/*! 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 \(4.1 \mathrm{~h}\) and \(4.5 \mathrm{~h}\), respectively. Suppose these are the population average lifetimes. a. Let \(X\) 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 \(X-Y\) (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 \(1.8 \mathrm{~h}\) for Duracell batteries and \(2.0 \mathrm{~h}\) 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 \(X-Y\) (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 \(X-Y\) is \(-0.4\) hours, variance is \(0.0724\), and standard deviation is \(0.269\) hours.

Step by step solution

01

Determine the mean of X - Y

The mean value of \( \bar{X} - \bar{Y} \) is given by the difference between the mean lifetimes of each type of battery since means are additive. Therefore, if the mean lifetime of Duracell batteries \( \mu_X = 4.1 \) hours and the mean lifetime of Eveready batteries \( \mu_Y = 4.5 \) hours, then the mean of \( \bar{X} - \bar{Y} = \mu_X - \mu_Y = 4.1 - 4.5 = -0.4 \) hours. The specified sample sizes do not change the difference in means; thus, it does not depend on the sample sizes.
02

Calculate the variance of X - Y

The variance of \( \bar{X} - \bar{Y} \) is the sum of the variances of \( \bar{X} \) and \( \bar{Y} \). Given the population standard deviations, \( \sigma_X = 1.8 \) and \( \sigma_Y = 2.0 \), for sample sizes of 100, the variances are \( \frac{\sigma_X^2}{100} = \frac{1.8^2}{100} = 0.0324 \) and \( \frac{\sigma_Y^2}{100} = \frac{2.0^2}{100} = 0.04 \). Therefore, the variance of \( \bar{X} - \bar{Y} \) is \( 0.0324 + 0.04 = 0.0724 \).
03

Determine the standard deviation of X - Y

The standard deviation of \( \bar{X} - \bar{Y} \) is the square root of the variance calculated in Step 2. Thus, \( \sqrt{0.0724} \approx 0.269 \) hours.
04

Sketch the distribution of X - Y

The distribution of \( \bar{X} - \bar{Y} \) is approximately normal due to the Central Limit Theorem. With a mean at \(-0.4\) and a standard deviation of approximately \(0.269\), the curve is centered at \(-0.4\) on the horizontal axis. If sample sizes of 10 are used rather than 100, the distribution would still be normal due to the Central Limit Theorem, but the variance would be larger (resulting in a wider, less concentrated curve) because less information is used to estimate the population means, increasing variability.

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Key Concepts

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

Population Mean
In statistics, the population mean is a key measure, often denoted by the Greek letter \( \mu \). It represents the average of a group of values, assuming you could measure the entire population. It's like finding the midpoint of everyone's duration on a battery. For example, the population mean for Duracell batteries is given as \(4.1\) hours, and for Eveready, it is \(4.5\) hours. These values provide a central point around which each battery's life is expected to hover, assuming no outliers.

Understanding the population mean helps in determining patterns and making predictions. It does not change due to sample sizes. Whether you pick 10 batteries or 100, the average lifespan derived from the entire batch remains consistent. In essence, the population mean is the foundation for comparing the central tendencies across different datasets, such as determining the difference in means, \( \bar{X}-\bar{Y} \), between Duracell and Eveready batteries.
Variance and Standard Deviation
Variance and standard deviation are critical when understanding data spread. Variance measures the degree to which each number in a set varies from the mean. The formula is the average of the squared differences from the mean. For individual battery lifespans, you may encounter values like variance \( \sigma^2 \), calculated as \( \frac{\text{population standard deviation}^2}{\text{sample size}} \).

The standard deviation is simply the square root of the variance. This gives you a clearer picture of the original units of measurement, such as hours. In the example, the variance for \( \bar{X}-\bar{Y} \) is calculated as \(0.0724\), and the standard deviation is \(\sqrt{0.0724} \approx 0.269\). This measures how much the average lifetimes vary in both types of batteries. Small standard deviations indicate that most values hover closely around the mean, while larger deviations suggest more spread out data.

