Math  /  Data & Statistics

QuestionWhen Brett Kavanaugh was nominated to be a Supreme Court justice, a survey of 1002 Americans showed that 50.9%50.9 \% of them disapproved of Kavanaugh. A newspaper published an article with this headline: "Majority of Americans Disapprove of Kavanaugh." Use a 0.10 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 pp denote the population proportion of all Americans who disapproved of Kavanaugh. Identify the null and alternative hypotheses. H0:p VH1:p V\begin{array}{l|l|l} \mathrm{H}_{0}: \mathrm{p} & \mathrm{~V} & \square \\ \mathrm{H}_{1}: \mathrm{p} & \mathrm{~V} & \square \end{array} (Type integers or decimals. Do not round.)

Studdy Solution

STEP 1

What is this asking? Can we *really* say *most* Americans disapproved of Kavanaugh based on this survey, or was it just close and could have been by chance? Watch out! Don't mix up the percentage from the survey with what we're actually testing about the *whole* US population!

STEP 2

1. Set up the Hypothesis Test
2. Calculate the Test Statistic
3. Find the P-value
4. Make a Decision

STEP 3

The newspaper headline claims a *majority* disapprove.
A majority means more than half, so more than 50%50\%.
We'll test that!

STEP 4

**Null Hypothesis (H0H_0):** The actual disapproval rate is 50%50\% or less.
In math terms: p0.50p \le 0.50. **Alternative Hypothesis (H1H_1):** The disapproval rate is *greater than* 50%50\%.
Mathematically: p>0.50p > 0.50.
This is what we're *really* trying to find evidence for!

STEP 5

Our **sample proportion** (p^\hat{p}) from the survey is 50.9%50.9\%, or 0.5090.509.
The **sample size** (nn) is **1002**.
Our **null hypothesis proportion** (p0p_0) is 0.500.50.

STEP 6

The **test statistic** formula for proportions is: z=p^p0p0(1p0)n z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} This basically measures how far our sample result is from what we'd expect if the null hypothesis were true, in terms of standard deviations.

STEP 7

Let's plug in our values: z=0.5090.500.50(10.50)1002=0.0090.2510020.0090.01580.57 z = \frac{0.509 - 0.50}{\sqrt{\frac{0.50(1 - 0.50)}{1002}}} = \frac{0.009}{\sqrt{\frac{0.25}{1002}}} \approx \frac{0.009}{0.0158} \approx 0.57 So, our test statistic is approximately **0.57**.

STEP 8

The **p-value** tells us the probability of getting a sample result as extreme as ours (or *more* extreme) *if* the null hypothesis were actually true.
Since our alternative hypothesis is p>0.50p > 0.50 (a one-tailed test), we want the area to the *right* of our test statistic (z=0.57z = 0.57) on the standard normal distribution.

STEP 9

Using a z-table or calculator, we find that the area to the *left* of z=0.57z = 0.57 is approximately 0.71570.7157.
Since the total area under the curve is 11, the area to the *right* (our p-value) is 10.7157=0.28431 - 0.7157 = 0.2843.

STEP 10

Our **significance level** (α\alpha) is 0.100.10.
This is like our "cutoff" for deciding if our result is unusual enough to reject the null hypothesis.

STEP 11

Our **p-value** (0.28430.2843) is *greater* than our significance level (0.100.10).
This means our result isn't unusual enough to reject the null hypothesis.

STEP 12

We *fail to reject* the null hypothesis.
There isn't enough evidence to support the claim that a majority of Americans disapproved of Kavanaugh at the 0.100.10 significance level.
Bummer for the newspaper!

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