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.