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Chapter 9: Q1 E (page 371)

1. 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 \({\bf{\bar X - \bar Y}}\) 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

\(\begin{array}{l}a. E(\bar X - \bar Y) = - 0.4; \\b. Variance = 0.0724; Standard deviation = 0.2691\end{array}\)

c.No

Step by step solution

01

To find the mean value

a.

The mean value is

\(\begin{array}{c}E(\bar X - \bar Y) = E(\bar X) - E(\bar Y)\\ = 4.1 - 4.5\\ = - 0.4\end{array}\)

02

To find the variance and standard deviation

b.

The variance of the statistic is

\(\begin{array}{c}V(\bar X - \bar Y) = V(\bar X) + {( - 1)^2}V(\bar Y)\\ = \frac{{\sigma _1^2}}{m} + \frac{{\sigma _2^2}}{n}\\ = \frac{{{{1.8}^2}}}{{100}} + \frac{{{2^2}}}{{100}}\\ = 0.0724\end{array}\)

The standard deviation of \(\bar X - \bar Y\) is

\(\sqrt {0.0724} = 0.2691\)

03

Drawing the graph

c.

Both random variables \(\bar X\) and \(\bar Y\) would be nearly normal because the sample sizes are large enough (both are 100), and so the\(\bar X - \bar Y\) would have an approximately normal distribution. When \(m = n = 10\), however, determining the distribution of \(\bar X - \bar Y\) becomes difficult. This is due to the fact that the Central Limit Theorem cannot be applied to such a large sample size.

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Most popular questions from this chapter

The following observations are on time (h) for a AA 1.5volt alkaline battery to reach a \(0.8\)voltage ("Comparing the Lifetimes of Two Brands of Batteries," J. of Statistical Educ., 2013, online):

\(\begin{array}{*{20}{l}}{ Energizer: }&{8.65}&{8.74}&{8.91}&{8.72}&{8.85}\\{ Ultracell: }&{8.76}&{8.81}&{8.81}&{8.70}&{8.73}\\{ Energizer: }&{8.52}&{8.62}&{8.68}&{8.86}&{}\\{ Ultracell: }&{8.76}&{8.68}&{8.64}&{8.79}&{}\end{array}\)

Normal probability plots support the assumption that the population distributions are normal. Does the data suggest that the variance of the Energizer population distribution differs from that of the Ultra cell population distribution? Test the relevant hypotheses using a significance level of .05. (Note: The two-sample \(t\)test for equality of population means gives a \(P - \)value of .763.) The Energizer batteries are much more expensive than the Ultra cell batteries. Would you pay the extra money?

Suppose \({\mu _1}\) and \({\mu _2}\) are true mean stopping distances at \(50mph\) for cars of a certain type equipped with two different types of braking systems. Use the two-sample t test at significance level

t test at significance level \(.01\) to test \({H_0}:{\mu _1} - {\mu _2} = - 10\) versus \({H_a}:{\mu _1} - {\mu _2} < - 10\) for the following data: \(m = 6,\;\;\;\bar x = 115.7,{s_1} = 5.03,n = 6,\bar y = 129.3,\;\)and \({s_2} = 5.38.\)

As the population ages, there is increasing concern about accident-related injuries to the elderly. The article "Age and Gender Differences in Single-Step Recovery from a Forward Fall" (J. of Gerontology, \(1999: M44 - M50\)) reported on an experiment in which the maximum lean angle-the farthest a subject is able to lean and still recover in one step-was determined for both a sample of younger females (\(21 - 29\)years) and a sample of older females (\(67 - 81\)years). The following observations are consistent with summary data given in the article:

YF:\(29,34,33,27,28,32,31,34,32,27\)

OF:\(18,15,23,13,12\)

Does the data suggest that true average maximum lean angle for older females is more than\(10\)degrees smaller than it is for younger females? State and test the relevant hypotheses at significance level . \(10\).

Recent incidents of food contamination have caused great concern among consumers. The article "How Safe Is That Chicken?" (Consumer Reports, Jan. 2010: 19-23) reported that 35 of 80 randomly selected Perdue brand broilers tested positively for either campylobacter or salmonella (or' both), the leading bacterial causes of food-borne disease, whereas 66 of 80 Tyson brand broilers tested positive.

  1. Does it appear that the true proportion of noncontaminated Perdue broilers differs from that for the Tyson brand? Carry out a test of hypotheses using a significance level .01.
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Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunnel syndrome (CTS). The article "A Gap Detection Tactility Test for Sensory Deficits Associated with Carpal Tunnel Syndrome" (Ergonomics, \(1995: 2588 - 2601\)) reported on a test that involved sensing a tiny gap in an otherwise smooth surface by probing with a finger; this functionally resembles many work-related tactile activities, such as detecting scratches or surface defects. When finger probing was not allowed, the sample average gap detection threshold for\(m = 8\)normal subjects was\(1.71\;mm\), and the sample standard deviation was\(.53\); for\(n = 10\)CTS subjects, the sample mean and sample standard deviation were\(2.53\)and\(.87\), respectively. Does this data suggest that the true average gap detection threshold for CTS subjects exceeds that for normal subjects? State and test the relevant hypotheses using a significance level of\(.01\).
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