A direct mail appeal for contributions from a university’s alumni and supporters is considered to be too costly if less than 22%
 of the alumni and supporters provide monetary contributions. To determine if a direct mail appeal is cost effective, the fundraising director sends the direct mail brochures to a simple random sample of 332
 people on the alumni and supporters mailing lists. They receive monetary contributions from 62
 people. Does this evidence demonstrate that the direct mail campaign is not cost effective? Use a 0.05
 level of significance.
Step 2 of 3 :  Compute the value of the test statistic. Round your answer to two decimal places.


We can use a **hypothesis test for a population proportion** to determine if the direct mail campaign is cost-effective. Here's the step-by-step approach:

### Step 1: State the hypotheses

We are testing whether the proportion of contributions (\(p\)) is less than 22% (i.e., 0.22), which would indicate the campaign is not cost-effective. The hypotheses are:

- **Null hypothesis (\(H_0\))**: \( p = 0.22 \)
- **Alternative hypothesis (\(H_a\))**: \( p < 0.22 \)

This is a **left-tailed test** because we are checking if the actual proportion is less than the expected proportion.

### Step 2: Compute the test statistic

We will use the formula for the **z-test statistic** for a population proportion:

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

Where:
- \(\hat{p}\) is the sample proportion of contributions.
- \(p_0\) is the population proportion (22%, or 0.22).
- \(n\) is the sample size.

Now, plug in the values:

- Sample size \(n = 332\)
- Number of contributions = 62
- \(\hat{p} = \frac{62}{332} \approx 0.1867\)
- \(p_0 = 0.22\)

Substitute into the z-test formula:

\[
z = \frac{0.1867 - 0.22}{\sqrt{\frac{0.22(1 - 0.22)}{332}}}
\]

First, calculate the denominator:

\[
\sqrt{\frac{0.22 \times (1 - 0.22)}{332}} = \sqrt{\frac{0.22 \times 0.78}{332}} \approx \sqrt{0.000516} \approx 0.0227
\]

Now, calculate the z-statistic:

\[
z = \frac{0.1867 - 0.22}{0.0227} = \frac{-0.0333}{0.0227} \approx -1.47
\]

### Final answer:

The value of the test statistic is approximately **-1.47**.

Would you like more details or have any questions?

Here are 5 related questions to consider:
1. How is the critical value determined for a one-tailed test?
2. What is the p-value corresponding to the test statistic?
3. How do we interpret the p-value in the context of this problem?
4. What assumptions must be satisfied for the z-test to be valid?
5. How does changing the sample size affect the test statistic?

**Tip**: When conducting hypothesis tests, always check if the sample size is large enough to justify the use of a normal approximation for proportions.

A direct mail appeal for contributions from a university’s alumni and supporters is considered to be too costly if less than 22% of the alumni and supporters provide monetary contributions. To determine if a direct mail appeal is cost-effective, the fundraising director sends the direct mail brochures to a simple random sample of 332 people on the alumni and supporters mailing lists. They receive monetary contributions from 62 people. Does this evidence demonstrate that the direct mail campaign is not cost-effective? Use a 0.05 level of significance. Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.

Analyze a direct mail campaign's cost-effectiveness by conducting a hypothesis test for population proportion. This problem uses a random sample of 332 alumni and supporters, where 62 contributed. Calculate the z-test statistic for a significance level of 0.05 to determine if the contribution rate is less than 22%.

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