In this article, we will discuss how to do a one- sample proportion z-test in R with some practical examples.

## What is One- Sample Proportion z-test ?

One-sample proportion z-test is used to conduct a hypothesis test about a population proportion(p). It is used to estimate the difference between the proportion of responses(or a number of successes) in a sample data and the actual proportion in the population data from which we draw the sample.

## Conditions required to conduct one sample proportion z test

**Assumptions for the one sampleproportion z-test are as follows:-**

- The sample should be drawn at random from the population.
- Population should follow a binomial distribution.
- np
_{0}>10 & n(1-p_{0})>10 where n is sample size and p is the hypothesized value for the population proportion. - The population size should be 10 times larger than the sample size.

**Hypothesis for the one sample proportion z-test**

Let p_{0} denote the hypothesized value for the proportion.

p denotes the population proportion

**Null Hypothesis:**

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

**Alternative Hypothesis:** Three forms of alternative hypothesis are as follows:

**H**Population proportion is less than the hypothesized proportion.It is called lower tail test (left-tailed test)._{a}: p < p_{0}**H**Population proportion is greater than the hypothesized proportion.It is called Upper tail test(right-tailed test)._{a}: p > p_{0}**H**Population proportion is not equal to hypothesized proportion.It is called two tail test._{a}: p ≠ p_{0}

**Formula for the test statistic one proportion Z test is:**

where :

**n: **sample size

**p ^{^}:** observed sample proportion

**p _{0}:** hypothesized population proportion

**Functions in R for one proportion z-test**

To perform one proportion z-test, we will use the following functions from the R stats library.

We will use binom.test() or prop.test() function from R stats library

**binom.test() :**

If n ≤ 30 i.e when sample size is small then we use binom.test() function to perform one -proportion z-test.

The **binom.test()** function uses the following basic syntax:

binom.test(x, n, p = 0.5, alternative = c("two.sided", "less", "greater"), conf.level = 0.95)

where:

**x: **The number of successes

**n:** The number of trials.

**p:** The hypothesized probability of success.

**alternative:** The alternative hypothesis for the test.It can be ‘greater’, ‘less’, ‘two.sided’ based on the alternative hypothesis.

**conf. level:** confidence level of the interval

**prop.test() :**

If n >30 i.e when sample size is large then we use prop.test() function to perform one -proportion z-test.

The **prop.test()** function uses the following basic syntax:

prop.test(x, n, p = NULL, alternative = c("two.sided", "less", "greater"), conf.level = 0.95, correct = TRUE)

where:

**x : **The number of successes

**n:** The number of trials.

**p:** The vector of probabilities of success.

**alternative:** The alternative hypothesis for the test.It can be ‘greater’, ‘less’, ‘two.sided’ based on the alternative hypothesis.

**conf. level:** confidence level of the interval.

**correct**: a logical indicating whether Yates’ continuity correction should be applied or not where it is possible.

### Summary for the one sample Proportion Z-test

Left-tailed Test | Right-tailed Test | Two-tailed Test | |

Null Hypothesis | H_{0} : p≥p_{0} | H_{0} : p≤p_{0} | H_{0} : p=p_{0} |

Alternate Hypothesis | H_{a} : p<p_{0} | H_{a} : p>p_{0} | H_{a} : p ≠ p_{0} |

Test Statistic | z= (p^{^} – p_{0)}/√(p_{0}(1- p_{0} )/n | z= (p^{^} – p_{0)}/√(p_{0}(1- p_{0} )/n | z= (p^{^} – p_{0)}/√(p_{0}(1- p_{0} )/n |

Decision Rule: p-value approach (where α is level of significance) | If p-value ≤α then Reject H _{0} | If p-value ≤α then Reject H _{0} | If p-value ≤α then Reject H _{0} |

Decision Rule: Critical-value approach | If z ≤ -z_{α}then Reject H _{0} | If z ≥ z_{α}then Reject H _{0} | If z ≤ -z_{α/2} or z ≥ z_{α/2} then Reject H_{0} |

## How to do one proportion z-test in R?

We will calculate the test statistic by using one proportion z-test.

**Procedure to perform One Proportion Z-Test in R**

**Step 1:** Define the Null Hypothesis and Alternate Hypothesis.

**Step 2:** Decide the level of significance α (alpha).

**Step 3:** Calculate the test statistic using the binom.test() or prop.test() depending upon the sample size.

**Step 4:** Interpret the one-proportion z-test results.

**Step 5:** Determine the rejection criteria for the given confidence level and conclude the results whether the test statistic lies in the rejection region or non-rejection region.

Let’s see practical examples that show how to use the binom.test() or prop.test() function in R.

## Examples of One Proportion z-test in R

### Example 1: Right-tailed one proportion test in R

An auditor for the Online Service wants to examine its special two-hour priority order delivery to determine the proportion of the orders that actually arrive within the promised two-hour period. A randomly selected sample of 1500 such orders is found to contain 1150 that were delivered on time. Does the sample data provide evidence to conclude that the percentage of on-time orders is more than 75%? Test at 5% level of significance.

**Solution:** Given data :

sample size (n) = 1500

number of success (x) = 1150

sample proportion (p^{^}) = x/n = 1150/1500 = 0.7666

hypothesized population proportion (p_{0}) = 0.75

level of significance (α) = 0.05

confidence level = 0.95

Let’s solve this example by the step-by-step procedure.

