Question

Internet tax: The Gallup Poll asked 1025 U.S. adults whether they believed that people should pay sales tax on items purchased over the internet. Of these, 472 said they supported such a tax. Does the survey provide convincing evidence that more than 42% of U.S. adults favor an internet sales tax? Use the =α0.05 level of significance and the P-value method with the TI-84 Plus calculator. Part: 0 / 50 of 5 Parts Complete Part 1 of 5 State the appropriate null and alternate hypotheses. :H0 =p0.42 :H1 >p0.42 This hypothesis test is a ▼right-tailed test. Part: 1 / 51 of 5 Parts Complete Part 2 of 5 Compute the value of the test statistic. Round the answer to at least two decimal places. =z

Answer 1

T he appropriate null and alternate hypotheses is
`\[ H_0: p \leq 0.42 \]\\H_1: p > 0.42 \]` and The value of the test statistic z is approximately 2.65.

Part 1: Hypotheses Setup

The null hypothesis
`\(H_0\)` is that the proportion of U.S. adults favoring an internet sales tax is equal to or less than 42%
`(\(p \leq 0.42\))`, while the alternate hypothesis
`\(H_1\)` is that the proportion is greater than 42% (p > 0.42).

This is a right-tailed test.

`\[ H_0: p \leq 0.42 \]\\H_1: p > 0.42 \]`

Part 2: Compute Test Statistic

Using the formula for the z-test for proportions:

`\[ z = \frac{\hat{p} - p_0}{\sqrt{(p_0(1-p_0))/(n)}} \]`

Where:

-
`\(\hat{p} = (x)/(n)\)` is the sample proportion,

-
`\(p_0 = 0.42\)` is the assumed population proportion,

- n = 1025 is the sample size.

Given that x = 472,
`\(\hat{p} = (472)/(1025) \approx 0.46098\)`, and
`\(p_0 = 0.42\)`, substitute these values into the formula:

`\[ z = \frac{0.46098 - 0.42}{\sqrt{(0.42(1-0.42))/(1025)}} \]`

Let's calculate the test statistic \(z\):

`\[ z = \frac{0.46098 - 0.42}{\sqrt{(0.42 * (1 - 0.42))/(1025)}} \]\\z = \frac{0.04098}{\sqrt{(0.42 * 0.58)/(1025)}} \]\\z \approx \frac{0.04098}{\sqrt{(0.2436)/(1025)}} \]`

`\[ z \approx (0.04098)/(√(0.0002379024)) \]`

`\[ z \approx (0.04098)/(0.01542) \]`

`\[ z \approx 2.6545 \]`

Rounded to two decimal places, the test statistic z is approximately 2.65.