Math Problem Statement

You wish to test the following claim (Ha) at a significance level of α = 0.002. H0: p1 = p2, Ha: p1 < p2. You obtain 4% successes in a sample of size n1 = 667 from the first population and 5.4% successes in a sample of size n2 = 408 from the second population. Round values for X to the nearest whole number.

Solution

This problem involves testing a claim about two population proportions at a significance level of α=0.002\alpha = 0.002. Let's work through the solution step by step:


Step 1: Extract Information

  • Null hypothesis (H0H_0): p1=p2p_1 = p_2
  • Alternative hypothesis (HaH_a): p1<p2p_1 < p_2 (a left-tailed test)
  • n1=667n_1 = 667, p1=4%100=0.04p_1 = \frac{4\%}{100} = 0.04
  • n2=408n_2 = 408, p2=5.4%100=0.054p_2 = \frac{5.4\%}{100} = 0.054

The combined sample proportion (pp) is calculated as: p=x1+x2n1+n2p = \frac{x_1 + x_2}{n_1 + n_2} where x1=n1p1x_1 = n_1 \cdot p_1 and x2=n2p2x_2 = n_2 \cdot p_2.


Step 2: Calculate the Combined Proportion (pp)

  1. x1=6670.04=26.6827x_1 = 667 \cdot 0.04 = 26.68 \approx 27 (rounded to the nearest whole number)
  2. x2=4080.054=22.03222x_2 = 408 \cdot 0.054 = 22.032 \approx 22
  3. Total successes = x1+x2=27+22=49x_1 + x_2 = 27 + 22 = 49
  4. Total sample size = n1+n2=667+408=1075n_1 + n_2 = 667 + 408 = 1075
  5. Combined proportion: p=4910750.0456p = \frac{49}{1075} \approx 0.0456

Step 3: Calculate the Standard Error (SE)

SE=p(1p)(1n1+1n2)SE = \sqrt{p \cdot (1 - p) \cdot \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}

  1. 1p=10.0456=0.95441 - p = 1 - 0.0456 = 0.9544
  2. 1n1+1n2=1667+14080.0015+0.00245=0.00395\frac{1}{n_1} + \frac{1}{n_2} = \frac{1}{667} + \frac{1}{408} \approx 0.0015 + 0.00245 = 0.00395
  3. SE=0.04560.95440.003950.00017170.0131SE = \sqrt{0.0456 \cdot 0.9544 \cdot 0.00395} \approx \sqrt{0.0001717} \approx 0.0131

Step 4: Calculate the Test Statistic

The test statistic (zz) is calculated as: z=p1p2SEz = \frac{p_1 - p_2}{SE}

z=0.040.0540.0131=0.0140.01311.069z = \frac{0.04 - 0.054}{0.0131} = \frac{-0.014}{0.0131} \approx -1.069


Step 5: Calculate the P-Value

Since this is a left-tailed test, we find the probability corresponding to z=1.069z = -1.069 using a standard normal table or software. The p-value is: P(z<1.069)0.142P(z < -1.069) \approx 0.142


Step 6: Compare the P-Value to α\alpha

  • α=0.002\alpha = 0.002
  • p-value=0.142>αp\text{-value} = 0.142 > \alpha

Thus, the p-value is greater than α\alpha.


Step 7: Decision

We fail to reject the null hypothesis because the p-value is greater than the significance level.


Final Answers

  1. Test statistic: 1.069-1.069 (rounded to 3 decimal places)
  2. P-value: 0.14200.1420 (rounded to 4 decimal places)
  3. Decision: Fail to reject the null hypothesis.

Would you like further clarification or details?


Related Questions:

  1. What is the significance of using a pooled proportion in this hypothesis test?
  2. How does the choice of significance level (α\alpha) affect the conclusion of a hypothesis test?
  3. How is a left-tailed test different from a right-tailed or two-tailed test in hypothesis testing?
  4. Why do we use the standard error in the calculation of the test statistic?
  5. How does the sample size influence the precision of the test statistic and the p-value?

Tip:

When calculating proportions and p-values, always ensure intermediate results are rounded only at the final step to maintain accuracy.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Two-Proportion Z-Test
Statistics

Formulas

Pooled Proportion: p = (x1 + x2) / (n1 + n2)
Standard Error: SE = sqrt(p(1-p) * (1/n1 + 1/n2))
Z-Statistic: z = (p1 - p2) / SE

Theorems

Central Limit Theorem
Standard Normal Distribution

Suitable Grade Level

Undergraduate (College-Level Statistics)