When conducting a hypothesis test that compares two independent population proportions, the following characteristics should be present:

- The two independent samples are simple random samples that are independent.
- The number of successes is at least five, and the number of failures is at least five, for each of the samples.
- Growing literature states that the population must be at least ten or 20 times the size of the sample. This keeps each population from being over-sampled and causing incorrect results.

Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populations or it may be due to chance. A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the population proportions.

The difference of two proportions follows an approximate normal distribution. Generally, the null hypothesis states that the two proportions are the same. That is, *H _{0}*:

*p*=

_{A}*p*. To conduct the test, we use a pooled proportion,

_{B}*p*.

_{c}*z*-score) is:

### Example 10.8

#### Problem

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.

#### Solution 1

The problem asks for a difference in proportions, making it a test of two proportions.

Let *A* and *B* be the subscripts for medication A and medication B, respectively. Then *p _{A}* and

*p*are the desired population proportions.

_{B}Random Variable:
*P′ _{A}* –

*P′*= difference in the proportions of adult patients who did not react after 30 minutes to medication A and to medication B.

_{B}*H _{0}*:

*p*=

_{A}*p*

_{B}*p _{A}* –

*p*= 0

_{B}*H _{a}*:

*p*≠

_{A}*p*

_{B}*p _{A}* –

*p*≠ 0

_{B}The words **"is a difference"** tell you the test is two-tailed.

**Distribution for the test:** Since this is a test of two binomial population proportions, the distribution is normal:

${p}_{c}=\frac{{x}_{A}+{x}_{B}}{{n}_{A}+{n}_{B}}=\frac{20+12}{200+200}=0.08\text{\hspace{1em}}1\u2013{p}_{c}=0.92$

${{P}^{\prime}}_{A}\u2013{{P}^{\prime}}_{B}~N\left[0,\sqrt{(0.08)(0.92)(\frac{1}{200}+\frac{1}{200})}\right]$

*P′ _{A}* –

*P′*follows an approximate normal distribution.

_{B}**Calculate the p-value using the normal distribution:**

*p*-value = 0.1404.

Estimated proportion for group A: ${{p}^{\prime}}_{A}=\frac{{x}_{A}}{{n}_{A}}=\frac{20}{200}=0.1$

Estimated proportion for group B: ${{p}^{\prime}}_{B}=\frac{{x}_{B}}{{n}_{B}}=\frac{12}{200}=0.06$

Graph:

*P′ _{A}* –

*P′*= 0.1 – 0.06 = 0.04.

_{B}Half the *p*-value is below –0.04, and half is above 0.04.

Compare *α* and the *p*-value: *α* = 0.01 and the *p*-value = 0.1404. *α* < *p*-value.

Make a decision: Since *α* < *p*-value, do not reject *H _{0}*.

**Conclusion:** At a 1% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the proportions of adult patients who did not react after 30 minutes to medication *A* and medication *B*.

### Using the TI-83, 83+, 84, 84+ Calculator

Press `STAT`

. Arrow over to `TESTS`

and press
`6:2-PropZTest`

. Arrow down and enter `20`

for x1, `200`

for n1, `12`

for x2,
and `200`

for n2. Arrow down to `p1`

: and arrow to `not equal p2`

. Press `ENTER`

. Arrow down to `Calculate`

and press `ENTER`

. The *p*-value is *p* = 0.1404 and the test statistic is 1.47. Do the procedure again, but instead of `Calculate`

do `Draw`

.

### Try It 10.8

Two types of valves are being tested to determine if there is a difference in pressure tolerances. Fifteen out of a random sample of 100 of Valve *A* cracked under 4,500 psi. Six out of a random sample of 100 of Valve *B* cracked under 4,500 psi. Test at a 5% level of significance.

### Example 10.9

#### Problem

A research study was conducted about gender differences in “sexting.” The researcher believed that the proportion of girls involved in “sexting” is less than the proportion of boys involved. The data collected in the spring of 2010 among a random sample of middle and high school students in a large school district in the southern United States is summarized in Table 10.10. Is the proportion of girls sending sexts less than the proportion of boys “sexting?” Test at a 1% level of significance.

Males | Females | |
---|---|---|

Sent “sexts” | 183 | 156 |

Total number surveyed | 2231 | 2169 |

#### Solution 1

This is a test of two population proportions. Let M and F be the subscripts for males and females. Then *p _{M}* and

*p*are the desired population proportions.

_{F}
Random Variable:
*p′ _{F}* −

*p′*= difference in the proportions of males and females who sent “sexts.”

