/*! 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 1 An article in the November 1983 ... [FREE SOLUTION] | 91Ó°ÊÓ

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An article in the November 1983 Consumer Reports compared various types of batteries. The average lifetimes of Duracell Alkaline AA batteries and Eveready Energizer Alkaline AA batteries were given as \(4.1\) hours and \(4.5\) hours, respectively. Suppose these are the population average lifetimes. a. Let \(\bar{X}\) be the sample average lifetime of 100 Duracell batteries and \(\bar{Y}\) be the sample average lifetime of 100 Eveready batteries. What is the mean value of \(\bar{X}-\bar{Y}\) (i.e., where is the distribution of \(\bar{X}-\bar{Y}\) centered)? How does your answer depend on the specified sample sizes? b. Suppose the population standard deviations of lifetime are \(1.8\) hours for Duracell batteries and \(2.0\) hours for Eveready batteries. With the sample sizes given in part (a), what is the variance of the statistic \(\bar{X}-\bar{Y}\), and what is its standard deviation? c. For the sample sizes given in part (a), draw a picture of the approximate distribution curve of \(\bar{X}-\bar{Y}\) (include a measurement scale on the horizontal axis). Would the shape of the curve necessarily be the same for sample sizes of 10 batteries of each type? Explain.

Short Answer

Expert verified
The mean of \(\bar{X}-\bar{Y}\) is \(-0.4\), its standard deviation is \(0.269\), and it is approximately normal with sample sizes of 100.

Step by step solution

01

Calculate the Mean of XÌ„ and YÌ„

For part (a), the population average lifetime of Duracell batteries is given as \(4.1\) hours, and for Eveready batteries, it is \(4.5\) hours. The mean of a sample average \(\bar{X}\) is equal to the population mean \(\mu_X = 4.1\) hours. Similarly, the mean of \(\bar{Y}\) is \(\mu_Y = 4.5\) hours. Therefore, the mean of \(\bar{X} - \bar{Y}\) is \(\mu_X - \mu_Y = 4.1 - 4.5 = -0.4\). The answer does not depend on the sample size because the mean difference is based on population means.
02

Calculate the Variance of XÌ„ - YÌ„

For part (b), given that the population standard deviations are \(\sigma_X = 1.8\) hours for Duracell and \(\sigma_Y = 2.0\) hours for Eveready, and both sample sizes are 100. The variance of the difference \(\bar{X} - \bar{Y}\) is calculated as:\[ \text{Var}(\bar{X} - \bar{Y}) = \frac{\sigma_X^2}{n_X} + \frac{\sigma_Y^2}{n_Y} = \frac{1.8^2}{100} + \frac{2.0^2}{100} = \frac{3.24}{100} + \frac{4.0}{100} = 0.0324 + 0.040 = 0.0724. \]
03

Calculate the Standard Deviation of XÌ„ - YÌ„

Continuing with part (b), the standard deviation of \(\bar{X} - \bar{Y}\) is the square root of the variance:\[ \text{SD}(\bar{X} - \bar{Y}) = \sqrt{0.0724} \approx 0.269\] hours.
04

Describe the Distribution Curve of XÌ„ - YÌ„

For part (c), since both sample sizes are 100, by the Central Limit Theorem, \(\bar{X} - \bar{Y}\) is approximately normally distributed with mean \(-0.4\) and standard deviation \(0.269\). If the sample sizes were 10, the curve might be less normal and wider due to increased variability. The distribution will still center at \(-0.4\) but will be less symmetric, which might deviate from a perfect normal distribution, making the tails heavier.

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Key Concepts

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

Population Standard Deviation
Population standard deviation is a measure that captures the amount of variability or dispersion in a set of data points in a population. In the context of the battery exercise, it's used to understand the variation in lifetime hours of Duracell and Eveready batteries. When calculating statistics like variance or standard deviation for sample means, knowing the population standard deviation (symbolized as \( \sigma \)) helps in assessing how much the sample means might fluctuate from the true population mean.For example, Duracell's population standard deviation is given as \( 1.8 \) hours and Eveready’s as \( 2.0 \) hours. By understanding these values, we can make informed predictions about sample data behavior or sampling error.
Central Limit Theorem
The Central Limit Theorem (CLT) is fundamental in statistics. It profoundly influences how we approach sample distributions in practice. According to the CLT, when you have a large enough sample size, the distribution of the sample mean will approximate a normal distribution, regardless of the original population distribution. In the battery lifetime example, the theorem assures us that the difference in sample means, \( \bar{X} - \bar{Y} \), will be approximately normally distributed if our sample sizes are adequately large (e.g., 100 batteries each). This makes it easier to make predictions and draw conclusions about the population mean differences with reliability.
Variance Calculation
Calculating variance is crucial because it provides a numerical value for the dispersion within data sets. For sample means, understanding variance helps infer how closely a sample mean might reflect the population mean.For our exercise, we calculate the variance of \( \bar{X} - \bar{Y} \), the difference in sample means of two battery types. This is done using the formula:\[\text{Var}(\bar{X} - \bar{Y}) = \frac{\sigma_X^2}{n_X} + \frac{\sigma_Y^2}{n_Y}\]where \( \sigma_X \) and \( \sigma_Y \) are the respective population standard deviations, and \( n_X \) and \( n_Y \) are the sample sizes. Understanding variance helps determine the consistency and reliability of data predictions.
Normal Distribution Approximation
Normal distribution approximation is a statistical method for describing data that clusters around a mean. Thanks to the Central Limit Theorem, even if population data is not normally distributed, the means of sufficiently large samples from that population will be normally distributed.In the discussed battery study, normal distribution approximation allows us to visualise the distribution of sample mean differences, \( \bar{X} - \bar{Y} \), with a graph resembling the bell curve. This approach is beneficial for examining how close our sample calculations are to what’s true for the entire population. If the sample sizes were a lot smaller (e.g., 10 batteries instead of 100), the approximation would be less accurate, resulting in a distribution that could be more skewed or "fatter" at the ends than a perfect normal distribution.

