Question

Test the claim that the proportion of people who own cats is larger than 90% at the 0.05 significance level.


Based on a sample of 300 people, 93% owned cats

The test statistic is: __________________ (to 2 decimals)

The p-value is: __________________ (to 2 decimals)

Answer 1

To test the claim that the proportion of people who own cats is larger than 90%, we can perform a one-sample proportion z-test.

### Step 1: Define the null and alternative hypotheses

- Null hypothesis \(H_0\): \(p \leq 0.90\) (the proportion of cat owners is 90% or less)
- Alternative hypothesis \(H_a\): \(p > 0.90\) (the proportion of cat owners is greater than 90%)

### Step 2: Define the sample statistics

- Sample size \(n = 300\)
- Sample proportion \(\hat{p} = 0.93\)
- Hypothesized population proportion \(p_0 = 0.90\)

### Step 3: Calculate the test statistic

The formula for the test statistic \(z\) when testing proportions is:

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

Substituting the values:

\[
\hat{p} = 0.93, \quad p_0 = 0.90, \quad n = 300
\]

First, calculate the standard error:

\[
\text{Standard Error} = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.90 \cdot (1 - 0.90)}{300}} = \sqrt{\frac{0.90 \cdot 0.10}{300}} = \sqrt{\frac{0.09}{300}} = \sqrt{0.0003} \approx 0.01732
\]

Now, substitute into the formula for \(z\):

\[
z = \frac{0.93 - 0.90}{0.01732} \approx \frac{0.03}{0.01732} \approx 1.73
\]

### Step 4: Calculate the p-value

Since this is a one-tailed test (we are testing if the proportion is greater than 0.90), we will look for the area to the right of our test statistic \(z = 1.73\) in the standard normal distribution.

Using a standard normal distribution table, or a calculator:

- The p-value for \(z = 1.73\) is approximately \(0.0418\).

### Step 5: Conclusion

Now, we can summarize the results:

- Test statistic \(z\): **1.73**
- p-value: **0.04**

### Final Answers:

- The test statistic is: **1.73**
- The p-value is: **0.04**