Math Problem Statement

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.

Solution

To compute the value of the test statistic, we follow these steps:

Step 1: Define the hypotheses

  • Null Hypothesis (H₀): p=0.93p = 0.93 (The proportion of college students who identify as procrastinators is 93%.)
  • Alternative Hypothesis (H₁): p<0.93p < 0.93 (The proportion of college students who identify as procrastinators is less than 93%.)

Step 2: Compute the sample proportion

The sample proportion p^\hat{p} is calculated as:

p^=Number of students identifying as procrastinatorsTotal number of students in the sample=174194\hat{p} = \frac{\text{Number of students identifying as procrastinators}}{\text{Total number of students in the sample}} = \frac{174}{194}

p^=0.8969\hat{p} = 0.8969

Step 3: Compute the standard error

The standard error SESE is given by:

SE=p(1p)n=0.93(10.93)194SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.93(1-0.93)}{194}}

SE=0.93×0.07194=0.06511940.0003350.0183SE = \sqrt{\frac{0.93 \times 0.07}{194}} = \sqrt{\frac{0.0651}{194}} \approx \sqrt{0.000335} \approx 0.0183

Step 4: Compute the test statistic

The test statistic zz is computed using the formula:

z=p^pSE=0.89690.930.01830.03310.01831.81z = \frac{\hat{p} - p}{SE} = \frac{0.8969 - 0.93}{0.0183} \approx \frac{-0.0331}{0.0183} \approx -1.81

Thus, the value of the test statistic is approximately -1.81.

Would you like details on how to proceed with this hypothesis test, or have any other questions?

Here are five related questions you might consider:

  1. How do you find the p-value corresponding to this test statistic?
  2. What does the significance level of 0.02 mean in this context?
  3. How do you determine whether to reject the null hypothesis?
  4. What assumptions must be met for the use of a z-test in this scenario?
  5. How would the conclusion change if the significance level were different?

Tip: Always ensure the sample size is large enough when performing hypothesis tests, as smaller samples can lead to less reliable results.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Proportion Estimation
Standard Error

Formulas

Standard Error formula: SE = sqrt((p * (1 - p)) / n)

Theorems

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Suitable Grade Level

Advanced College