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Message mania! A professional employee in a large corporation receives an average of \(\mu=41.7\) e-mails per day. Most of these e-mails are from other employees in the company. Because of the large number of e-mails, employees find themselves distracted and are unable to concentrate when they return to their tasks (Reference: The Wall Street Journal). In an effort to reduce distraction caused by such interruptions, one company established a priority list that all employees were to use before sending an e-mail. One month after the new priority list was put into place, a random sample of 45 employees showed that they were receiving an average of \(\bar{x}=36.2 \mathrm{e}\) -mails per day. The computer server through which the e-mails are routed showed that \(\sigma=18.5\). Has the new policy had any effect? Use a \(5 \%\) level of significance to test the claim that there has been a change (either way) in the average number of e-mails received per day per employee.

Step by step solution

01

Define the Hypotheses

We need to set up the null hypothesis and alternative hypothesis. The null hypothesis, denoted as \(H_0\), states that there is no change in the average number of emails received (\(\mu = 41.7\)). The alternative hypothesis, denoted as \(H_1\), states that the number of emails has changed (\(\mu eq 41.7\)).

02

Determine the Test Statistic

Since we know the population standard deviation (\(\sigma = 18.5\)) and the sample size is \(n = 45\), we will use a Z-test for the mean. The formula for the test statistic is \( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \). Substituting the values, we have \( z = \frac{36.2 - 41.7}{\frac{18.5}{\sqrt{45}}} \).

03

Calculate the Test Statistic

Substituting the values into the Z-test formula gives us: \[ z = \frac{36.2 - 41.7}{\frac{18.5}{\sqrt{45}}} = \frac{-5.5}{2.757} \approx -1.994 \].

04

Determine the Critical Value(s)

For a two-tailed Z-test at a significance level of \(0.05\), the critical Z-values are \(-1.96\) and \(1.96\). These are the points beyond which the null hypothesis will be rejected.

05

Compare Test Statistic with Critical Value(s)

The calculated test statistic \(-1.994\) is less than the lower critical value \(-1.96\). Therefore, it falls into the rejection region of the null hypothesis.

06

Draw a Conclusion

Since the test statistic falls outside the range of \(-1.96\) to \(1.96\), we reject the null hypothesis \(H_0\). This provides evidence at the 5% significance level that the email policy change has reduced the amount of emails the employees receive.

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Key Concepts

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

Z-test

The Z-test is a statistical method used to determine whether there is a significant difference between the means of two groups when the variances are known and the sample size is large. In this context, the Z-test is employed to check if the priority list established by the corporation has effectively changed the average number of emails employees receive each day.

The formula used for the Z-test is \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]where

• \(\bar{x}\) is the sample mean,
• \(\mu\) is the population mean,
• \(\sigma\) is the population standard deviation, and
• \(n\) is the sample size.

The sample mean in the exercise is 36.2, while the population mean is 41.7, and the standard deviation is known as 18.5. This test helps us understand whether the observed change is statistically significant or due to random chance.

null hypothesis

The null hypothesis (\(H_0\)) is a statement of no effect or no difference, serving as a starting point for statistical testing. It presumes that any kind of difference or significance you see in your data is due to random chance, rather than a real effect.

In the context of our email example, the null hypothesis posits that the new priority list has had no impact on the average number of emails received per employee per day. Mathematically, it is expressed as \[H_0: \mu = 41.7\]This implies that the mean number of emails remains the same as before the policy implementation. The null hypothesis is what we attempt to either reject or fail to reject based on our statistical test.

alternative hypothesis

In contrast to the null hypothesis, the alternative hypothesis (\(H_1\)) suggests that there is an effect or a difference. It is what a researcher aims to prove through evidence and statistical testing.

For our exercise, the alternative hypothesis suggests that the change in the email policy has indeed affected the average number of emails received per day. It is mathematically represented as:\[H_1: \mu eq 41.7\]This hypothesis indicates that there is a difference in the average emails received due to the policy, but it does not specify the direction of the change. It could be either an increase or a decrease in the number of emails.

significance level

The significance level, often denoted by \(\alpha\), is a threshold chosen by the researcher to judge whether the results observed in a study are statistically significant. It represents the probability of rejecting the null hypothesis when it is actually true, known as a Type I error.

In our email policy case, the significance level was set at 5% or 0.05. This means there is a 5% chance of concluding that the policy change affected the email count when, in reality, it did not. It is common to use \(\alpha = 0.05\) in many scientific studies as a standard signifying reasonable confidence in the results.

At this level, a two-tailed test examines whether the Z-value falls below -1.96 or above 1.96, which are the critical values for a 5% significance level. Since our test statistic is -1.994, it lies in the rejection region, indicating that the policy change might have significantly affected the email traffic at this confidence level.