Problem 25 - Basic Statistics for Business and Economics with Student

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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 \( \alpha = 0.05 \).

Step 3: Calculate Standard Error

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

Step 4: Compute Test Statistic

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

Step 5: 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.

Step 6: 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.

Key Concepts

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.