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