/*! 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 138 Variation in the life of a batte... [FREE SOLUTION] | 91Ó°ÊÓ

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Variation in the life of a battery is expected, but too much variation would be of concern to the consumer, who would never know if the purchased battery might have a very short life. A random sample of 30 AA batteries of a particular brand produced a standard deviation of 350 hours. If a standard deviation of 288 hours (12 days) is considered acceptable, does this sample provide sufficient evidence that this brand of battery has greater variation than what is acceptable at the 0.05 level of significance? Assume battery life is normally distributed.

Short Answer

Expert verified
No, based on the sample data, we don't have sufficient evidence at the 0.05 level of significance to say that this brand of battery has greater variation than what is considered acceptable.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis \(H_{0}\) is that the variance \(σ^{2}\) is equal to \(288^{2} = 82944\) hours. The alternative hypothesis \(H_{1}\) is that the variance \(σ^{2}\) is greater than 82944 hours.
02

Calculate the Test Statistic

Since we are dealing with variances, we use the Chi-Square test. The formula for the test statistic is \(X^{2} = (n - 1) * s^{2} / σ_0^{2} = (30 - 1) * 350^{2} / 82944 = 13.91\) where n is the number of observations, s is the sample standard deviation, and \(σ_{0}\) is the value of standard deviation under the null hypothesis.
03

Determine the Rejection Region

At the 0.05 level of significance, the critical value \(χ_{critical}^{2}\) can be found in the Chi-Square distribution table with \(n - 1 = 29\) degrees of freedom. It turns out to be 42.557. If our \(X^{2}\) value falls in the rejection region (greater than \(χ_{critical}^{2}\)), then we reject the null hypothesis.
04

Make a Decision

Our test statistic \(X^{2} = 13.91\) does not fall in the rejection region, since \(13.91 < 42.557\). So, we do not reject the null hypothesis.

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Key Concepts

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

Understanding Standard Deviation
Standard deviation is a crucial concept in statistics that measures the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean (also known as the expected value), while a high standard deviation indicates that the values are spread out over a wider range.

In the context of our exercise, the standard deviation is used to express the variability in the lifespan of batteries. Knowing that the consumer expects a certain consistency in the product, the standard deviation gives us insight into how much inconsistency there actually is. A consumer would find a standard deviation of 350 hours concerning if they expect the deviation to be 288 hours or less, which translates to less variability and more predictability in the battery life.

The sample standard deviation is calculated as follows:
\[ s^2 = \frac{1}{N-1} \sum_{i=1}^{N} (x_i - \bar{x})^2 \]
Where:\[ s^2 \] is the sample variance, \( N \) is the number of data points, \( x_i \) represents each value, and \( \bar{x} \) is the mean of the values. It's important to ensure that the dataset represents a random sample and that the sample size is adequate enough to reflect the population's behavior.
The Role of the Chi-Square Test in Hypothesis Testing
The Chi-Square test is a statistical method used to compare observed data with expected data according to a specific hypothesis. It's particularly useful when assessing the goodness-of-fit or testing for independence in categorical data. However, in our exercise, the Chi-Square test is used in the context of variance and standard deviation, which makes it slightly less intuitive.

The test statistic formula in our exercise is:
\[ X^2 = \frac{(n - 1) \cdot s^2}{\sigma_0^2} \]
Here, the sample variance (\( s^2 \)) is used to gauge how much the measured sample deviates from what is considered normal or acceptable (in this case, the standard deviation of 288 hours). The test statistic itself follows a Chi-Square distribution, which is where degrees of freedom come into play. These degrees of freedom, which in our exercise are \( n - 1 \) (the sample size minus one), determine the shape of the Chi-Square distribution we use to find our critical value and assess whether our observed test statistic is significant.
Degrees of Freedom's Impact on Hypothesis Tests
Degrees of freedom in statistics represent the number of values that can vary within a data set while still adhering to certain constraints, often related to the sample size. In the context of a Chi-Square test, the degrees of freedom dictate the shape of the Chi-Square distribution, which is critical when determining the critical value.

In our textbook exercise, the degrees of freedom are calculated by subtracting one from the sample size (\( n - 1 \)). This is tied to the idea that once we have calculated our sample's mean, only \( n - 1 \) values are free to vary, with the last one being determined to meet the mean.

With 29 degrees of freedom in our case, it indicates the number of independent ways the battery life's variance can be distributed. This impacts the rejection region—the area under the Chi-Square distribution curve that corresponds with our level of significance (in this example, 0.05). As the degrees of freedom increase, the shape of the distribution changes, affecting where this rejection region lies and thereby influencing the conclusion of the hypothesis test.

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Most popular questions from this chapter

Today's newspapers and magazines often report the findings of survey polls about various aspects of life. The Pew Internet \& American Life Project (January 13-February 9,2005 ) found that "63\% of cell phone users ages \(18-27\) have used text messaging within the past month." Other information obtained from the project included "random telephone survey of 1,460 cell phone users" and "has a margin of sampling error of plus or minus 3 percentage points." Relate this information to the statistical inferences you have been studying in this chapter. a. Is a percentage of people a population parameter, and if so, how is it related to any of the parameters that we have studied? b. Based on the information given, find the \(95 \%\) confidence interval for the true proportion of cell phone users who have used text messaging. c. Explain how the terms "point estimate," "level of confidence," "maximum error of estimate," and "confidence interval" relate to the values reported in the article and to your answers in part b.

a. Calculate the standard deviation for each set. A: 5,6,7,7,8,10 B: 5,6,7,7,8,15 b. What effect did the largest value changing from 10 to 15 have on the standard deviation? c. Why do you think 15 might be called an outlier?

The National Highway Traffic Safety Administration found that, among the crashes with recorded times, EMS notification times exceeded 10 minutes in \(19.4 \%\) of rural fatal crashes. A random sample of 500 reported fatal crashes in Kentucky showed \(21.8 \%\) of the notification times exceeded 10 minutes. Construct the \(95 \%\) confidence interval for the true proportion of fatal crashes in Kentucky whose elapsed notification time exceeded 10 minutes.

To test the hypothesis that the standard deviation on a standard test is \(12,\) a sample of 40 randomly selected students' exams was tested. The sample variance was found to be \(155 .\) Does this sample provide sufficient evidence to show that the standard deviation differs from 12 at the 0.05 level of significance?

The full-time student body of a college is composed of \(50 \%\) males and \(50 \%\) females. Does a random sample of students \((30 \text { male, } 20\) female) from an introductory chemistry course show sufficient evidence to reject the hypothesis that the proportion of male and of female students who take this course is the same as that of the whole student body? Use \(\alpha=0.05\). a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

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