**Step 1:** Define the Null Hypothesis and Alternate Hypothesis.

Let p_{0} denote the hypothesized value for the proportion.

p denotes the population proportion

**Null Hypothesis:** The population proportion is equal to 0.75 (i.e. 75%).

**H _{0} : p=**

**0.75**(right-tailed test)

**Alternate Hypothesis**: The population proportion is greater than 0.75 (i.e. 75%).

**H _{a} : p > 0.75**

**Step 2:** level of significance (α) = 0.05

**Step 3:** Calculate the test statistic using the below code.

Since here sample size = 1500 >30.

So we use prop.test() in this example.

# Perform one-proportion z-test prop.test(x=1150, n=1500, p=0.75, alternative="greater")

Specify the alternative hypothesis as “greater” because we are performing a right-tailed test. The results for the one-proportion z-test are as follows.

#Results 1-sample proportions test with continuity correction data: 1150 out of 1500, null probability 0.75 X-squared = 2.1342, df = 1, p-value = 0.07202 alternative hypothesis: true p is greater than 0.75 95 percent confidence interval: 0.7478919 1.0000000 sample estimates: p 0.7666667

**Step 4:** Interpret the one-proportion test results.

**How to interpret one proportion z-test results in R?**

Let’s see the interpretation of one-proportion z-test results in R.

**data**: This gives information about the data used in the one-proportion z-test.

**X-squared:** the value of Pearson’s chi-squared test statistic.

**df:** the degree of freedom of the approximate chi-squared distribution of the test statistic.

**p-value**: This is the p-value corresponding to a statistic. In our case, the p-value is 0.07202.

**alternative**: It is the alternative hypothesis used for the z-test. In our case, an alternative hypothesis is population proportion is greater than 0.75 i.e. right-tailed.

**95 percent confidence interval:** This gives us a 95% confidence interval for the true proportion. Here the 95% confidence interval is [0.7478919,1.0000000].

**sample estimates**: It gives the sample proportion. In our case, the sample proportion is 0.7666667.

**Step 5:** Determine the rejection criteria for the given confidence level and conclude the results whether the test statistic lies in the rejection region or non-rejection region.

**Conclusion:**

Since the p-value[ 0.07202 ] is greater than the level of significance (α) = 0.05, we fail to reject the null hypothesis.

This means we have sufficient evidence to say that the population proportion is equal to 0.75.

### Example 2: Two-tailed one proportion test in R

In a sample of 20 students in College,11 are tea-drinkers and the rest are coffee-drinkers. Can we assume that both tea and coffee drinkers are equally popular in college at the 5% level of significance?

**Solution:** Given data :

sample size (n) = 20

Number of tea-drinkers (x)= 11

sample proportion (p^{^}) = number of tea-drinkers /sample size

p^{^} = 11/20 = 0.55

Let’s solve this example by the step-by-step procedure.

**Step 1:** Define the Null Hypothesis and Alternate Hypothesis.

let p be the population proportion for the tea drinkers.

**Null Hypothesis**: Both tea and coffee drinkers are equally popular in the college

**H _{0} : p = 0.5**

**Alternate Hypothesis**: Tea and Coffee drinkers are not equal in college.

**H _{a} : p ≠ 0.5**

**Step 2:** level of significance (α) = 0.05

**Step 3:** Calculate the test statistic using the below code.

Since here, sample size = 20 <30.

So we use binom.test() in this example.

# Perform one-proportion z-test binom.test(x=11,n=20,p=0.5,alternative = "two.sided")

Specify the alternative hypothesis as “two.sided” because we are performing a two-tailed test. The results for the one-proportion z-test are as follows.

#Results Exact binomial test data: 11 and 20 number of successes = 11, number of trials = 20, p-value = 0.8238 alternative hypothesis: true probability of success is not equal to 0.5 95 percent confidence interval: 0.3152781 0.7694221 sample estimates: probability of success 0.55

**Step 4:** Interpret the one-proportion test results.

**How to interpret one proportion z-test results in R?**

Let’s see the interpretation of one-proportion z-test results in R.

**data**: This gives information about the data used in the one-proportion z-test. It tells the number of successes and the number of trials.

**p-value**: This is the p-value corresponding to a statistic. In our case, the p-value is 0.8238.

**alternative**: It is the alternative hypothesis used for the z-test. In our case, an alternative hypothesis is a Tea and Coffee drinkers are not equal in college, i.e. two-tailed.

**95 percent confidence interval:** This gives us a 95% confidence interval for the true proportion. Here the 95% confidence interval is [0.3152781,0.7694221].

**sample estimates**: It gives the probability of success. In our case, the probability of success is 0.55.

**Step 5:** Determine the rejection criteria for the given confidence level and conclude the results whether the test statistic lies in the rejection region or non-rejection region.

**Conclusion:**

Since the p-value[0.8238] is greater than the level of significance (α) = 0.05, we fail to reject the null hypothesis.

This means we have sufficient evidence to say that tea and coffee drinkers are equally popular in college.

**What package is needed for one proportion z-test in R?**

The R Stats Package is needed to do a z-test in R.

**Summary**

I hope you found the above article on One Proportion z-test in R with Examples informative and educational.