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Chapter 16: Problem 619

The makers of a certain brand of car mufflers claim that the life of the mufflers has a variance of 8 year. A random sample of 16 of these mufflers showed a variance of 1 year. Using a \(5 \%\) level of significance, test whether the variance of all the mufflers of this manufacturer exceeds \(.8\) year.

Short Answer

Expert verified
The null hypothesis (\( H_0 \)) states that the population variance is 8 years, while the alternative hypothesis (\( H_a \)) claims that it exceeds 0.8 years. At a 5% level of significance and a sample size of 16, we use the Chi-square test for variance and calculate the test statistic (\( \chi^2 = 1.875 \)). Comparing this to the critical value (\( \chi^2_{0.95,15} = 24.996 \)), we cannot reject the null hypothesis. Therefore, there is insufficient evidence to conclude that the variance of this manufacturer's muffler life exceeds 0.8 years.

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis (\( H_0 \)) and alternative hypothesis (\( H_a \)) are as follows: \( H_0 \): \(\sigma^2 = 8\), the population variance is 8 years. \( H_a \): \(\sigma^2 > 0.8\), the population variance exceeds 0.8 years.
02

Find the critical value

Given that we have a 5% level of significance, we need to find the critical value for the chi-square distribution that corresponds to the 5% probability in the right tail. The degrees of freedom for the chi-square test for variance is equal to the sample size minus 1, which is 16 - 1 = 15. Using a chi-square table or calculator for the 95% confidence level, we get a critical value of \( \chi^2_{0.95,15} = 24.996 \).
03

Compute the test statistic

Now we need to compute the test statistic using the given information. The test statistic for the chi-square test for variance is: \( \chi^2 = \frac{(n - 1)S^2}{\sigma^2_0} \) Where: - \(n\) is the sample size (16), - \(S^2\) is the sample variance (1), - \(\sigma^2_0\) is the population variance under the null hypothesis (8). Plugging in these values, we get: \( \chi^2 = \frac{(16 - 1) \cdot 1}{8} = \frac{15}{8} = 1.875 \)
04

Compare the test statistic with the critical value and make a decision

Since the test statistic (\( \chi^2 = 1.875 \)) is less than the critical value (\( \chi^2_{0.95,15} = 24.996 \)), we cannot reject the null hypothesis. Therefore, at a 5% level of significance, we do not have enough evidence to conclude that the variance of all the mufflers of this manufacturer exceeds 0.8 years.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method that allows us to make inferences about the population using sample data. In hypothesis testing, we start by assuming a null hypothesis (\( H_0 \) ), which is a statement of no effect or no difference. In contrast, we have an alternative hypothesis (\( H_a \) ), which represents what we are trying to prove or evidence counter to the null hypothesis.

For the car mufflers exercise, the null hypothesis asserts that the population variance of the mufflers' life is 8 years (\( H_0: \) \( \(\sigma^2 = 8\) \)), while the alternative hypothesis suggests that the variance exceeds 0.8 years (\( H_a: \) \( \(\sigma^2 > 0.8\) \)). The testing process involves calculating a test statistic from the sample data and comparing it to a critical value to determine whether to reject the null hypothesis.
Population Variance
Population variance (\( \sigma^2 \) ) is a measure of the spread or dispersion of a set of data points in a population. It quantifies how much the individual data points in the entire population deviate from the population mean. Variance is calculated as the average of the squared differences from the mean.

In our muffler example, we are testing whether the true variance of the mufflers' life is different from the claimed variance of 8 years. The population variance is a parameter that typically needs to be estimated from the sample because we rarely have access to data from the entire population.
Degrees of Freedom
Degrees of freedom (\( df \) ) in statistics is a concept related to the number of independent values or quantities which can vary in the analysis without breaking any constraints. In the context of the chi-square test for variance, the degrees of freedom equal the number of observations in the sample minus one (\( df = n - 1 \) ).

