to 1, ensuring that every possible outcome has been considered. In this setting, we associate each box size with its respective probability to calculate key statistical measures of the distribution.
Understanding these probabilities lays the foundation for further analysis, such as expected values and variance.
Sampling Distribution
The concept of sampling distribution focuses on the distribution of a statistic, such as the sample mean, calculated from all possible samples of the same size from a population. In this exercise, we examine the sampling distribution of the mean package size selected by two independent purchasers.
Each sample mean, denoted as \(\bar{X}\), is calculated by taking the average of two observed package sizes. Given the linearity of expectations, the expected value of the sample mean \(E(\bar{X})\) is equal to the expected value of a single selection, \(E(X)\). This makes it straightforward to determine the expected value for \(\bar{X}\), which turns out to be 44.5 oz.
The sampling distribution helps to understand and anticipate the behavior of sample means when drawing samples from a population. It's crucial in fields like statistics and data analysis to make generalizations from sample data alone.
Variance Calculation
Calculating variance is a key step in understanding the spread or variability in a dataset. For the probability distribution defined by the soap box sizes, variance gives us insight into how much the box sizes deviate from the average size.
To calculate the variance \(\sigma^2\) for the given distribution, use the formula:\[\sigma^2 = \sum (x_i - E(X))^2 p_i\]This involves computing the squared difference between each box size and the mean size, multiplied by the probability of that size. Summing these values gives a variance of 64.125, indicating the extent of variability in box size selections.
Understanding variance is vital as it provides a measure of dispersion, helping to make informed decisions based on data variability.
Unbiased Estimator
In statistics, an unbiased estimator is a statistic used to estimate a population parameter that, on average, equals the actual parameter value. For the dish soap sizes, we use the sample variance \(S^2\) as an unbiased estimator for the population variance \(\sigma^2\).
The formula for sample variance in a two-sample case simplifies due to the degrees of freedom being \(n-1\); hence, it's calculated as:\[S^2 = \frac{1}{n-1} \sum (x_i - \bar{x})^2\]Here, \(E(S^2)\) equals the population variance \(\sigma^2\), indicating that the sample variance is reliable as an unbiased estimator. Our calculations confirmed this as both \(E(S^2)\) and \(\sigma^2\) equal 64.125.
This quality of being unbiased is crucial in ensuring that sample statistics reliably represent their respective population parameters, facilitating accurate conclusions from statistical analyses.

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Most popular questions from this chapter

A restaurant serves three fixed-price dinners costing \(\$ 12\), \(\$ 15\), and \(\$ 20\). For a randomly selected couple dining at this restaurant, let \(X=\) the cost of the man's dinner and \(Y=\) the cost of the woman's dinner. The joint pmf of \(X\) and \(Y\) is given in the following table: \begin{tabular}{cc|ccc} \(p(x, y)\) & & 12 & 15 & 20 \\ \hline & 12 & \(.05\) & \(.05\) & \(.10\) \\ \multirow{x}{*}{} & 15 & \(.05\) & \(.10\) & \(.35\) \\ & 20 & 0 & \(.20\) & \(.10\) \end{tabular} a. Compute the marginal pmf's of \(X\) and \(Y\). b. What is the probability that the man's and the woman's dinner cost at most \(\$ 15\) each? c. Are \(X\) and \(Y\) independent? Justify your answer. d. What is the expected total cost of the dinner for the two people? e. Suppose that when a couple opens fortune cookies at the conclusion of the meal, they find the message "You will receive as a refund the difference between the cost of the more expensive and the less expensive meal that you have chosen." How much would the restaurant expect to refund?

A binary communication channel transmits a sequence of "bits" (0s and 1s). Suppose that for any particular bit transmitted, there is a \(10 \%\) chance of a transmission error (a 0 becoming a 1 or a 1 becoming a 0 ). Assume that bit errors occur independently of one another. a. Consider transmitting 1000 bits. What is the approximate probability that at most 125 transmission errors occur? b. Suppose the same 1000 -bit message is sent two different times independently of one another. What is the approximate probability that the number of errors in the first transmission is within 50 of the number of errors in the second?

You have two lightbulbs for a particular lamp. Let \(X=\) the lifetime of the first bulb and \(Y=\) the lifetime of the second bulb (both in 1000 s of hours). Suppose that \(X\) and \(Y\) are independent and that each has an exponential distribution with parameter \(\lambda=1\). a. What is the joint pdf of \(X\) and \(Y\) ? b. What is the probability that each bulb lasts at most 1000 hours (i.e., \(X \leq 1\) and \(Y \leq 1\) )? c. What is the probability that the total lifetime of the two bulbs is at most 2 ? [Hint: Draw a picture of the region \(A=\\{(x, y): x \geq 0, y \geq 0, x+y \leq 2\\}\) before integrating.] d. What is the probability that the total lifetime is between 1 and 2 ?

A box contains ten sealed envelopes numbered \(1, \ldots, 10\). The first five contain no money, the next three each contains \(\$ 5\), and there is a \(\$ 10\) bill in each of the last two. A sample of size 3 is selected with replacement (so we have a random sample), and you get the largest amount in any of the envelopes selected. If \(X_{1}, X_{2}\), and \(X_{3}\) denote the amounts in the selected envelopes, the statistic of interest is \(M=\) the maximum of \(X_{1}, X_{2}\), and \(X_{3}\). a. Obtain the probability distribution of this statistic. b. Describe how you would carry out a simulation experiment to compare the distributions of \(M\) for various sample sizes. How would you guess the distribution would change as \(n\) increases?

a. Show that \(\operatorname{Cov}(X, Y+Z)=\operatorname{Cov}(X, Y)+\operatorname{Cov}(X, Z)\). b. Let \(X_{1}\) and \(X_{2}\) be quantitative and verbal scores on one aptitude exam, and let \(Y_{1}\) and \(Y_{2}\) be corresponding scores on another exam. If \(\operatorname{Cov}\left(X_{1}, Y_{1}\right)=5\), \(\operatorname{Cov}\left(X_{1}, Y_{2}\right)=1, \operatorname{Cov}\left(X_{2}, Y_{1}\right)=2\), and \(\operatorname{Cov}\left(X_{2}, Y_{2}\right)=\) 8 , what is the covariance between the two total scores \(X_{1}+X_{2}\) and \(Y_{1}+Y_{2} ?\)

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