Questions: One study claimed that 87% 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 143 college students and finds that 117 of them identify themselves as procrastinators. Does this evidence support the professor's claim that fewer than 87% of college students are procrastinators? Use a 0.01 level of significance. Compute the value of the test statistic. Round your answer to two decimal places.

One study claimed that 87% 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 143 college students and finds that 117 of them identify themselves as procrastinators. Does this evidence support the professor's claim that fewer than 87% of college students are procrastinators? Use a 0.01 level of significance. Compute the value of the test statistic. Round your answer to two decimal places.

Transcript text: One study claimed that $87 \%$ 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 143 college students and finds that 117 of them identify themselves as procrastinators. Does this evidence support the professor's claim that fewer than $87 \%$ of college students are procrastinators? Use a 0.01 level of significance. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.

Solution

Solution Steps

Step 1: State the Hypotheses

We want to test the professor's claim that fewer than $ 87\% $ of college students identify themselves as procrastinators. The hypotheses are defined as follows:

Null Hypothesis ($ H_0 $): $ p = 0.87 $
Alternative Hypothesis ($ H_a $): $ p < 0.87 $

Step 2: Calculate the Test Statistic

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

\[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]

Where:

$ \hat{p} = \frac{117}{143} \approx 0.8196 $ (sample proportion)
$ p_0 = 0.87 $ (hypothesized population proportion)
$ n = 143 $ (sample size)

Substituting the values:

\[ Z = \frac{0.8196 - 0.87}{\sqrt{\frac{0.87(1 - 0.87)}{143}}} \approx -1.84 \]

Step 3: Calculate the P-value

The P-value associated with the test statistic $ Z = -1.84 $ is calculated to determine the probability of observing a sample proportion as extreme as $ \hat{p} $ under the null hypothesis. The P-value is found to be:

\[ \text{P-value} = 0.03 \]

Step 4: Conclusion

At a significance level of $ \alpha = 0.01 $, we compare the P-value to $ \alpha $:

Since $ 0.03 > 0.01 $, we fail to reject the null hypothesis.

This indicates that there is not enough evidence to support the professor's claim that fewer than $ 87\% $ of college students identify themselves as procrastinators.