9.5 Statistical Inference for Two Population Proportions

A marketing company places an advertisement for a new brand of deodorant on two different platforms:  television and social media.  The company wants to study the proportion of people who remembered seeing the advertisement two hours later.  In a sample of 200 people who saw the advertisement on television, 74 remembered seeing it two hours later.  In a sample of 300 people who saw the advertisement on social media, 129 remembered seeing it two hours later.

1. Construct a 98% confidence interval for the difference in the proportion of people from the two different platforms that remember seeing the advertisement two hours later.
2. Interpret the confidence interval found in part 1.
3. Is there evidence to suggest that the proportion of people from social media who remember seeing the advertisement two hours later is greater than the proportion of people from television?  Explain.

Solution:

1. Let television be population 1 and social media be population 2.  From the question we have the following information:
   

   Television	Social Media
   [latex]n_1=200[/latex]	[latex]n_2=300[/latex]
   [latex]\hat{p}_1=\frac{74}{200}=0.37[/latex]	[latex]\hat{p}_2=\frac{129}{300}=0.43[/latex]

   Before constructing the confidence interval, we check that the normal distribution applies:

   [latex]\begin{eqnarray*} n_1 \times \hat{p}_1 & = & 200 \times 0.37=74 \geq 5 \\ n_1 \times (1-\hat{p}_1) & = & 200 \times (1-0.37)=126 \geq 5 \\ n_2 \times \hat{p}_2 & = & 300 \times 0.43=129 \geq 5 \\ n_2 \times (1-\hat{p}_1) & = & 300 \times (1-0.37)=171 \geq 5 \end{eqnarray*}[/latex]

   To find the confidence interval, we need to find the [latex]z[/latex]-score for the 98% confidence interval.  This means that we need to find the [latex]z[/latex]-score so that the entire area to the left of [latex]z[/latex] is [latex]\displaystyle{0.98+\frac{1-0.98}{2}=0.99}[/latex].

   

   Function	norm.s.inv	Answer
   Field 1	0.99	2.3263…

   So [latex]z=2.3263...[/latex]. The 98% confidence interval is

   [latex]\begin{eqnarray*} \\ \mbox{Lower Limit} & = & \hat{p}_1-\hat{p}_2-z \times \sqrt{\frac{\hat{p}_1 \times (1-\hat{p}_1)}{n_1}+\frac{\hat{p_2} \times (1-\hat{p}_2)}{n_2}}\\ & = & 0.37-0.43-2.3263... \times \sqrt{\frac{0.37 \times (1-0.37)}{200}+\frac{0.43 \times (1-0.43)}{300}} \\ & = & -0.1636  \\ \\ \mbox{Upper Limit} & = & \hat{p}_1-\hat{p}_2+z \times \sqrt{\frac{\hat{p}_1 \times (1-\hat{p}_1)}{n_1}+\frac{\hat{p_2} \times (1-\hat{p}_2)}{n_2}}\\ & = & 0.37-0.43+2.3263... \times \sqrt{\frac{0.37 \times (1-0.37)}{200}+\frac{0.43 \times (1-0.43)}{300}} \\ & = & 0.0436  \\ \\ \end{eqnarray*}[/latex]

2. We are 98% confident that the difference in the proportion of people from the two platforms that remember seeing the advertisement two hours later is between -16.36% and 4.36%.
3. Because 0 is inside the confidence interval, it suggests that the difference in the proportions [latex]\displaystyle{p_1-p_2}[/latex] is 0.  That is, [latex]\displaystyle{p_1-p_2=0}[/latex].  This suggests that the two proportions are equal.  So the proportion of people from social media who remember seeing the advertisement two hours is not greater than the proportion of people from television.

A cell phone company claimed that iPhones are more popular with adults 30 years old or younger than with adults over 30 years old.  A consumer advocacy group wants to test this claim.  In a sample of 1340 adults 30 years old or younger, 134 own an iPhone.  In a sample of 250 adults over the age of 30, 15 own an iPhone.  At the 5% significance level, is the proportion of adults 30 years old or younger who own an iPhone greater than the proportion of adults over the age of 30 who own an iPhone?

Solution:

Let adults 30 years old or younger be population 1 and adults over 30 years old be population 2.  From the question, we have the following information:

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & p_1-p_2=0 \\ H_a: & & p_1-p_2 \gt 0  \end{eqnarray*}[/latex]

p-value: 

Before calculating the p-value, we check that the normal distribution applies:

[latex]\begin{eqnarray*} n_1 \times \hat{p}_1 & = & 1340 \times 0.1=134 \geq 5 \\ n_1 \times (1-\hat{p}_1) & = & 1340 \times (1-0.1)=1206 \geq 5 \\ n_2 \times \hat{p}_2 & = & 250 \times 0.05=15 \geq 5 \\ n_2 \times (1-\hat{p}_1) & = & 250 \times (1-0.05)=235 \geq 5 \end{eqnarray*}[/latex]

Because [latex]n_1 \times \hat{p}_1 \geq 5[/latex], [latex]n_1 \times (1-\hat{p}_1) \geq 5[/latex], [latex]n_2 \times \hat{p}_2 \geq 5[/latex] and [latex]n_2 \times (1-\hat{p}_2) \geq 5[/latex], the normal distribution applies and so we use a normal distribution to calculate the p-value.  Because the alternative hypothesis is a [latex]\gt[/latex], the p-value is the area in the right tail of the distribution.

The pooled sample proportion is:

[latex]\begin{eqnarray*} \overline{p} & = & \frac{x_1+x_2}{n_1+n_2} \\ & = & \frac{134+15}{1340+250} \\ & = & \frac{149}{1590} \\ & = & 0.09371... \end{eqnarray*}[/latex]

So the p-value[latex]=0.0232[/latex].

