Testing Claims About Proportions. In Exercises 7–22, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.

Cell Phones and Handedness A study was conducted to investigate the association between cell phone use and hemispheric brain dominance. Among 216 subjects who prefer to use their left ear for cell phones, 166 were right-handed. Among 452 subjects who prefer to use their right ear for cell phones, 436 were right-handed (based on data from “Hemi- spheric Dominance and Cell Phone Use,” by Seidman et al., JAMA Otolaryngology—Head & Neck Surgery, Vol. 139, No. 5). We want to use a 0.01 significance level to test the claim that the rate of right-handedness for those who prefer to use their left ear for cell phones is less than the rate of right-handedness for those who prefer to use their right ear for cell phones. (Try not to get too confused here.)

a. Test the claim using a hypothesis test.

b. Test the claim by constructing an appropriate confidence interval.

The statistics for the two groups:

• Among the 216 who prefer the left ear for cell phones, 166 are right-handed.
• Among the 452 who prefer the right ear for cell phones, 436 are right-handed.

The significance level is \(\alpha = 0.01\) to test the proportion of right-handed individuals who prefer the left ear less than the ones who prefer the right ear for cellphones.

Let\({p_1},{p_2}\)be the actual proportion of subjects who are right-handed among the subjects who prefer the left and right hands for cellphones, respectively.

Using the claim, the hypotheses are as follows.

\(\begin{array}{l}{H_0}:{p_1} = {p_2}\\\,{H_1}:{p_1} < {\rm{ }}{p_2}\end{array}\)

From the given information, summarize as follows.

\(\begin{array}{c}{{\rm{n}}_1} = 216\,\\{{\rm{x}}_1} = 166\\\,{{\rm{n}}_2} = 452\,\\{{\rm{x}}_2} = 436\end{array}\)

The sample proportions are

\(\begin{array}{c}{{\hat p}_1} = \frac{{{x_1}}}{{{n_1}}}\\ = \frac{{166}}{{216}}\\ = 0.7685\end{array}\)

and

\(\begin{array}{c}{{\hat p}_2} = \frac{{{x_2}}}{{{n_2}}}\\ = \frac{{436}}{{452}}\\ = 0.9646\end{array}\).

The sample pooled proportions are calculated as

\(\begin{array}{c}\bar p = \frac{{\left( {{x_1} + {x_2}} \right)}}{{\left( {{n_1} + {n_2}} \right)}}\,\\ = \frac{{\left( {166 + 436} \right)}}{{\left( {216 + 452} \right)}}\\ = 0.9012\end{array}\)

And

\(\begin{array}{c}{\rm{\bar q}} = 1 - {\rm{\bar p}}\\ = 1 - 0.9012\\ = 0.0988\end{array}\).

To conduct a hypothesis test of two proportions, the test statistic is computed as follows.

\(z = \frac{{\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - \left( {{p_1} - {p_2}} \right)}}{{\sqrt {\left( {\frac{{\bar p\bar q}}{{{n_1}}} + \frac{{\bar p\bar q}}{{{n_2}}}} \right)} }}\,\,\)

Substitute the values. So,

\(\begin{array}{c}z = \frac{{\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - \left( {{p_1} - {p_2}} \right)}}{{\sqrt {\left( {\frac{{\bar p\bar q}}{{{n_1}}} + \frac{{\bar p\bar q}}{{{n_2}}}} \right)} }}\\ = \frac{{\left( {0.7685 - 0.9646} \right) - 0}}{{\sqrt {\left( {\frac{{0.9012 \times 0.0988}}{{216}} + \frac{{0.9012 \times 0.0988}}{{452}}} \right)} }}\\ = - 7.944\end{array}\).

The value of the test statistic is -7.94.

Referring to the standard normal table for the negative z-score of 0.0001, the cumulative probability of -7.94 is obtained from the cell intersection for rows -3.50 and above and the column value of 0.00.

For the left-tailed test, the p-value is the area to the left of the test statistic. That is,

\(P\left( {z < - 7.94} \right) = 0.0001\).

Thus, the p-value is 0.0001.

As the\(P - value = 0.0001 < \alpha = 0.01\), it is concluded that the null hypothesis is rejected.

Thus, it is concluded that there is enough evidence to support the claim that the proportion of right-handed individuals who prefer the left ear is less than that of the ones who prefer the right ear for cell phones.

b.

The general formula for the confidence interval of the difference of proportions is as follows.

\({\rm{Confidence}}\,\,{\rm{Interval}} = \left( {\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E\,\,,\,\,\left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E} \right)\)\(\)

Here, E is the margin of error, which is calculated as follows.

\(E = {z_{\frac{\alpha }{2}}} \times \sqrt {\left( {\frac{{{{\hat p}_1} \times {{\hat q}_1}}}{{{n_1}}} + \frac{{{{\hat p}_2} \times {{\hat q}_2}}}{{{n_2}}}} \right)} \)

The confidence level is 98% if the level of significance used in the one-tailed test is 0.01.

Thus, the value of the level of significance for the confidence interval becomes\(\alpha = 0.02\).

Hence,

\(\begin{array}{c}\frac{\alpha }{2} = \frac{{0.02}}{2}\\ = 0.01\end{array}\).

The value of\({z_{\frac{\alpha }{2}}}\)from the standard normal table is equal to 2.33.

The margin of error E is computed as follows.

\(\begin{array}{c}E = {z_{\frac{\alpha }{2}}} \times \sqrt {\left( {\frac{{{{\hat p}_1} \times {{\hat q}_1}}}{{{n_1}}} + \frac{{{{\hat p}_2} \times {{\hat q}_2}}}{{{n_2}}}} \right)} \\ = 2.33 \times \sqrt {\left( {\frac{{0.7685 \times 0.2315}}{{216}} + \frac{{0.9646 \times 0.0354}}{{452}}} \right)} \\ = 0.0699\end{array}\).

Substitute the value of E as follows.

\(\begin{array}{c}{\rm{Confidence}}\,\,{\rm{Interva}}l = \left( {\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E\,\,,\,\,\left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E} \right)\\ = \left( {\left( {0.7685 - 0.9646} \right) - 0.0698\,,\,\left( {0.7685 - 0.9646} \right) + 0.0698} \right)\\ = \left( { - 0.266\,\,,\,\, - 0.126} \right)\end{array}\).

Thus, the 98% confidence interval for two proportions is \( - 0.266 < \left( {{{\rm{p}}_1} - {{\rm{p}}_2}} \right) < - 0.126\) .

The confidence interval does not include 0, and thus, the null hypothesis is rejected.

Thus, there is enough evidence to conclude that the claim is supported.