Question

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.
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.