I created a problem where I am having my students complete a hypothesis test and confidence interval for proportions to highlight the connection between the two. However, in doing the problem I am getting conflicting results.

With a random sample of n = 300, X = 235, and a confidence level of 99%, I obtained the following CI for the population proportion p: ( 0.72207 , 0.8446 ). I used the TI-84 calculator to get this interval and I checked using various online calculators as well. They all agreed with the interval.

The problem is set up such that the null hypothesis is Ho: p = 0.72 and the alternative hypothesis is Ha: p =/= 0.72. I am using a 1% significance level. Using the TI-84 calculator, I got a test statistic of Z = 2.443 and a P-value of 0.0145 (being 2(0.0072)). Because this P-value exceeds 0.01, I would fail to reject Ho. However, my null value of 0.72 is not within my 99% confidence interval. It is my understanding that the hypothesis test and the confidence interval must always agree as long as I have a two tailed test and the significance level and confidence level sum to 1.

Does anyone see where the contradiction occurs? Thanks much!


1 Answer 1


The confidence interval is $$\hat p \pm z^*_{\alpha/2} \sqrt{\frac{\hat p (1 - \hat p)}{n}},$$ where $\hat p = x/n$ is the sample proportion, and $z^*_{\alpha/2}$ is the $\alpha/2$ upper quantile of the standard normal distribution. In your case $\alpha = 0.01$, so $z^*_{\alpha/2} \approx 2.32635$.

Note here that the standard error of the sample proportion is computed from the sample, because the confidence interval is a statistic. If we are to perform the corresponding hypothesis test, we must also use the corresponding variance estimate: that is to say, if $$H_0 : p = p_0 \quad \text{vs.} \quad H_1 : p \ne p_0$$ for a two sided test, where in your case the proportion under the null hypothesis is $p_0 = 0.72$, then the test statistic is $$Z \mid H_0 = \frac{\hat p - p_0}{\sqrt{\hat p (1 - \hat p)/n}} \sim \operatorname{Normal}(0,1),$$ and not $$Z \mid H_0 = \frac{\hat p - p_0}{\sqrt{p_0 (1 - p_0)/n}}.$$ The second one uses the null variance. To be clear, there are two versions of the one-sample proportion test; only the one that uses the standard error calculated from the sample will match the confidence interval.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .