10.5 One-Proportion z Test for p
The assumptions and steps of a one-proportion [latex]z[/latex] test are as follows.
Assumptions:
- A simple random sample.
- Both [latex]np_0[/latex] and [latex]n(1-p_0)[/latex] are at least 5, where [latex]p_0[/latex] is the hypothesized value of [latex]p[/latex] under the null [latex]H_0[/latex].
Steps to perform a one-proportion z test:
- Set up the hypotheses:
Two-tailedRight-tailedLeft-tailed[latex]H_0: p = p_0[/latex][latex]H_0: p \leq p_0[/latex][latex]H_0: p \geq p_0[/latex][latex]H_a: p \neq p_0[/latex][latex]H_a: p \: \gt \:p_0[/latex][latex]H_a: p \: \lt \:p_0[/latex]
- State the significance level [latex]\alpha[/latex].
- Compute the value of the test statistic: [latex]z_o = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n} }}[/latex], with [latex]\hat{p} = \frac{x}{n}[/latex].
- Find the P-value or rejection region.
Two-tailedRight-tailedLeft-tailedNull [latex]H_0: p = p_0[/latex][latex]H_0: p \leq p_0[/latex][latex]H_0: p \geq p_0[/latex]Alternative [latex]H_a: p \neq p_0[/latex][latex]H_a: p \: \gt \:p_0[/latex][latex]H_a: p \: \lt \:p_0[/latex]P-value [latex]2P(Z \geq |z_o|)[/latex][latex]P(Z \geq z_o)[/latex][latex]P(Z \leq z_o)[/latex]Rejection region [latex]Z \geq z_{\alpha / 2}[/latex] or [latex]Z \leq - z_{\alpha / 2}[/latex] [latex]Z \geq z_{\alpha }[/latex][latex]Z \leq - z_{\alpha }[/latex] - Reject the null [latex]H_0[/latex] if the P-value [latex]\leq \alpha[/latex] or [latex]z_o[/latex] falls in the rejection region.
- Conclusion.
Example: One-Proportion z Test
Let [latex]p[/latex] be the proportion of athletes wearing blue suits who win a judo match. Randomly select n = 100 Olympic judo matches and suppose 55 winners wore a blue suit. The other 45 wore a white suit.
- Test at the 5% significance level whether a color bias exists.
If there is no colour bias, the proportions of blue and white winners should be 0.5 and 0.5. Therefore, letting p be the proportion of winners in blue, the hypotheses are [latex]H_0: p = 0.5[/latex] versus [latex]H_a: p \neq 0.5[/latex].
Check the assumptions:- We have a simple random sample (SRS).
- Both [latex]np_0 = 100 \times 0.5 = 50[/latex] and [latex]n (1 - p_0) = 100 \times (1-0.5) = 50[/latex] are at least 5.
Steps:
- Set up the hypotheses. [latex]H_0: p = 0.5[/latex] versus [latex]H_a: p \neq 0.5[/latex]
- State the significance level is [latex]\alpha = 0.05[/latex].
- Compute the test statistic:
[latex]\hat{p} = \frac{x}{n} = \frac{55}{100} = 0.55, z_o = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} = \frac{0.55 - 0.5}{ \sqrt{ \frac{0.5(1-0.5)}{100} }} = 1.[/latex]
- Find the P-value. For a two-tailed test, the P-value is twice the area to the right of the absolute value of the observed test statistic [latex]z_o[/latex]. That is:
P-value = [latex]2P(Z \geq |z_o|)[/latex] [latex]= 2P(Z \geq 1)[/latex] [latex]= 2P(Z \leq -1)[/latex] [latex]= 2 \times 0.1587[/latex] [latex]= 0.3174[/latex]. - Decision: Since the P-value [latex]= 0.3174 \: \gt \: 0.05(\alpha)[/latex], we cannot reject the null [latex]H_0[/latex].
- Conclusion: At the 5% significance level, we do not have sufficient evidence that color bias exists in judging Olympic Judo matches.
- Obtain a confidence interval corresponding to the test in part (a).
For a two-tailed test at the 5% significance level, we obtain a 95% two-tailed interval.
The sample proportion is [latex]\hat{p} = \frac{x}{n} = \frac{55}{100} = 0.55[/latex].[latex]1 - \alpha = 0.95 \Longrightarrow \alpha = 0.05 \Longrightarrow z_{\alpha / 2} = z_{0.025} = 1.96[/latex].
A 95% confidence interval for the proportion of winners in blue is
[latex]\hat{p} \pm z_{\alpha / 2} \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}} = 0.55 \pm 1.96 \times \sqrt{\frac{0.55(1-0.55)}{100}} = (0.4525, 0.6475)[/latex].
Interpretation: We are 95% confident that the proportion of winners in blue is somewhere between 0.4525 and 0.6475, i.e., we are 95% confident that the percentage of winners in blue is somewhere between 45.25% and 64.75%.
- Does this interval support the conclusion of the one-proportion z test?
Yes. In part a), we failed to reject [latex]H_0: p=0.5[/latex] at the 5% significance level. Similarly, in part b), the 95% confidence interval [latex](0.4525, 0.6475)[/latex] contains the hypothesized value [latex]p_0 = 0.5[/latex], meaning we don’t have sufficient evidence to claim that [latex]p[/latex] is significantly different from 0.5.
