One study claimed that 88% of college students identify themselves as procrastinators. A professor believes that the claim regarding college students is too high. The professor conducts a simple random sample of 272 college students and finds that 231 of them identify themselves as procrastinators. Does this evidence support the professor’s claim that fewer than 88% of college students are procrastinators? Use a 0.02 level of significance.
Step 2 of 3 :  Compute the value of the test statistic. Round your answer to two decimal places.


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!

One study claimed that 88% of college students identify themselves as procrastinators. A professor believes that the claim regarding college students is too high. The professor conducts a simple random sample of 272 college students and finds that 231 of them identify themselves as procrastinators. Does this evidence support the professor’s claim that fewer than 88% of college students are procrastinators? Use a 0.02 level of significance. Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.

Learn how to perform hypothesis testing for a population proportion with this example, where a professor questions if fewer than 88% of college students identify as procrastinators. This problem involves computing the z-statistic based on a sample of 272 students, with 231 identifying as procrastinators, at a 0.02 significance level.

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