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Chapter 8: Problem 20

The labor charge for repairs at an automobile service center are based on a standard time specified for each type of repair. The time specified for replacement of universal joint in a drive shaft is one hour. The manager reviews a sample of 30 such repairs. The average of the actual repair times is 0.86 hour with standard deviation 0.32 hour. a. Test at the \(1 \%\) level of significance the null hypothesis that the actual mean time for this repair differs from one hour. b. The sample mean is less than one hour, suggesting that the mean actual time for this repair is less than one hour. Perform this test, also at the \(1 \%\) level of significance. (The computation of the test statistic done in part (a) still applies here.)

Short Answer

Expert verified
a. Do not reject the null hypothesis; insufficient evidence for difference. b. Reject the null hypothesis; evidence suggests mean is less than 1 hour.

Step by step solution

01

Understand the Hypotheses

In part (a), we want to test if the actual mean repair time differs from the specified time of 1 hour. This is a two-tailed test. For part (b), we want to check if the actual mean time is less than 1 hour, which is a one-tailed test.
02

State the Null and Alternative Hypotheses for Part (a)

For part (a): \[H_0: \mu = 1 \text{ hour} \H_a: \mu eq 1 \text{ hour}\]where \( \mu \) is the true mean repair time.
03

State the Null and Alternative Hypotheses for Part (b)

For part (b): \[H_0: \mu = 1 \text{ hour} \H_a: \mu < 1 \text{ hour}\]This one-sided alternative hypothesis suggests a mean repair time less than 1 hour.
04

Calculate the Test Statistic

Use the sample mean \( \bar{x} = 0.86 \), sample standard deviation \( s = 0.32 \), and sample size \( n = 30 \). The test statistic for both parts is calculated as follows:\[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\]Substituting the values, we have:\[t = \frac{0.86 - 1}{0.32/\sqrt{30}} \approx -2.543\]
05

Determine Critical Values for Part (a)

At a \(1\%\) significance level for a two-tailed test, and \(29\) degrees of freedom (\(n-1\)), the critical t-values are approximately \(\pm 2.756\). These values are found using a t-distribution table or calculator.
06

Make a Decision for Part (a)

Since the test statistic \(-2.543\) is not less than \(-2.756\) or greater than \(2.756\), we do not reject the null hypothesis for part (a). There is not enough evidence to say the mean differs from 1 hour.
07

Determine Critical Value for Part (b)

At a \(1\%\) significance level for a one-tailed test, the critical t-value is approximately \(-2.462\) for 29 degrees of freedom.
08

Make a Decision for Part (b)

The test statistic \(-2.543\) is less than the critical value \(-2.462\). We reject the null hypothesis for part (b). There is sufficient evidence at the \(1\%\) level to conclude that the mean repair time is less than 1 hour.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis (\( H_0 \)) is a statement that suggests no effect or no difference in the population parameter we are investigating. It acts like a default position that we assume to be true unless enough evidence suggests otherwise. In our automobile repair example, for part (a), the null hypothesis is that the mean repair time \( \mu \) equals 1 hour. This implies that any deviation from the mean time of 1 hour is only due to random sampling error.
  • Null hypothesis: \( H_0: \mu = 1 \text{ hour} \)
Rejecting the null hypothesis suggests there is enough statistical evidence to support the alternative claim that the actual mean repair time differs from 1 hour.
T-Test
The t-test is a statistical procedure used to determine whether there is a significant difference between the mean of a sample and the mean of a population. In simpler terms, it tells us if the observed sample mean can be explained by random chance alone.
The t-test is especially useful when dealing with small sample sizes and unknown population variances, which is the case here. Specifically, we use a one-sample t-test because we are comparing the sample mean to the known population mean of 1 hour.
  • Formula for t-test: \[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]
  • Where \( \bar{x} \) is the sample mean, \( \mu_0 \) is the hypothesized population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size.
Two-Tailed Test
A two-tailed test is used when we are interested in determining if the population parameter is simply different from the hypothesized value, without specifying a direction. It tests for any significant deviation on both sides of the hypothesized parameter. In the context of our exercise, the null hypothesis states that the mean repair time is equal to 1 hour.
  • Set for part (a): \( H_a: \mu eq 1 \text{ hour} \)
The two-tailed test will check if the mean is either greater or less than 1 hour. We use t-distribution critical values to conclude if observed deviations could occur by chance.
One-Tailed Test
A one-tailed test is used when we want to see if a parameter is either specifically greater than or less than a certain value. In our exercise, part (b) asks us to check if the mean time is less than 1 hour. This means we are only interested in detecting deviation on one side of the hypothesis.
  • Setup for part (b): \( H_a: \mu < 1 \text{ hour} \)
The one-tailed test provides a more precise evaluation when the direction of the deviation is assumed, offering more statistical power to detect effects in a specific direction.
Significance Level
The significance level, denoted by \( \alpha \), represents the probability of rejecting the null hypothesis when it is actually true. This is essentially the risk of committing a Type I error. Typical significance levels are 5%, 1%, or 0.1%, representing decreasing tolerance for such errors. In this exercise, we use a 1% significance level, indicating tolerance for fairly low risk.
  • The critical threshold for deciding whether to reject the null hypothesis.
  • Used to determine the critical value for our test statistic.
  • Lower significance level requires stronger evidence to reject the null hypothesis.
In practice, if the p-value derived from the test is less than \( \alpha \), the null hypothesis is rejected, confirming a significant result.