Conclusion:

Because p-value[latex]=0.0232 \lt 0.05=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis.  At the 5% significance level there is enough evidence to suggest that the proportion of adults 30 years old or younger who own an iPhone is greater than the proportion of adults over the age of 30 who own an iPhone.

Two types of medication for hives are tested to determine if there is a difference in the proportions of adult patient reactions.  In a sample of 200 adults given medication A, 20 still had hives 30 minutes after taking the medication.  In a sample of 200 adults given medication B, 12 still had hives 30 minutes after taking the medication.  At the 1% significance level, is there a difference in the proportion of adults who still have hives 30 minutes after taking medications?

Solution:

Let medication A be population 1 and medication B be population 2.  From the question, we have the following information:

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & p_1-p_2=0 \\ H_a: & & p_1-p_2 \neq 0  \end{eqnarray*}[/latex]

p-value: 

Before calculating the p-value, we check that the normal distribution applies:

[latex]\begin{eqnarray*} n_1 \times \hat{p}_1 & = & 200 \times 0.1=20 \geq 5 \\ n_1 \times (1-\hat{p}_1) & = & 200 \times (1-0.1)=180 \geq 5 \\ n_2 \times \hat{p}_2 & = & 200 \times 0.06=12 \geq 5 \\ n_2 \times (1-\hat{p}_1) & = & 200 \times (1-0.06)=188 \geq 5 \end{eqnarray*}[/latex]

Because [latex]n_1 \times \hat{p}_1 \geq 5[/latex], [latex]n_1 \times (1-\hat{p}_1) \geq 5[/latex], [latex]n_2 \times \hat{p}_2 \geq 5[/latex] and [latex]n_2 \times (1-\hat{p}_2) \geq 5[/latex], the normal distribution applies and so we use a normal distribution to calculate the p-value. Because the alternative hypothesis is a [latex]\neq[/latex], the p-value is the sum of the area in the two tails of the distribution.

We need to know if the sample information relates to the left or right tail because that will determine how we calculate out the area of that tail using the normal distribution.  In this case, [latex]\hat{p}_1 \gt \hat{p}_2[/latex] ([latex]0.1 \gt 0.06[/latex]), so the sample information relates to the right tail of the normal distribution.  This means that we will calculate out the area in the right tail using 1-norm.dist.  However, this is a two-tailed test where the p-value is the sum of the area in the two tails and the area in the right tail is only one half of the p-value.  The area in the right tail equals the area in the left tail and the p-value is the sum of these two areas.

The pooled sample proportion is:

[latex]\begin{eqnarray*} \overline{p} & = & \frac{x_1+x_2}{n_1+n_2} \\ & = & \frac{20+12}{200+200} \\ & = & \frac{32}{400} \\ & = & 0.08 \end{eqnarray*}[/latex]

So the area in the right tail is 0.0702, which means [latex]\frac{1}{2}[/latex](p-value)[latex]=0.0702[/latex].  This is also the area in the left tail, so

p-value[latex]=0.0702+0.0702=0.1404[/latex]

Conclusion:

Because p-value[latex]=0.1404 \gt 0.01=\alpha[/latex], we do not reject the null hypothesis.  At the 1% significance level there is not enough evidence to suggest that the there is a difference in the proportion of adults who still have hives 30 minutes after taking medication.

A valve manufacturer recently launched a new valve, Valve A, and they want to claim that the proportion of their valves that fail under 4500 psi is the smallest of all the other valves on the market.  The manufacturer decides to compare Valve A with the most popular valve on the market, Valve B.  In a sample of 100 Valve A’s, 6 failed at 4500 psi.  In a sample of 150 Valve B’s, 16 failed at 4500 psi.  At the 5% significance level, is the proportion of Valve As that fail under 4500 psi less than the proportion of Valve Bs that fail under 4500 psi?

Solution:

Let Valve A be population 1 and Valve B be population 2.  From the question, we have the following information:

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & p_1-p_2=0 \\ H_a: & & p_1-p_2 \lt 0  \end{eqnarray*}[/latex]

p-value: 

Before calculating the p-value, we check that the normal distribution applies:

[latex]\begin{eqnarray*} n_1 \times \hat{p}_1 & = & 100 \times 0.06=6 \geq 5 \\ n_1 \times (1-\hat{p}_1) & = & 100 \times (1-0.06)=94 \geq 5 \\ n_2 \times \hat{p}_2 & = & 150 \times 0.1066....=16 \geq 5 \\ n_2 \times (1-\hat{p}_1) & = & 150 \times (1-0.1066...)=134 \geq 5 \end{eqnarray*}[/latex]

Because [latex]n_1 \times \hat{p}_1 \geq 5[/latex], [latex]n_1 \times (1-\hat{p}_1) \geq 5[/latex], [latex]n_2 \times \hat{p}_2 \geq 5[/latex] and [latex]n_2 \times (1-\hat{p}_2) \geq 5[/latex], the normal distribution applies and so we use a normal distribution to calculate the p-value.  Because the alternative hypothesis is a [latex]\lt[/latex], the p-value is the area in the left tail of the distribution.

The pooled sample proportion is:

[latex]\begin{eqnarray*} \overline{p} & = & \frac{x_1+x_2}{n_1+n_2} \\ & = & \frac{6+16}{100+150} \\ & = & \frac{22}{250} \\ & = & 0.088 \end{eqnarray*}[/latex]

So the p-value[latex]=0.1010[/latex].

Conclusion:

Because p-value[latex]=0.1010 \gt 0.05=\alpha[/latex], we do not reject the null hypothesis.  At the 5% significance level there is not enough evidence to suggest that the proportion of Valve As that fail under 4500 psi less than the proportion of Valve Bs that fail under 4500 psi.