Let's go through each part step-by-step.

Part 1: Hypotheses

To test whether more than 41% of U.S. adults favor an internet sales tax, we can set up the following hypotheses:

• Null Hypothesis (H₀): $p = 0.41$
• Alternative Hypothesis (H₁): $p > 0.41$

This is a one-tailed test since we are testing if the percentage is greater than 41%.

Part 2: Critical Value

Since the level of significance is $\alpha = 0.01$ for a one-tailed test, we need to find the z-score that corresponds to the 99th percentile.

Looking up the z-score for a one-tailed test at the 0.01 significance level:

• The critical value is approximately 2.326.

Part 3: Test Statistic

Now, we calculate the test statistic $z$ using the formula for a proportion test:

$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$

Where:

• $\hat{p} = \frac{440}{1023} \approx 0.4302$
• $p_0 = 0.41$
• $n = 1023$

Substituting in the values:

1. Calculate $\hat{p}$: $\hat{p} = \frac{440}{1023} \approx 0.4302$

2. Calculate the standard error: $\text{SE} = \sqrt{\frac{0.41 \times (1 - 0.41)}{1023}} \approx 0.0154$

3. Calculate the z-score: $z = \frac{0.4302 - 0.41}{0.0154} \approx 1.31$

Thus, the test statistic $z \approx 1.31$.

Part 4: Decision Rule

We compare the test statistic to the critical value:

• Since $z = 1.31$ is less than the critical value of 2.326, we do not reject the null hypothesis at the 0.01 significance level.

Part 5: Conclusion

At the $\alpha = 0.01$ level of significance,