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Most popular questions from this chapter

A mechanical engineer wishes to compare strength properties of steel beams with similar beams made with a particular alloy. The same number of beams, \(n\), of each type will be tested. Each beam will be set in a horizontal position with a support on each end, a force of 2500 lb will be applied at the center, and the deflection will be measured. From past experience with such beams, the engineer is willing to assume that the true standard deviation of deflection for both types of beam is \(.05\) in. Because the alloy is more expensive, the engineer wishes to test at level \(.01\) whether it has smaller average deflection than the steel beam. What value of \(n\) is appropriate if the desired type II error probability is .05 when the difference in true average deflection favors the alloy by 04 in.?

Olestra is a fat substitute approved by the FDA for use in snack foods. Because there have been anecdotal reports of gastrointestinal problems associated with olestra consumption, a randomized, double-blind, placebo- controlled experiment was carried out to compare olestra potato chips to regular potato chips with respect to Gl symptoms ("Gastrointestinal Symptoms Following Consumption of Olestra or Regular Triglyceride Potato Chips, J. of the Amer. Med. Assoc., 1998: 150-152). Among 529 individuals in the TG control group, 17.6\% experienced an adverse GI event, whereas among the 563 individuals in the olestra treatment group, \(15.8 \%\) experienced such an event. a. Carry out a test of hypotheses at the \(5 \%\) significance level to decide whether the incidence rate of GI problems for those who consume olestra chips according to the experimental regimen differs from the incidence rate for the TG control treatment. b. If the true percentages for the two treatments were \(15 \%\) and \(20 \%\), respectively, what sample sizes \((m=n)\) would be necessary to detect such a difference with probability \(90 ?\)

McNemar's test, developed in Exercise 54, can also be used when individuals are paired (matched) to yield \(n\) pairs and then one member of each pair is given treatment 1 and the other is given treatment 2 . Then \(X_{1}\) is the number of pairs in which both treatments were successful, and similarly for \(X_{2}, X_{3}\), and \(X_{4}\). The test statistic for testing equal efficacy of the two treatments is given by \(\left(X_{2}-X_{3}\right) / \sqrt{\left(X_{2}+X_{3}\right)}\), which has approximately a standard normal distribution when \(H_{0}\) is true. Use this to test whether the drug ergotamine is effective in the treatment of migraine headaches. \begin{tabular}{cc|cc} & & Ergotamine \\ \hline & & \(\boldsymbol{S}\) & \(\boldsymbol{F}\) \\ \hline Placebo & \(\boldsymbol{S}\) & 44 & 34 \\ & \(\boldsymbol{F}\) & 46 & 30 \\ \hline \end{tabular} The data is fictitious, but the conclusion agrees with that in the article "Controlled Clinical Trial of Ergotamine Tartrate" (British Med. J., 1970: 325-327).

The article "Fatigue Testing of Condoms" cited in Exercise \(7.32\) reported that for a sample of 20 natural latex condoms of a certain type, the sample mean and sample standard deviation of the number of cycles to break were 4358 and 2218 , respectively, whereas a sample of 20 polyisoprene condoms gave a sample mean and sample standard deviation of 5805 and 3990 , respectively. Is there strong evidence for concluding that true average number of cycles to break for the polyisoprene condom exceeds that for the natural latex condom by more than 1000 cycles? Carry out a test using a significance level of .01. [Note: The cited paper reported \(P\)-values of \(t\) tests for comparing means of the various types considered.]

An experiment to determine the effects of temperature on the survival of insect eggs was described in the article "Development Rates and a Temperature- Dependent Model of Pales Weevil" (Environ. Entomology, 1987:956-962). At \(11^{\circ} \mathrm{C}, 73\) of 91 eggs survived to the next stage of development. At \(30^{\circ} \mathrm{C}, 102\) of 110 eggs survived. Do the results of this experiment suggest that the survival rate (proportion surviving) differs for the two temperatures? Calculate the \(P\)-value and use it to test the appropriate hypotheses.

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