A professional employee in a large corporation receives an average of = 45 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. 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 50 employees showed that they were receiving an average of = 39.1 e-mails per day. The computer server through which the e-mails are routed showed that = 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. Find (or estimate) the P-value.

To test the claim that there has been a change in the average number of emails received per day per employee, we can use a hypothesis test.

Let's set up our hypotheses:

Null Hypothesis (H0): The new policy has no effect, the average number of emails received per day per employee is equal to 45. µ = 45.

Alternative Hypothesis (Ha): The new policy has an effect, the average number of emails received per day per employee is not equal to 45. µ ≠ 45.

Given that we have a sample size (n) of 50, the sample mean (x̄) of 39.1, and the population standard deviation (σ) of 18.5, we can use a two-sample t-test to compare the sample mean to the population mean.

The formula to calculate the t-score is:

t = (x̄ - µ) / (σ / √n)

Plugging in the values, we get:

t = (39.1 - 45) / (18.5 / √50)

Simplifying, t = -5.9 / (18.5 / √50)

Next, we calculate the degrees of freedom (df), which is n - 1:
df = 50 - 1 = 49

Using the t-distribution with 49 degrees of freedom, we can find the critical t-value for a 5% level of significance (α = 0.05). In a two-tailed test, we divide α by 2, so α/2 = 0.05/2 = 0.025.

Looking up the critical t-value using a t-table or a t-distribution calculator with α/2 = 0.025 and df = 49, we find that the critical t-value is approximately ±2.0096.

Now we can compare the calculated t-score to the critical t-value to determine if we reject or fail to reject the null hypothesis.

If the calculated t-score falls outside the range of ±2.0096, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

To calculate the p-value, we need to find the area under the t-distribution curve that corresponds to the calculated t-score. The p-value represents the probability of obtaining a sample mean as extreme or more extreme than the observed value (assuming the null hypothesis is true).

Using the t-distribution table or a t-distribution calculator, we find the p-value associated with the calculated t-score.

If the p-value is less than the significance level (α = 0.05), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Note: The t-score and p-value calculations shown here assume that the population standard deviation (σ) is known. If the population standard deviation is unknown and estimated from the sample, we would use a t-test with n-1 degrees of freedom and estimated standard deviation (s).

By calculating the t-score, comparing it to the critical t-value, and determining the p-value, we can evaluate whether the new policy has had any effect on the average number of emails received per day per employee.