Question

According to MessageLabs Ltd., 80% of all email sent in July 2010 was spam. A system manager at a large corporation believes that the percentage at his company may be greater than 80%, and he wants to do a hypothesis test to see if there is evidence to support his belief. He examines a random sample of 500 emails received at an email server and finds that 420 of the messages are spam.

1. What is the parameter for this hypothesis test (choose one)?

?μ ?¯y¯ ?p ?̂ p^

2. What is the null hypothesis? ?0H0: ____

   What is the alternative hypothesis? ??HA: ____

3. This test is (Choose one):

   One-sided
   Two-sided

4. What proportion of emails were spam in the sample? Round to two decimal places.

5. Calculate the test statistic. Round to two decimal places.

1. Your test statistic is a z-score. Draw a line at that z-score on this standard normal distribution, and shade the area corresponding for the p-value for the hypothesis test.

2. To calculate the p-value, you would need to look up the z-score on a z-table (like the one provided in class and posted on eCampus). For convenience, you are being provided with this fact from the z-table: P(Z<2.24) = .9875. What is the p-value for the hypothesis test?

3. Compare your p-value to the significance level α=0.05. What is your decision based on the p-value? (Choose one letter.)

   A. Since the p-value <α, reject the null hypothesis (H_0).

   B. Since the p-value <α, do not reject the null hypothesis (H_0).

   C. Since the p-value >α, reject the null hypothesis (H_0).

   D. Since the p-value >α, do not reject the null hypothesis (H_0).

4. Based on your decision, what is the conclusion? (Choose one letter.)

   A. At the 0.05 level of significance, we do have sufficient evidence to say that the true proportion of spam email at the company is greater than 0.80. There is evidence to support the manager’s belief.

   B. At the 0.05 level of significance, we do not have sufficient evidence to say that the true proportion of spam email at the company is greater than 0.80. There is not enough evidence to support the manager’s belief.

5. Suppose your alternative hypothesis had been H_A:p≠0.80. What would have been the p-value for that hypothesis test?

Answer 1

p

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p = 0.8
Alternative Hypothesis, Ha: p > 0.8

one sided

sample proportion = 420/500 = 0.84

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.84 - 0.8)/sqrt(0.8*(1-0.8)/500)
z = 2.24

P-value Approach
P-value = 0.0125

A. Since the p-value <α, reject the null hypothesis (H_0).

A. At the 0.05 level of significance, we do have sufficient evidence to say that the true proportion of spam email at the company is greater than 0.80. There is evidence to support the manager’s belief

P-value = 0.0251 for 2 tailed test