A researcher reported that 71.8% of all email sent in a recent month was spam. A system manager at a large corporation believes that the percentage at his company may be 69%. He examines a random sample of 500 emails received at an email server, and finds that 365 of the messages are spam. Can you conclude that greater than 69% of emails are spam? Use both a=0.01 and a=0.05 levels of significance and the critical value method with the table.

Question

Hv0: [?]
Hv1: [?]
two, right or left tailed test?

Answer 1

To determine whether greater than 69% of emails are spam at the corporation, we will set up the hypothesis test as follows:

Hypotheses

• Null Hypothesis (H0): \( p \leq 0.69 \) (The proportion of spam emails is 69% or less)
• Alternative Hypothesis (H1): \( p > 0.69 \) (The proportion of spam emails is greater than 69%)

This is a right-tailed test since we are testing if the proportion is greater than a certain value.

Sample Data

• Sample size (\( n \)) = 500
• Number of spam emails in the sample (\( X \)) = 365
• Sample proportion (\( \hat{p} \)) = \( \frac{X}{n} = \frac{365}{500} = 0.73 \)

Test Statistic

We can use the following formula for the z-test statistic when testing proportions:

\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]

where:

• \( \hat{p} \) is the sample proportion (0.73),
• \( p_0 \) is the hypothesized population proportion (0.69),
• \( n \) is the sample size (500).

Substituting the values we know:

\[ z = \frac{0.73 - 0.69}{\sqrt{\frac{0.69(1 - 0.69)}{500}}} \]

Calculating the standard error:

\[ \sqrt{\frac{0.69(0.31)}{500}} = \sqrt{\frac{0.2139}{500}} = \sqrt{0.0004278} \approx 0.0207 \]

Calculating the z-test statistic:

\[ z = \frac{0.73 - 0.69}{0.0207} \approx \frac{0.04}{0.0207} \approx 1.932 \]

Critical Values

Now we need to determine the critical value for the significance levels \( \alpha = 0.01 \) and \( \alpha = 0.05 \).

1. For \( \alpha = 0.05 \):

   • The critical value (from z-table) corresponding to 0.05 in the upper tail is approximately \( z_{0.05} = 1.645 \).

2. For \( \alpha = 0.01 \):

   • The critical value for \( \alpha = 0.01 \) in the upper tail is approximately \( z_{0.01} = 2.33 \).

Conclusion

Now we can compare our calculated z-value against the critical values.

• For \( \alpha = 0.05 \):

  • If \( z > 1.645 \): reject \( H_0 \)

• For \( \alpha = 0.01 \):

  • If \( z > 2.33 \): reject \( H_0 \)

From our calculation, \( z \approx 1.932 \).

• At \( \alpha = 0.05 \): Since \( 1.932 > 1.645\), we reject \( H_0 \).
• At \( \alpha = 0.01 \): Since \( 1.932 < 2.33\), we fail to reject \( H_0 \).

Summary

• At \( \alpha = 0.05\): We conclude that there is enough evidence to suggest that the proportion of spam emails is greater than 69%.
• At \( \alpha = 0.01\): We do not have enough evidence to make the same conclusion.

These results suggest that while the evidence is somewhat strong at the 0.05 level, it isn't conclusive at the 0.01 level.