This is a hypothesis test for a population proportion. Here's the step-by-step process to compute the test statistic.

Step 1: Identify the null and alternative hypotheses

The null hypothesis ($H_0$) and alternative hypothesis ($H_a$) are:

• $H_0$: $p = 0.88$ (the population proportion of procrastinators is 88%)
• $H_a$: $p < 0.88$ (the professor believes the proportion is less than 88%)

Step 2: Calculate the test statistic

The test statistic for a population proportion is calculated using the formula:

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

where:

• $\hat{p}$ = sample proportion
• $p_0$ = population proportion under the null hypothesis (0.88)
• $n$ = sample size (272)

First, calculate the sample proportion $\hat{p}$: $\hat{p} = \frac{231}{272} \approx 0.8493$

Now, plug in the values into the formula: $z = \frac{0.8493 - 0.88}{\sqrt{\frac{0.88 \times (1 - 0.88)}{272}}}$ $z = \frac{-0.0307}{\sqrt{\frac{0.88 \times 0.12}{272}}}$ $z = \frac{-0.0307}{\sqrt{0.0003882353}}$ $z = \frac{-0.0307}{0.0197}$ $z \approx -1.56$

Test Statistic:

The value of the test statistic is approximately -1.56.

Would you like details on how to interpret this result?

Here are 5 related questions:

1. What are the critical values for a 0.02 significance level in a one-tailed test?
2. How do we compute the p-value based on the z-statistic?
3. What is the conclusion based on the test statistic?
4. How does sample size affect the accuracy of the test?
5. Can the test be applied to other proportions, like students in different majors?

Tip: When performing hypothesis tests, always check assumptions like sample size and the normality of the sampling distribution!