Math Problem Statement

A Poll asked 1025. US adults. 472 SAID THEY SUPPORT THE TAX. Does the survey provide convincing evidence that more than 42% of U.S. adults favor an internet sales tax? Use the 0.01 level of significance and the P-value method. WHAT IS THE TEST STATISTIC?

Solution

To determine the test statistic and analyze whether the survey provides convincing evidence that more than 42% of U.S. adults favor an internet sales tax, we can follow these steps:

Step 1: State the hypotheses

  • Null Hypothesis (H0H_0): p=0.42p = 0.42 (The proportion of U.S. adults who favor the tax is 42%).
  • Alternative Hypothesis (HaH_a): p>0.42p > 0.42 (The proportion of U.S. adults who favor the tax is more than 42%).

This is a one-tailed test since we are testing if the proportion is greater than 42%.

Step 2: Identify the sample proportion

The sample proportion (p^\hat{p}) is: p^=xn=47210250.4605\hat{p} = \frac{x}{n} = \frac{472}{1025} \approx 0.4605

Step 3: Compute the test statistic

The test statistic for a proportion is calculated using the formula: z=p^p0p0(1p0)nz = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} where:

  • p^=0.4605\hat{p} = 0.4605 (sample proportion),
  • p0=0.42p_0 = 0.42 (null hypothesis proportion),
  • n=1025n = 1025 (sample size).

Substitute the values into the formula: z=0.46050.420.42(10.42)1025z = \frac{0.4605 - 0.42}{\sqrt{\frac{0.42 \cdot (1 - 0.42)}{1025}}} First, calculate the denominator: 0.420.581025=0.24361025=0.00023760.0154\sqrt{\frac{0.42 \cdot 0.58}{1025}} = \sqrt{\frac{0.2436}{1025}} = \sqrt{0.0002376} \approx 0.0154 Now compute the numerator and the test statistic: z=0.46050.420.0154=0.04050.01542.63z = \frac{0.4605 - 0.42}{0.0154} = \frac{0.0405}{0.0154} \approx 2.63

Step 4: Interpret the test statistic

The test statistic is: z2.63z \approx 2.63

At a significance level of α=0.01\alpha = 0.01, compare the p-value to determine if we reject H0H_0.

Step 5: Compute the P-value

Using standard normal tables or a z-score calculator, the p-value for z=2.63z = 2.63 in a one-tailed test is approximately: P(z>2.63)0.0043P(z > 2.63) \approx 0.0043

Since the p-value (0.00430.0043) is less than α=0.01\alpha = 0.01, we reject H0H_0.

Conclusion

There is convincing evidence at the 0.01 level of significance to conclude that more than 42% of U.S. adults favor an internet sales tax.

Would you like a breakdown of these steps or further clarification?


  1. What does the test statistic represent in hypothesis testing?
  2. Why do we use the P-value method in this test?
  3. How is the sample proportion (p^\hat{p}) calculated?
  4. What are the steps to interpret a z-test in hypothesis testing?
  5. Can you explain why we chose a one-tailed test?

Tip: Always ensure the sample size is sufficiently large for proportions to approximate a normal distribution, as done here!

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Math Problem Analysis

Mathematical Concepts

Statistics
Hypothesis Testing
Proportions
P-value Analysis

Formulas

Sample Proportion: p̂ = x / n
Test Statistic for Proportion: z = (p̂ - p₀) / sqrt(p₀(1 - p₀) / n)

Theorems

Central Limit Theorem
Standard Normal Distribution

Suitable Grade Level

Undergraduate (Introductory Statistics)