_{M}*H _{0}*:

*p*=

_{F}*p*

_{M}*H*:

_{0}*p*–

_{F}*p*= 0

_{M}*H _{a}*:

*p*<

_{F}*p*

_{M}*H*:

_{a}*p*–

_{F}*p*< 0

_{M}The words **"less than"** tell you the test is left-tailed.

**Distribution for the test:** Since this is a test of two population proportions, the distribution is normal:

${p}_{c}=\frac{{x}_{F}+{x}_{M}}{{n}_{F}+{n}_{M}}=\frac{156+183}{2169+2231}=\text{0}\text{.077}$

$1-{p}_{c}=0.923$

Therefore,

${{p}^{\prime}}_{F}\u2013{{p}^{\prime}}_{M}\sim N\left(0,\sqrt{(0.077)(0.923)\left(\frac{1}{2169}+\frac{1}{2231}\right)}\right)$
*p′ _{F}* –

*p′*follows an approximate normal distribution.

_{M}**Calculate the p-value using the normal distribution:**

*p*-value = 0.1045

Estimated proportion for females: 0.0719

Estimated proportion for males: 0.082

Graph:

**Decision:** Since *α* < *p*-value, Do not reject *H _{0}*

**Conclusion:** At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the proportion of girls sending “sexts” is less than the proportion of boys sending “sexts.”

### Using the TI-83, 83+, 84, 84+ Calculator

Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 156 for x1, 2169 for n1, 183 for x2, and 2231 for n2. Arrow down to p1: and arrow to less than p2. Press `ENTER`

. Arrow down to Calculate and press ENTER. The *p*-value is *P* = 0.1045 and the test statistic is *z* = -1.256.

### Example 10.10

#### Problem

Researchers conducted a study of smartphone use among adults. A cell phone company claimed that iPhone smartphones are more popular with White people (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?

#### Solution 1

This is a test of two population proportions. Let W and A be the subscripts for the White people and African American people. Then *p _{W}* and

*p*are the desired population proportions.

_{A}Random Variable: *p′ _{W}* –

*p′*= difference in the proportions of White and African American cell phone users who own iPhones.

_{A}*H _{0}*:

*p*=

_{W}*p*

_{A}*H*:

_{0}*p*–

_{W}*p*= 0

_{A}*H _{a}*:

*p*>

_{W}*p*

_{A}*H*:

_{a}*p*–

_{W}*p*> 0

_{A}The words "more popular" indicate that the test is right-tailed.

Distribution for the test: The distribution is approximately normal:

${p}_{c}=\frac{{x}_{W}+{x}_{A}}{{n}_{W}+{n}_{A}}=\frac{134+12}{1343+232}=\text{}0.0927$

$1-{p}_{c}=0.9073$

Therefore,

${{p}^{\prime}}_{W}\u2013{{p}^{\prime}}_{A}\backsim N\left(0,\sqrt{\left(0.0927\right)\left(0.9073\right)\left(\frac{1}{1343}+\frac{1}{232}\right)}\right)$

${{p}^{\prime}}_{W}\u2013{{p}^{\prime}}_{A}$ follows an approximate normal distribution.

Calculate the *p*-value using the normal distribution:*p*-value = 0.0077

Estimated proportion for group A: 0.10

Estimated proportion for group B: 0.05

**Graph:**

**Decision:** Since *α* > *p*-value, reject the *H _{0}*.

**Conclusion:** At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that a larger proportion of White cell phone owners use iPhones compared to African Americans.

### Using the TI-83, 83+, 84, 84+ Calculator

TI-83+ and TI-84: Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 135 for x1, 1343 for n1, 12 for x2, and 232 for n2. Arrow down to p1: and arrow to greater than p2. Press ENTER. Arrow down to Calculate and press ENTER. The P-value is P = 0.0092 and the test statistic is Z = 2.33.

### Try It 10.10

A concerned group of citizens wanted to know if the proportion of forcible rapes in Texas was different in 2011 than in 2010. Their research showed that of the 113,231 violent crimes in Texas in 2010, 7,622 of them were forcible rapes. In 2011, 7,439 of the 104,873 violent crimes were in the forcible rape category. Test at a 5% significance level. Answer the following questions:

a. Is this a test of two means or two proportions?

b. Which distribution do you use to perform the test?

c. What is the random variable?

d. What are the null and alternative hypothesis? Write the null and alternative hypothesis in symbols.

e. Is this test right-, left-, or two-tailed?

f. What is the *p*-value?

g. Do you reject or not reject the null hypothesis?

h. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ____________.