Math Problem Statement

A survey conducted five years ago by the health center at a university showed that 18% of the students smoked at the time. This year a new survey was conducted on a random sample of 200 students from this university, and it was found that 44 of them smoke. Do these data provide convincing evidence to suggest that the percentage of students who smoke has changed over the last five years? Limit the type I error rate to 5%.

Calculate the standardized test statistic that would be used during the hypothesis test. Round your answer to 3 decimal places.

Solution

To determine if the percentage of students who smoke has changed over the last five years, we need to conduct a hypothesis test for the population proportion. This involves comparing the observed sample proportion with the historical population proportion (from five years ago) and calculating a standardized test statistic. Here's the step-by-step process:

Step 1: Define the Hypotheses

  • Null Hypothesis (H₀): The percentage of students who smoke has not changed. H0:p=0.18H_0: p = 0.18
  • Alternative Hypothesis (H₁): The percentage of students who smoke has changed. H1:p0.18H_1: p \neq 0.18 This will be a two-tailed test since we're checking for any change (increase or decrease) in the proportion of smokers.

Step 2: Calculate the Sample Proportion

From the sample of 200 students, 44 of them smoke. The sample proportion (p^\hat{p}) is: p^=44200=0.22\hat{p} = \frac{44}{200} = 0.22

Step 3: Calculate the Standard Error (SE)

The standard error for the proportion is calculated using the formula: SE=p(1p)nSE = \sqrt{\frac{p(1 - p)}{n}} where p=0.18p = 0.18 (the population proportion from five years ago) and n=200n = 200 (the sample size).

SE=0.18(10.18)200=0.18×0.82200=0.1476200=0.0007380.0272SE = \sqrt{\frac{0.18(1 - 0.18)}{200}} = \sqrt{\frac{0.18 \times 0.82}{200}} = \sqrt{\frac{0.1476}{200}} = \sqrt{0.000738} \approx 0.0272

Step 4: Calculate the Test Statistic (z)

The standardized test statistic (z) is calculated as: z=p^pSEz = \frac{\hat{p} - p}{SE} Substituting the values: z=0.220.180.0272=0.040.02721.471z = \frac{0.22 - 0.18}{0.0272} = \frac{0.04}{0.0272} \approx 1.471

Conclusion:

The standardized test statistic is approximately 1.471 (rounded to 3 decimal places).

Would you like further details or clarifications on any step?


Here are 5 related questions to explore:

  1. What is the p-value associated with a z-score of 1.471 in a two-tailed test?
  2. How would the conclusion of the hypothesis test change if the sample size were larger?
  3. What does a Type I error mean in the context of this hypothesis test?
  4. What is the critical value for a 5% significance level in a two-tailed z-test?
  5. How does changing the significance level affect the outcome of the hypothesis test?

Tip: The standard error decreases as the sample size increases, making the test more sensitive to small differences.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Proportions
Z-Test

Formulas

Sample Proportion: \( \hat{p} = \frac{x}{n} \)
Standard Error for Proportion: \( SE = \sqrt{\frac{p(1 - p)}{n}} \)
Test Statistic: \( z = \frac{\hat{p} - p}{SE} \)

Theorems

Central Limit Theorem

Suitable Grade Level

Grades 11-12, Undergraduate