Let's solve the problem step-by-step: --- **Step 2: Compute the value of the test statistic.** ### **Given:** - Claimed population proportion: \( p_0 = 0.86 \) - Sample size: \( n = 166 \) - Number of successes (students identifying as procrastinators): \( x = 136 \) - Sample proportion: \( \hat{p} = \frac{136}{166} \) - Significance level: \( \alpha = 0.02 \) - We are testing if the true proportion is **less than** 0.86: \( H_1: p < 0.86 \) (left-tailed test) ### **Test Statistic Formula:** \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \] ### **Step 1: Calculate \(\hat{p}\):** \[ \hat{p} = \frac{136}{166} \approx 0.8193 \] ### **Step 2: Calculate the standard error:** \[ SE = \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.86 \times 0.14}{166}} \] Calculate inside the square root: - \( 0.86 \times 0.14 = 0.1204 \) - \( \frac{0.1204}{166} \approx 0.000725 \) - \( \sqrt{0.000725} \approx 0.0269 \) ### **Step 3: Compute \( z \):** \[ z = \frac{0.8193 - 0.86}{0.0269} \] \[ z = \frac{-0.0407}{0.0269} \approx -1.51 \] --- ### **Final Answer:** \[ \boxed{-1.51} \] This is the value of the test statistic, rounded to two decimal places.

**Step 1: Introduction** We are testing whether fewer than 86% of college students are procrastinators. The hypotheses are: - Null hypothesis (\(H_0\)): \( p = 0.86 \) - Alternative hypothesis (\(H_1\)): \( p < 0.86 \) Given data: - Sample size \( n = 166 \) - Number of students identifying as procrastinators \( x = 136 \) - Sample proportion \( \hat{p} = \frac{136}{166} \) The significance level is \( \alpha = 0.02 \). --- **Step 2: Compute the test statistic** Calculate the sample proportion: \[ \hat{p} = \frac{136}{166} \approx 0.8193 \] Calculate the standard error under the null hypothesis: \[ SE = \sqrt{\frac{p_0 (1 - p_0)}{n}} = \sqrt{\frac{0.86 \times 0.14}{166}} \] Compute numerator: \[ 0.86 \times 0.14 = 0.1204 \] Divide by \( n \): \[ \frac{0.1204}{166} \approx 0.000725 \] Take the square root: \[ SE \approx \sqrt{0.000725} \approx 0.0269 \] Compute the z-statistic: \[ z = \frac{\hat{p} - p_0}{SE} = \frac{0.8193 - 0.86}{0.0269} = \frac{-0.0407}{0.0269} \approx -1.51 \] --- **Final Answer:** \[ \boxed{-1.51} \] The value of the test statistic, rounded to two decimal places, is **-1.51**.