When Brett Kavanaugh was nominated to be a Supreme Court justice, a survey of 1036 Americans showed that of them disapproved of Kavanaugh. A newspaper published an article with this headline.
“Majority of Americans Disapprove of Kavanaugh.” Use a 0.05 significance level to test the claim made in that headline. Use the P-value method. Use the normal distribution as an approximation to the binomial
distribution.
Let p denote the population proportion of all Americans who disapproved of Kavanaugh. Identify the null and alternative hypotheses

:
(Type integers or decimals. Do not round.)
Identify the test statistic.

(Round to two decimal places as needed.)
Identify the P-value.
P-value
(Round to three decimal places as needed.)

Ask by Graham Wells. in the United States

Mar 21,2025

Question

When Brett Kavanaugh was nominated to be a Supreme Court justice, a survey of 1036 Americans showed that of them disapproved of Kavanaugh. A newspaper published an article with this headline.
“Majority of Americans Disapprove of Kavanaugh.” Use a 0.05 significance level to test the claim made in that headline. Use the P-value method. Use the normal distribution as an approximation to the binomial
distribution.
Let p denote the population proportion of all Americans who disapproved of Kavanaugh. Identify the null and alternative hypotheses

:
(Type integers or decimals. Do not round.)
Identify the test statistic.

(Round to two decimal places as needed.)
Identify the P-value.
P-value
(Round to three decimal places as needed.)

Ask by Graham Wells. in the United States

Mar 21,2025

UpStudy AI · Accepted Answer

**Step 1. State the hypotheses.**  
The null and alternative hypotheses are:  
$$
H_{0}: p = 0.5 \quad \text{and} \quad H_{1}: p > 0.5 
$$

**Step 2. Compute the test statistic.**  
Given the sample proportion $$ \hat{p} = 0.514 $$, sample size $$ n = 1036 $$, and hypothesized proportion $$ p_0 = 0.5 $$, the test statistic $$ z $$ is computed using the formula:  
$$
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}
$$
Substitute the values:  
$$
z = \frac{0.514 - 0.5}{\sqrt{\frac{0.5 \times 0.5}{1036}}} = \frac{0.014}{\sqrt{\frac{0.25}{1036}}}
$$
Calculate the standard error:  
$$
\sqrt{\frac{0.25}{1036}} \approx \sqrt{0.00024155} \approx 0.01554
$$
Thus,  
$$
z \approx \frac{0.014}{0.01554} \approx 0.90
$$

**Step 3. Determine the P-value.**  
Since the alternative hypothesis is $$ p > 0.5 $$, we are interested in the right-tail of the normal distribution. The P-value is given by:  
$$
\text{P-value} = P(Z > 0.90)
$$
Using standard normal distribution tables or a calculator, we find:  
$$
P(Z \leq 0.90) \approx 0.8159
$$
Thus,  
$$
\text{P-value} = 1 - 0.8159 \approx 0.1841
$$
Rounded to three decimals, the P-value is:  
$$
\boxed{0.184}
$$
**Hypotheses:** - $$ H_{0}: p = 0.5 $$ - $$ H_{1}: p > 0.5 $$ **Test Statistic:** $$ z = 0.90 $$ **P-value:** 0.184

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