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Most popular questions from this chapter

In a recent year the fuel economy of all passenger vehicles was \(19.8 \mathrm{mpg}\). A trade organization sampled 50 passenger vehicles for fuel economy and obtained a sample mean of 20.1 mpg with standard deviation \(2.45 \mathrm{mpg}\). The sample mean 20.1 exceeds \(19.8,\) but perhaps the increase is only a result of sampling error. a. Perform the relevant test of hypotheses at the \(20 \%\) level of significance using the critical value approach. b. Compute the observed significance of the test. c. Perform the test at the \(20 \%\) level of significance using the \(p\) -value approach. You need not repeat the first three steps, already done in part (a).

State the null and alternative hypotheses for each of the following situations. (That is, identify the correct number \(\mu_{0}\) and write \(H_{0} \mu=\mu_{0}\) and the appropriate analogous expression for \(H_{a}\).) a. The average time workers spent commuting to work in Verona five years ago was 38.2 minutes. The Verona Chamber of Commerce asserts that the average is less now. b. The mean salary for all men in a certain profession is \(\$ 58,291\). A special interest group thinks that the mean salary for women in the same profession is different. c. The accepted figure for the caffeine content of an 8 -ounce cup of coffee is \(133 \mathrm{mg}\). A dietitian believes that the average for coffee served in a local restaurants is higher. d. The average yield per acre for all types of corn in a recent year was 161.9 bushels. An economist believes that the average yield per acre is different this year. e. An industry association asserts that the average age of all self-described fly fishermen is 42.8 years. A sociologist suspects that it is higher.

At every setting a high-speed packing machine delivers a product in amounts that vary from container to container with a normal distribution of standard deviation 0.12 ounce. To compare the amount delivered at the current setting to the desired amount 64.1 ounce, a quality inspector randomly selects five containers and measures the contents of each, obtaining sample mean 63.9 ounces and sample standard deviation 0.10 ounce. Test whether the data provide sufficient evidence, at the \(5 \%\) level of significance, to conclude that the mean of all containers at the current setting is less than 64.1 ounces.

The government of a particular country reports its literacy rate as \(52 \% .\) A nongovernmental organization believes it to be less. The organization takes a random sample of 600 inhabitants and obtains a literacy rate of \(42 \% .\) Perform the relevant test at the \(0.5 \%\) (one-half of \(1 \%)\) level of significance.

A medical laboratory claims that the mean turn-around time for performance of a battery of tests on blood samples is 1.88 business days. The manager of a large medical practice believes that the actual mean is larger. A random sample of 45 blood samples yielded mean 2.09 and sample standard deviation 0.13 day. Perform the relevant test at the \(10 \%\) level of significance, using these data.
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