Math Problem Statement

You wish to test the following claim (π»π‘Ž) at a significance level of π›Ό=0.05.       π»π‘œ:𝑝1=𝑝2       π»π‘Ž:𝑝1≠𝑝2

The 1st population's sample has 539 successes and a sample size = 754. The 2nd population's sample has 483 successes and a sample size = 596. What is the test statistic (z-score) for this sample? (Round to 3 decimal places.) test statistic = 

What is the p-value for this sample? (Round to 3 decimal places.) p-value = 

The p-value is...

less than (or equal to) π›Ό

greater than π›Ό

This test statistic leads to a decision to...

do not reject the null

accept the null

reject the null

As such, the final conclusion is that...

There is not sufficient sample evidence to support the alternate hypothesis claim that p1 is not equal to p2.

The sample data support the alternate hypothesis claim that p1 is not equal to p2.

Solution

To solve this problem, we will follow these steps:

  1. Define the null and alternative hypotheses.
  2. Calculate the pooled proportion.
  3. Calculate the test statistic.
  4. Find the p-value.
  5. Make a decision based on the p-value and the significance level.

Step 1: Define the Hypotheses

  • H0:p1=p2H_0: p_1 = p_2
  • Ha:p1β‰ p2H_a: p_1 \neq p_2

Step 2: Calculate the Pooled Proportion

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

p^=x1+x2n1+n2\hat{p} = \frac{x_1 + x_2}{n_1 + n_2}

where:

  • x1x_1 = number of successes in the first sample = 539
  • n1n_1 = total number in the first sample = 754
  • x2x_2 = number of successes in the second sample = 483
  • n2n_2 = total number in the second sample = 596

So,

p^=539+483754+596=10221350β‰ˆ0.7563\hat{p} = \frac{539 + 483}{754 + 596} = \frac{1022}{1350} \approx 0.7563

Step 3: Calculate the Test Statistic

The test statistic for the difference in proportions is calculated using the formula:

z=(p^1βˆ’p^2)p^(1βˆ’p^)(1n1+1n2)z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\hat{p} (1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}}

where:

  • p^1=539754β‰ˆ0.7141\hat{p}_1 = \frac{539}{754} \approx 0.7141
  • p^2=483596β‰ˆ0.8104\hat{p}_2 = \frac{483}{596} \approx 0.8104

So,

z=0.7141βˆ’0.81040.7563β‹…(1βˆ’0.7563)(1754+1596)z = \frac{0.7141 - 0.8104}{\sqrt{0.7563 \cdot (1 - 0.7563) \left( \frac{1}{754} + \frac{1}{596} \right)}}

First, calculate the standard error (SE):

SE=0.7563β‹…0.2437(1754+1596)β‰ˆ0.7563β‹…0.2437β‹…0.00251β‰ˆ0.000462β‰ˆ0.0215SE = \sqrt{0.7563 \cdot 0.2437 \left( \frac{1}{754} + \frac{1}{596} \right)} \approx \sqrt{0.7563 \cdot 0.2437 \cdot 0.00251} \approx \sqrt{0.000462} \approx 0.0215

Now, calculate zz:

z=0.7141βˆ’0.81040.0215β‰ˆβˆ’0.09630.0215β‰ˆβˆ’4.481z = \frac{0.7141 - 0.8104}{0.0215} \approx \frac{-0.0963}{0.0215} \approx -4.481

Step 4: Find the P-Value

Using the standard normal distribution, we find the p-value for z=βˆ’4.481z = -4.481.

The p-value is twice the area to the right of ∣z∣|z|:

p-value=2β‹…P(Z>4.481)β‰ˆ2β‹…0.0000037β‰ˆ0.0000074p\text{-value} = 2 \cdot P(Z > 4.481) \approx 2 \cdot 0.0000037 \approx 0.0000074

Step 5: Decision and Conclusion

The significance level Ξ±\alpha is 0.05.

  • Since p-value=0.0000074p\text{-value} = 0.0000074 is less than 0.05, we reject the null hypothesis.

Summary of Results

  • Test statistic: zβ‰ˆβˆ’4.481z \approx -4.481
  • P-value: β‰ˆ0.0000074\approx 0.0000074
  • The p-value is less than the significance level (0.05).
  • This test statistic leads to a decision to reject the null.
  • Conclusion: The sample data support the alternate hypothesis claim that p1p_1 is not equal to p2p_2.

Do you want more details or have any questions?

Here are 8 relative questions that expand previous information:

  1. How would the results change if the significance level were 0.01 instead of 0.05?
  2. What are Type I and Type II errors in the context of this hypothesis test?
  3. How can the power of this test be increased?
  4. What assumptions must be met for this test to be valid?
  5. How does sample size affect the results of a hypothesis test for proportions?
  6. What is the interpretation of the p-value in this context?
  7. How would you conduct a confidence interval for the difference in proportions?
  8. What are alternative methods for comparing two population proportions?

Tip: Always check the assumptions of your statistical test to ensure valid results.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Proportions
Z-score

Formulas

Pooled Proportion formula
Test Statistic formula
P-value calculation

Theorems

-

Suitable Grade Level

Advanced High School or College