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91Ó°ÊÓ

Fry Brothers Heating and Air Conditioning, Inc. employs Larry Clark and George Murnen to make service calls to repair furnaces and air conditioning units in homes. Tom Fry, the owner, would like to know whether there is a difference in the mean number of service calls they make per day. Assume the population standard deviation for Larry Clark is 1.05 calls per day and 1.23 calls per day for George Murnen. A random sample of 40 days last year showed that Larry Clark made an average of 4.77 calls per day. For a sample of 50 days George Murnen made an average of 5.02 calls per day. At the .05 significance level, is there a difference in the mean number of calls per day between the two employees? What is the \(p\) -value?

Short Answer

Expert verified
No significant difference in mean daily calls; \( p \)-value = 0.364.

Step by step solution

01

Define Hypotheses

We need to determine if there is a statistical difference in the mean number of service calls made per day by Larry Clark and George Murnen. The null hypothesis \( H_0 \) states that the means are equal: \( \mu_1 = \mu_2 \). The alternative hypothesis \( H_a \) states that the means are not equal: \( \mu_1 eq \mu_2 \). This will be a two-tailed test.
02

Identify Variables

We know the samples and population standard deviations: \( \sigma_1 = 1.05 \), \( \sigma_2 = 1.23 \), sample means \( \bar{x}_1 = 4.77 \), \( \bar{x}_2 = 5.02 \), and sample sizes \( n_1 = 40 \), \( n_2 = 50 \). Alpha level \( \alpha = 0.05 \).
03

Calculate Standard Error

The standard error for the difference in sample means is calculated using the formula: \[ SE = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} = \sqrt{\frac{1.05^2}{40} + \frac{1.23^2}{50}} \approx 0.276. \]
04

Compute Test Statistic

The test statistic \( z \) is found by: \[ z = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{4.77 - 5.02}{0.276} \approx -0.91. \]
05

Determine P-Value

Using standard normal distribution tables, or calculator, find the \( p \)-value for \( z = -0.91 \). Since it is a two-tailed test, we must consider both ends of the distribution which gives a \( p \)-value of approximately 0.364.
06

Decision Rule and Conclusion

If \( p \)-value \( < \alpha \), reject \( H_0 \). Here, \( p = 0.364 \) and \( \alpha = 0.05 \). Since 0.364 is greater than 0.05, we fail to reject \( H_0 \). Thus, there is not enough evidence to suggest a difference in the mean number of service calls per day between Larry Clark and George Murnen.

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Key Concepts

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

Significance Level
The significance level, often denoted by the symbol \( \alpha \), is a threshold used in hypothesis testing to determine whether the null hypothesis can be rejected. Think of it as the probability of incorrectly rejecting the null hypothesis when it is, in fact, true. This is also known as a Type I error. The significance level is set before conducting a test and is usually 0.05 or 5%, meaning there is a 5% risk of concluding that a difference exists when there is no actual difference.

In our scenario, Tom Fry is trying to determine if there is a difference in the number of service calls made by Larry Clark and George Murnen. He uses a significance level of 0.05. This means Tom is willing to accept a 5% chance of wrongly stating that there is a difference in performance when there isn’t one.

Setting the right significance level is crucial. A lower \( \alpha \) reduces the chance of a Type I error, but increases the risk of a Type II error (failing to reject the null hypothesis when it is false). Therefore, an appropriate balance must be struck based on the context and importance of the decision.
P-value
The \( p \)-value is a statistic that helps determine the significance of results from a hypothesis test. It measures the probability of obtaining a test statistic as extreme as, or more extreme than, the observed value, assuming the null hypothesis is true.

The \( p \)-value allows us to gauge the strength of the evidence against the null hypothesis. A smaller \( p \)-value suggests stronger evidence against the null hypothesis. If the \( p \)-value is lower than the predetermined significance level (\( \alpha \)), it implies that the observed result is unlikely under the null hypothesis, and we reject \( H_0 \).

