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Chapter 9: Problem 25

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
a. The p-value is 0.017, which is less than the level of significance \(\alpha = 0.025\), so the null hypothesis is rejected. b. The computed z-score (-2.13) is less than the critical z-value (-1.96), hence the null hypothesis is rejected.

Step by step solution

01

Calculate the Z-score

The Z-score is calculated using the formula: \(Z = \frac{{\bar{X} - \mu}}{{\sigma / \sqrt{n}}}\) where \(\bar{X}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. Substituting the respective values as \(Z = \frac{{43.05 - 45}}{{4.5 / \sqrt{24}}}\), we get \(Z = -2.13\)
02

Find P-value

The p-value is the probability that a z-score is less than -2.13 under the null hypothesis. This can be found using the standard normal table, or using a statistical software. In this case, the P-value is found to be 0.017.
03

Decision Based on p-Value

If the p-value is less than the level of significance, the null hypothesis is rejected. Here, the p-value (0.017) is less than the significance level (\(\alpha = 0.025\)), so the null hypothesis is rejected.
04

Decision Based on Critical Value

The z-test is a left-tailed test since the test is for the mean being less than 45 months. The critical z-value for a left-tailed test with a significance level of 0.025 can be found in the z-table or by using a statistical software. It is -1.96. Since the z-score computed (-2.13) is lesser than the critical value (-1.96), the null hypothesis is rejected.

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Most popular questions from this chapter

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