Knowing these values allows you to compare with known standards and determine probabilities and expectations about future sets.
Central Limit Theorem
The Central Limit Theorem (CLT) is a powerful tool in statistics that suggests that the distribution of sample means approximates a normal distribution, even if the original population distribution is not normal, provided the sample size is large enough. This explains why, in our battery example, the distribution of \( \bar{X}-\bar{Y} \) is normal. Despite individual battery lives potentially not adhering to a normal curve, the average of multiple samples will roughly create one.

The CLT is significant because it justifies using the normal distribution as a basis for inferential statistics. When you have a sufficiently large sample size - typically considered 30 or more - the theorem holds, which reassures statisticians and analysts that their predictions through mean values are reliable. The battery example uses 100 samples, which comfortably meets this requirement.

In this context, regardless of the underlying distribution of lifespans among single batteries, predicting or analyzing mean differences like \( X-Y \) is reasonable and based on strong statistical foundations.
Sample Size Effects
Sample size is a crucial factor in statistical analysis, impacting both the accuracy and the reliability of estimates. Its effect is clearly illustrated in the battery lifespan example. Larger sample sizes, such as the 100 batteries used for each type in the given problem, lead to more precise and reliable estimates of population parameters. This is because they reduce the variance, resulting in a more concentrated normal distribution curve.

Conversely, if only 10 batteries were sampled instead of 100, the variance increases due to fewer data points being averaged out. This results in a wider and less concentrated distribution curve for \( X-Y \). The smaller the sample, the greater the chance that the sample mean will differ from the true population mean, leading to greater variability and potentially less accurate results.

Understanding sample size effects allows statisticians to choose appropriate sample sizes needed for their studies, ensuring that they gather enough data to make confident and accurate inferences about the populations being studied.

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Most popular questions from this chapter

Can the right diet help us cope with diseases associated with aging such as Alzheimer's disease? A study ("Reversals of Age-Related Declines in Neuronal Signal Transduction, Cognitive, and Motor Behavioral Deficits with Blueberry, Spinach, or Strawberry Dietary Supplement," \(J\). Neurosci., 1999; 8114-8121) investigated the effects of fruit and vegetable supplements in the diet of rats. The rats were 19 months old, which is aged by rat standards. The 40 rats were randomly assigned to four diets, of which we will consider just the blueberry diet and the control diet here. After 8 weeks on their diets, the rats were given a number of tests. We give the data for just one of the tests, which measured how many seconds they could walk on a rod. Here are the times for the ten control rats (C) and ten blueberry rats (B): The objective is to obtain a \(95 \%\) confidence interval for the difference of population means. a. Determine a \(95 \%\) confidence interval for the difference of population means using the method based on the Theorem of Section 10.2. b. Obtain a bootstrap sample of 999 differences of means. Check the bootstrap distribution for normality using a normal probability plot. c. Use the standard deviation of the bootstrap distribution along with the mean and \(t\) critical value from (a) to get a \(95 \%\) confidence interval for the difference of means. d. Use the bootstrap sample and the percentile method to obtain a \(95 \%\) confidence interval for the difference of means. e. Compare your three confidence intervals. If they are very similar, why do you think this is the case? If you had used a critical value from the normal table rather than the \(t\) table, would the result of (c) agree better with the result of (d)? Why? f. Interpret your results. Do the blueberries make a substantial difference?

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Give as much information as you can about the \(P\)-value of the \(F\) test in each of the following situations: a. \(v_{1}=5, v_{2}=10\), upper-tailed test, \(f=4.75\) b. \(v_{1}=5, v_{2}=10\), upper-tailed test, \(f=2.00\) c. \(v_{1}=5, v_{2}=10\), two-tailed test, \(f=5.64\) d. \(v_{1}=5, v_{2}=10\), lower-tailed test, \(f=.200\) e. \(v_{1}=35, v_{2}=20\), upper-tailed test, \(f=3.24\)

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