# Practice 12 Conducting One and Two-proportion Tests in R

## 12.1 Directions

In this practice exercise, you will conduct conduct one and two-proportion tests in R.

## 12.2 A closer look at the code

### 12.2.1 One-sample Proportion test

About 42.3% of Californians and 19.6% of all Americans over age five speak a language other than English. A researcher conducts a survey of 1,500 Wisconsin residents of which 7.3% could speak a language other than English. Conduct a hypothesis test to determine if the percent of the Wisconsin residents who speak a language other than English is different from 42.3%.

#### 12.2.1.1 What do we know?

The following is given in the question:

- The null hypothesis is \(P_{WI} = 0.423\)
- The sample proportion for WI is 0.073
- The sample size is 1,500

Let’s put this into r:

```
1500
n <- 0.073
p.WI <- 0.423 p.null <-
```

#### 12.2.1.2 Conduct the test

To conduct a test, we need the to use the `prop.test()`

.

`prop.test()`

needs the following arguments

`x`

the number of successes`n`

the number of trials`alternative`

which can be set to “two.sided” for \(=/\ne\) hypothesis tests, “greater” for \(=/>\) hypothesis tests, or “less” for \(=/<\) hypothesis tests.`conf.level`

the confidence level of the returned confidence interval

`prop.test(x=p.WI*n,n=n,p=p.null,alternative = "two.sided", conf.level = 0.95)`

```
##
## 1-sample proportions test with continuity correction
##
## data: p.WI * n out of n, null probability p.null
## X-squared = 751.42, df = 1, p-value < 2.2e-16
## alternative hypothesis: true p is not equal to 0.423
## 95 percent confidence interval:
## 0.06059296 0.08764350
## sample estimates:
## p
## 0.073
```

With a p-value of \(2.2e-16\), we **reject** the null hypothesis.

*BONUS* if `alternative = "two.sided"`

then the `95 percent confidence interval:`

returned by `prop.test()`

is the confidence interval for the sample proportion.

### 12.2.2 Two-sample Proportion test

Two types of medication for hives are being tested to determine if there is a difference in the proportions of adult patient reactions. Twenty out of a random sample of 200 adults given medication A still had hives 30 minutes after taking the medication. Twelve out of another random sample of 200 adults given medication B still had hives 30 minutes after taking the medication. Test at a 1% level of significance.

#### 12.2.2.1 What do we know?

The following is given in the question:

- The null hypothesis is \(P_{WI} = 0.423\)
- The sample proportion for WI is 0.073
- The sample size is 1,500

Let’s put this into r:

```
20 # Number of adults given medication A who still had hives 30 minutes after taking the medication.
p.A <- 200 # Number of adults given medication A
n.A <- 12 # Number of adults given medication B who still had hives 30 minutes after taking the medication.
p.B <- 200 # Number of adults given medication B
n.B <- 0.01 # 1% level of significance alpha <-
```

#### 12.2.2.2 Conduct the test

To conduct a test, we need the to use the `prop.test()`

and the `c()`

command

`prop.test(x=c(p.A,p.B),n=c(n.A,n.B),alternative = "two.sided", conf.level = 1-alpha)`

```
##
## 2-sample test for equality of proportions with continuity correction
##
## data: c(p.A, p.B) out of c(n.A, n.B)
## X-squared = 1.6644, df = 1, p-value = 0.197
## alternative hypothesis: two.sided
## 99 percent confidence interval:
## -0.03469035 0.11469035
## sample estimates:
## prop 1 prop 2
## 0.10 0.06
```

With a p-value of 0.197, we fail to reject the null hypothesis and conclude that there is insufficient evidence to conclude there is a significant difference between the sample proportions.
## Now you try
Use R to complete the following activities (this is just for practice you do not need to turn anything in).

Researchers conducted a study of smartphone use among adults. A cell phone company claimed that iPhone smartphones are more popular with whites (non-Hispanic) than with African Americans. The results of the survey indicate that of the 232 African American cell phone owners randomly sampled, 5% have an iPhone. Of the 1,343 white cell phone owners randomly sampled, 10% own an iPhone. Test at the 5% level of significance. Is the proportion of white iPhone owners greater than the proportion of African American iPhone owners? (See EXAMPLE 10.10)