One study claimed that $93 \%$ 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 194 college students and finds that 174 of them identify themselves as procrastinators. Does this evidence support the professor's claim that fewer than $93 \%$ 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.

Step 1 ：The problem is asking for a hypothesis test for a proportion. The null hypothesis is that the proportion of college students who identify as procrastinators is 0.93, and the alternative hypothesis is that the proportion is less than 0.93.

Step 2 ：The test statistic for a hypothesis test for a proportion is a z-score, which is calculated as \((\hat{p} - p_0) / \sqrt{(p_0 * (1 - p_0)) / n}\), where \(\hat{p}\) is the sample proportion, \(p_0\) is the hypothesized population proportion, and n is the sample size.

Step 3 ：In this case, \(\hat{p} = 174 / 194\), \(p_0 = 0.93\), and n = 194.

Step 4 ：Substituting these values into the formula gives a z-score of -1.81. This is the number of standard deviations that the sample proportion is below the hypothesized population proportion.

Step 5 ：Final Answer: The value of the test statistic is \(\boxed{-1.81}\).