To test whether the claim of 85% of college students being procrastinators is too high, we perform a hypothesis test for proportions. In this case, we are testing if fewer than 85% of college students identify as procrastinators.

Step 1: Set up the hypotheses

The null hypothesis ($H_0$) and the alternative hypothesis ($H_1$) are as follows:

• Null hypothesis ($H_0$): $p = 0.85$ (the true proportion of procrastinators is 85%).
• Alternative hypothesis ($H_1$): $p < 0.85$ (the true proportion of procrastinators is less than 85%).

Step 2: Calculate the test statistic

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

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

Where:

• $\hat{p}$ is the sample proportion,
• $p_0$ is the population proportion (in this case, 0.85),
• $n$ is the sample size.

Given:

• Sample size ($n$) = 164,
• Number of students who identify as procrastinators = 132,
• Sample proportion ($\hat{p}$) = $\frac{132}{164}$,
• Population proportion ($p_0$) = 0.85.

Step 3: Compute the sample proportion

$\hat{p} = \frac{132}{164} = 0.8049$

Step 4: Plug the values into the z-test formula

$z = \frac{0.8049 - 0.85}{\sqrt{\frac{0.85(1 - 0.85)}{164}}}$

First, calculate the denominator:

$\sqrt{\frac{0.85(1 - 0.85)}{164}} = \sqrt{\frac{0.85 \times 0.15}{164}} = \sqrt{\frac{0.1275}{164}} = \sqrt{0.000776} \approx 0.02786$

Now, compute the z-value:

$z = \frac{0.8049 - 0.85}{0.02786} = \frac{-0.0451}{0.02786} \approx -1.62$

Step 5: Conclusion

The value of the test statistic is -1.62.

Would you like to proceed with the next steps to make a conclusion about the hypothesis test or need further clarification?