In the exercise example, the test statistic yields a \( p \)-value of approximately 0.364. Given that this \( p \)-value is significantly higher than the significance level of 0.05, it indicates the evidence is not strong enough to reject the null hypothesis. Hence, there is no statistical support for a difference in the average number of calls made by Larry Clark and George Murnen.
Two-tailed Test
A two-tailed test is a statistical test used to determine if there are differences in two directions, both higher and lower than the expected outcome. It checks for any significant deviation in both directions from a stated hypothesis.

In hypothesis testing, the two-tailed test is crucial when we want to assess if the parameter of interest is significantly different from the null hypothesis value, either greater or less. This is indicated when we're specifically interested in whether an effect can occur in either direction.

In Tom Fry's scenario, since we want to know if the mean number of calls is unequal (either Larry makes more or George does), a two-tailed test is appropriate. The formulation of the hypotheses was \( H_0: \mu_1 = \mu_2 \) (mean difference = 0) and \( H_a: \mu_1 eq \mu_2 \). This setup aligns with the two-tailed approach because we are looking at both possibilities: Larry could make more or George could make more calls.

This type of test requires considering the possibility of deviations on both ends of the distribution. As seen, the two-tailed nature involves the \( p \)-value calculated covering both extremes, leading to a conclusion that there isn't enough statistical evidence to declare a difference.

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Most popular questions from this chapter

A coffee manufacturer is interested in whether the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers. Assume the population standard deviation for those drinking regular coffee is 1.20 cups per day and 1.36 cups per day for those drinking decaffeinated coffee. A random sample of 50 regular-coffee drinkers showed a mean of 4.35 cups per day. A sample of 40 decaffeinated-coffee drinkers showed a mean of 5.84 cups per day. Use the .01 significance level. Compute the \(p\) -value.

A financial analyst wants to compare the turnover rates, in percent, for shares of oil-related stocks versus other stocks, such as GE and IBM. She selected 32 oil-related stocks and 49 other stocks. The mean turnover rate of oil-related stocks is 31.4 percent and the population standard deviation 5.1 percent. For the other stocks, the mean rate was computed to be 34.9 percent and the population standard deviation 6.7 percent. Is there a significant difference in the turnover rates of the two types of stock? Use the .01 significance level.

The null and alternate hypotheses are: $$ \begin{array}{l} H_{0}: \pi_{1} \leq \pi_{2} \\ H_{1}: \pi_{1}>\pi_{2} \end{array} $$ A sample of 100 observations from the first population indicated that \(X_{1}\) is \(70 .\) A sample of 150 observations from the second population revealed \(X_{2}\) to be \(90 .\) Use the .05 significance level to test the hypothesis. a. State the decision rule. b. Compute the pooled proportion. c. Compute the value of the test statistic. d. What is your decision regarding the null hypothesis? e. The null and alternate hypotheses are: $$ \begin{array}{l} H_{0}: \pi_{1}=\pi_{2} \\ H_{1}: \pi_{1} \neq \pi_{2} \end{array} $$

Each month the National Association of Purchasing Managers publishes the NAPM index. One of the questions asked on the survey to purchasing agents is: Do you think the economy is expanding? Last month, of the 300 responses, 160 answered yes to the question. This month, 170 of the 290 responses indicated they felt the economy was expanding. At the .05 significance level, can we conclude that a larger proportion of the agents believe the economy is expanding this month?

Mary Jo Fitzpatrick is the vice president for Nursing Services at St. Luke's Memorial Hospital. Recently she noticed in the job postings for nurses that those that are unionized seem to offer higher wages. She decided to investigate and gathered the following information. $$ \begin{array}{|lccc|} \hline & & \text { Population } & \\ \text { Group } & \text { Mean Wage } & \text { Standard Deviation } & \text { Sample Size } \\ \hline \text { Union } & \$ 20.75 & \$ 2.25 & 40 \\ \text { Nonunion } & \$ 19.80 & \$ 1.90 & 45 \\ \hline \end{array} $$ Would it be reasonable for her to conclude that union nurses earn more? Use the .02 significance level. What is the \(p\) -value?

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