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Chapter 9: Problem 35

The manufacturer of a certain brand of auto batteries claims that the mean life of these batteries is 45 months. A consumer protection agency that wants to check this claim took a random sample of 24 such batteries and found that the mean life for this sample is \(43.05\) months. The lives of all such batteries have a normal distribution with the population standard deviation of \(4.5\) months. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the mean life of these batteries is less than 45 months. Will you reject the null hypothesis at \(\alpha=.025\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.025\).

Short Answer

Expert verified
For a, the p-value of the test is approximately 0.0179 and thus, we reject the null hypothesis because 0.0179 < 0.025. For b, using the critical-value approach, we again reject the null hypothesis as the calculated Z value (-2.1) is less than the critical Z (-1.96). The results of both tests suggest that the manufacturer's claim about average battery life is not valid.

Step by step solution

01

Identify Null and Alternative Hypotheses

For a. and b., first establish the null hypothesis (H0) and the alternative hypothesis (Ha). Given that manufacturer claims the average battery life is 45 months, \(H_0: \mu = 45\) is our null hypothesis. The alternative hypothesis - \(Ha: \mu < 45\) - is proposing that the battery life is less than 45 months, which is what the consumer protection agency is trying to prove.
02

Calculate Test Statistic

It's important calculate the Test Statistic using the formula: \(Z = \frac{{\bar{x} - \mu_0}}{{\sigma / \sqrt{n}}}\), where \(\bar{x}\) is the sample mean, \(\mu_0\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. Substituting our values in, \(Z = \frac{{43.05 - 45}}{{4.5 / \sqrt{24}}}\) which equals approximately -2.1 after computation.
03

Calculate the p-value

With our Z-value, we can find the corresponding p-value from the Z-distribution table or using software tools. The p-value for Z=-2.1 is around 0.0179.
04

Compare p-value with \(\alpha\)

Now compare the p-value we obtained against the \(\alpha\) value given in the problem (\(\alpha = 0.025\)). Since 0.0179 < 0.025, we will reject the null hypothesis.
05

Critical-Value approach

For part b, we can use the critical-value approach. In this case, find the critical value of Z for \(\alpha =0.025\) in a one-tailed test, which gives us -1.96. The decision rule here is to reject \(H_0\) if the test statistic Z is less than -1.96. Our calculated Z (-2.1) is less than -1.96, thus we again reject the null hypothesis, which is consistent with part a.

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Most popular questions from this chapter

Consider \(H_{0}: \mu=40\) versus \(H_{1}: \mu>40\) a. A random sample of 64 observations taken from this population produced a sample mean of 43 and a standard deviation of \(5 .\) Using \(\alpha=.025\), would you reject the null hypothesis? b, Another random sample of 64 observations taken from the same population produced a sample mean of 41 and a standard deviation of 7 . Using \(\alpha=.025\), would you reject the null hypothesis?

According to the Magazine Publishers of America (www.magazine.org), the average visit at the magazines' Web sites during the fourth quarter of 2007 lasted \(4.145\) minutes. Forty-six randomly selected visits to magazine's Web sites during the fourth quarter of 2009 produced a sample mean visit of \(4.458\) minutes, with a standard deviation of \(1.14\) minutes. Using the \(10 \%\) significance level and the critical value approach, can you conclude that the length of an average visit to these Web sites during the fourth quarter of 2009 was longer than \(4.145\) minutes? Find the range for the \(p\) -value for this test. What will your conclusion be using this \(p\) -value range and \(\alpha=.10\) ?

The mean consumption of water per household in a city was 1245 cubic feet per month. Due to a water shortage because of a drought, the city council campaigned for water use conservation by households. A few months after the campaign was started, the mean consumption of water for a sample of 100 households was found to be 1175 cubic feet per month. The population standard deviation is given to be 250 cubic feet. a. Find the \(p\) -value for the hypothesis test that the mean consumption of water per household has decreased due to the campaign by the city council. Would you reject the null hypothesis at \(\alpha=.025 ?\) b. Make the test of part a using the critical-value approach and \(\alpha=.025\).

Alpha Airlines claims that only \(15 \%\) of its flights arrive more than 10 minutes late. Let \(p\) be the proportion of all of Alpha's flights that arrive more than 10 minutes late. Consider the hypothesis test $$ H_{0}: p \leq .15 \text { versus } H_{1}: p>.15 $$ Suppose we take a random sample of 50 flights by Alpha Airlines and agree to reject \(H_{0}\) if 9 or more of them arrive late. Find the significance level for this test.

The Environmental Protection Agency (EPA) recommends that the sodium content in public water supplies should be no more than \(20 \mathrm{mg}\) per liter (http://www.disabledworld.com/artman/publish/ sodiumwatersupply.shtml). Forty samples were taken from a large reservoir, and the amount of sodium in each sample was measured. The sample average was \(23.5 \mathrm{mg}\) per liter. Assume that the population standard deviation is \(5.6 \mathrm{mg}\) per liter. The EPA is interested in knowing whether the average sodium content for the entire reservoir exceeds the recommended level. If so, the communities served by the reservoir will have to be made aware of the violation. a. Find the \(p\) -value for the test of hypothesis. Based on this \(p\) -value, would the communities need to be informed of an excessive average sodium level if the maximum probability of a Type I error is to be \(.05 ?\) What if the maximum probability of a Type I error is to be \(.01 ?\) b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.05 .\) Would the communities need to be notified? What if \(\alpha=.01 ?\) What if \(\alpha\) is zero?
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