A one proportion z-test is used to compare an observed proportion to a theoretical one.

This test uses the following null hypotheses:

**H**p = p_{0}:_{0}(population proportion is equal to hypothesized proportion p_{0})

The alternative hypothesis can be either two-tailed, left-tailed, or right-tailed:

**H**p ≠ p_{1}(two-tailed):_{0}(population proportion is not equal to some hypothesized value p_{0})**H**p < p_{1}(left-tailed):_{0}(population proportion is less than some hypothesized value p_{0})**H**p > p_{1}(right-tailed):_{0}(population proportion is greater than some hypothesized value p_{0})

The test statistic is calculated as:

z = (p-p_{0}) / √p_{0}(1-p_{0})/n

where:

**p:**observed sample proportion**p**hypothesized population proportion_{0}:**n:**sample size

If the p-value that corresponds to the test statistic z is less than your chosen significance level (common choices are 0.10, 0.05, and 0.01) then you can reject the null hypothesis.

**One Proportion Z-Test in R**

To perform a one proportion z-test in R, we can use one of the following functions:

**If n ≤ 30:**binom.test(x, n, p = 0.5, alternative = “two.sided”)**If n> 30:**prop.test(x, n, p = 0.5, alternative = “two.sided”, correct=TRUE)

where:

**x:**The number of successes**n:**The number of trials**p:**The hypothesized population proportion**alternative:**The alternative hypothesis**correct:**Whether or not to apply Yates’ continuity correction

The following example shows how to carry out a one proportion z-test in R.

**Example: One Proportion Z-Test in R**

Suppose we want to know whether or not the proportion of residents in a certain county who support a certain law is equal to 60%. To test this, we collect the following data on a random sample:

**p**hypothesized population proportion = 0.60_{0}:**x:**residents who support law: 64**n:**sample size = 100

Since our sample size is greater than 30, we can use the **prop.test() **function to perform a one sample z-test:

prop.test(x=64, n=100, p=0.60, alternative="two.sided") 1-sample proportions test with continuity correction data: 64 out of 100, null probability 0.6 X-squared = 0.51042, df = 1, p-value = 0.475 alternative hypothesis: true p is not equal to 0.6 95 percent confidence interval: 0.5372745 0.7318279 sample estimates: p 0.64

From the output we can see that the p-value is **0.475**. Since this value is not less than α = 0.05, we fail to reject the null hypothesis. We do not have sufficient evidence to say that the proportion of residents who support the law is different from 0.60.

The 95% confidence interval for the true proportion of residents in the county that support the law is also found to be:

**95% C.I. = [0.5373, 7318]**

Since this confidence interval contains the proportion **0.60**, we do not have evidence to say that the true proportion of residents who support the law is different from 0.60. This matches the conclusion we came to using just the p-value of the test.

**Additional Resources**

An Introduction to the One Proportion Z-Test

One Proportion Z-Test Calculator

How to Perform a One Proportion Z-Test in Excel

prop.test(x=64, n=100, p=0.60, alternative=”two.sided”) DOES NOT use the test statistics z = (p-p0) / sqrt(p0(1-p0)/n).

Using your example, we have z=(0.64-0.6)/sqrt(0.6*0.4/100) = 0.8165.

The p-value is 2P(Z >= 0.8165), where Z is N(0,1), i.e.,

p-value=0.4142.

The corresponding 95% confidence interval is

[0.64-1.959964*sqrt(0.64*0.36/100), 0.64+1.959964*sqrt(0.64*0.36/100) ]=

[0.54592173, 0.7340783].

I used 1.959964 instead of 1.96, for better accuracy.

prop.test(x=64, n=100, p=0.60, alternative=”two.sided”) USES a X2 test statistics, wich value for your example (using Yates continuity correction), is given by

X-squared=((|36-40|-0.5)^2)/40+((|64-60|-0.5)^2)/60=0.51041(6) . This is the X-squared in the output: 0.51042.

In this case, the p-value is given by P(Q >= 0.51042)=0.4749571, which is the p-value in the output: 0.475 (Q is distributed according to a chi-squared distribution with 1 df).

I don’t know how the corresponding confidence interval is obtained (but I would very much like to know!) and and also don’t know if there is a function in R that allow us to perform the test for proportions using the Z statistic. I also would very much like to know this!