In our example with the sample of 16 mufflers, we calculate the degrees of freedom as 15 (\( df = 16 - 1 \) ). This is crucial because the degrees of freedom determine the shape of the chi-square distribution which in turn is used to find the critical value.
Critical Value
The critical value in a statistical test is a threshold to which the test statistic is compared to determine whether to reject the null hypothesis. It is dependent on the chosen significance level (also known as the alpha level), which is the probability of rejecting the null hypothesis when it is true.

For the given level of significance of 5%, and using the degrees of freedom determined for our muffler life example, we refer to a chi-square distribution table or use a calculator to find the critical value (\( \(\chi^2_{0.95,15} = 24.996\) \)). If our test statistic exceeds this critical value, we would reject the null hypothesis.
Test Statistic
The test statistic is a calculated value used in hypothesis testing to compare against the critical value to decide whether to reject the null hypothesis. It incorporates sample data and under the chi-square test for variance, it is calculated as:\[ \chi^2 = \frac{(n - 1)S^2}{\sigma^2_0} \]where \(n\) is the sample size, \(S^2\) is the sample variance, and \(\sigma^2_0\) is the population variance under the null hypothesis. In the muffler example, the test statistic was found to be 1.875 using the given sample variance and hypothesized population variance. As this value is less than the critical value, we do not reject the null hypothesis as there is insufficient evidence that the population variance differs from what has been claimed.

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Most popular questions from this chapter

A college dean wants to determine if the students entering the college in a given year have higher IQ's than the students entering the same college in the previous year. The IQ's of the college entrants in the two years are known to be normally distributed and to have equal variances. A random sample of four students from this year's entering class gives IQ scores of \(110,113,116\), and 117. Another random sample of four students from last year's entering class gives IQ scores of \(109,111,112\), and 112 . At the \(5 \%\) level of significance, can we conclude that this year's students have a higher IQ than last years?

In a recent science fiction novel, aliens invented a death ray that killed 3 Earthman. The lethal doses were 10,11 , and 12 units of the ray. The mean lethal dose for Moonmen is \(13.30\) units. For this sample of 3 Earthmen \(\underline{\mathrm{X}}=11\) and \(\mathrm{S}=1.0\) and a test of significance of the difference between the mean for Earthmen and Moonmen yields \(\mathrm{t}=3.98\), not significant at the \(\alpha=.05\) level of significance. Then 6 more Earthmen were killed with lethal doses of \(9,10,11,12,13\) and \(14 .\) For this sample, \(\underline{\mathrm{X}}=11.5\), and \(\mathrm{S}=1.87\), and a test of significance of the difference between the mean for Earthmen and Moonmen yields \(\mathrm{t}=2.36\), also not significant at \(\alpha=.05\). What can be said if we combine the results of these two samples?

Suppose we have a binomial distribution for which \(\mathrm{H}_{0}\) is \(p=(1 / 2)\) where \(p\) is the probability of success on a single trial. Suppose the type I error, \(\alpha=.05\) and \(\mathrm{n}=100\). Calculate the power of this test for each of the following alternate hypotheses, \(\mathrm{H}_{1}: \mathrm{p}=.55, \mathrm{p}=.60, \mathrm{p}=.65, \mathrm{p}=.70\), and \(\mathrm{p}=.75\). Do the same when \(\alpha=.01\)

A man has just purchased a trick die which was advertised as not yielding the proper proportion of sixes. He wonders whether the advertising was correct, and tests the advertising claim by rolling the die 100 times. The 100 rolls yielded ten sixes. Should he conclude that the advertising was legitimate?

Suppose that for the preceding problem, we have taken a sample of 15 pairs of specimens. The differences in the average maximum pits for each coating, \(\underline{d}\) is 8 with a standard deviation for the differences of \(11.03\). Test whether there is a significant difference between the corrosion preventing abilities of coatings \(\mathrm{A}\) and \(\mathrm{B}\) at a \(.05\) level of significance.
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