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.

Are Seat Belts Effective? A simple random sample of front-seat occupants involved in car crashes is obtained. Among 2823 occupants not wearing seat belts, 31 were killed. Among 7765 occupants wearing seat belts, 16 were killed (based on data from “Who Wants Airbags?” by Meyer and Finney, Chance, Vol. 18, No. 2). We want to use a 0.05 significance level to test the claim that seat belts are effective in reducing fatalities.

a. Test the claim using a hypothesis test.

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

c. What does the result suggest about the effectiveness of seat belts?

Of the simple random sample of car crashes, the number of occupants wearing seat belts is 7765, and 2823 were not wearing seat belts.

In the group wearing seat belts, 31 were killed, and 16 were killed in the other groups.

The significance level is \(\alpha = 0.05\) to test the claim that seat belts are effective in reducing fatalities.

a.

Let\({p_1},{p_2}\)be the actual proportion of fatalities in the population of occupants not wearing the seat belts and wearing the seat belts, respectively.

To test the effectiveness of wearing seat belts, the hypotheses are formulated as follows.

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

Among 2823 occupants not wearing seat belts, 31 were killed.

Among 7765 occupants wearing seat belts,16 were killed.

From the above information, we get the following:

\(\begin{array}{l}{n_1} = 2823\,\\{x_1} = 31\,\,\\{n_2} = 7765\\{x_2} = 16\end{array}\)

The sample proportions are as follows.

\(\begin{array}{c}{{\hat p}_1} = \frac{{{x_1}}}{{{n_1}}}\\ = \frac{{31}}{{2823}}\\ = 0.011\end{array}\)

and

\(\begin{array}{c}{{\hat p}_2} = \frac{{{x_2}}}{{{n_2}}}\\ = \frac{{16}}{{7765}}\\ = 0.0021\end{array}\).

The sample pooled proportions are calculated as follows.

\(\begin{array}{c}\bar p = \frac{{\left( {{x_1} + {x_2}} \right)}}{{\left( {{n_1} + {n_2}} \right)}}\,\\ = \frac{{\left( {31 + 16} \right)}}{{\left( {2823 + 7765} \right)}}\\ = 0.0044\end{array}\)

and

\(\begin{array}{c}\bar q = 1 - \bar p\\ = 1 - 0.00444\\ = 0.9956\end{array}\).

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

\(z = \frac{{\left( {{{{\rm{\hat p}}}_1} - {{{\rm{\hat p}}}_2}} \right) - \left( {{{\rm{p}}_{\rm{1}}} - {{\rm{p}}_{\rm{2}}}} \right)}}{{\sqrt {\left( {\frac{{{\rm{\bar p\bar q}}}}{{{{\rm{n}}_{\rm{1}}}}} + \frac{{{\rm{\bar p\bar q}}}}{{{{\rm{n}}_{\rm{2}}}}}} \right)} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)\)

Here, \({\rm{\bar p}}\)is the pooled sample proportion, and\({\rm{\bar q}} = 1 - {\rm{\bar p}}\).

Substitute the obtained value in equation (2) as follows.

\(\begin{array}{c}z = \frac{{\left( {{{{\rm{\hat p}}}_1} - {{{\rm{\hat p}}}_2}} \right) - \left( {{{\rm{p}}_{\rm{1}}} - {{\rm{p}}_{\rm{2}}}} \right)}}{{\sqrt {\left( {\frac{{{\rm{\bar p\bar q}}}}{{{{\rm{n}}_{\rm{1}}}}} + \frac{{{\rm{\bar p\bar q}}}}{{{{\rm{n}}_{\rm{2}}}}}} \right)} }}\\ = \frac{{\left( {0.011 - 0.0021} \right) - 0}}{{\sqrt {\left( {\frac{{0.0044 \times 0.9956}}{{2823}} + \frac{{0.0044 \times 0.9956}}{{7765}}} \right)} }}\\ = 6.11\end{array}\)

The value of the test statistic is 6.11.

Referring to the standard normal table for a positive z-score of 0.9999, the cumulative probability of 6.11 is obtained from the cell intersection for rows 3.50 and above and the column value of 0.00.

That is,\(P\left( {Z < 6.11} \right) = 0.9999\).

For the right-tailed test, the p-value is the area to the right of the test statistic, which is

\(\begin{array}{c}P\left( {Z > 6.11} \right) = 1 - P\left( {Z < 6.11} \right)\\ = 1 - 0.9999\\ = 0.0001\end{array}\).

Thus, the p-value is 0.0001.

As the p-value is less than 0.05, the null hypothesis is rejected at a 0.05 significance level. It can be concluded that there is sufficient evidence to support the claim that seat belts are effective as the proportion of fatalities is higher when occupants are not wearing seatbelts.

b.

The 0.05 significance level for a one-tailed test implies a confidence level of 90%.

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

\({\rm{Confidence}}\,\,{\rm{interval}} = \left( {\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E\,\,,\,\,\left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E} \right)\,\,\,\,\,\,\,\,...\left( 1 \right)\)\(\)

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

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

For computing the confidence interval, first, find the critical value\({z_{\frac{\alpha }{2}}}\).

Thus, the value of the level of significance for the confidence interval becomes\(\alpha = 0.10\).

Hence,

\(\begin{array}{c}\frac{\alpha }{2} = \frac{{0.10}}{2}\\ = 0.05\end{array}\).

The value of\({z_{\frac{\alpha }{2}}}\)from the standard normal table is equal to 1.645.

The margin of error is computed as follows.

\(\begin{array}{c}E = {z_{\frac{\alpha }{2}}} \times \sqrt {\left( {\frac{{{{\hat p}_1} \times {{\hat q}_1}}}{{{n_1}}} + \frac{{{{\hat p}_2} \times {{\hat q}_2}}}{{{n_2}}}} \right)} \\ = 1.64 \times \sqrt {\left( {\frac{{0.011 \times 0.989}}{{2823}} + \frac{{0.0021 \times 0.9979}}{{7765}}} \right)} \\ = 0.00333\end{array}\).

Substitute the value of E in equation (1) as follows.

\(\begin{array}{c}{\rm{Confidence}}\,\,{\rm{interval}} = \left( {\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E\,\,,\,\,\left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E} \right)\\ = \left( {\left( {0.011 - 0.0021} \right) - 0.00399\,,\,\,\left( {0.011 - 0.0021} \right) + 0.00399} \right)\\ = \left( {0.0089 - 0.00333\,\,,\,0.0089 + 0.00333} \right)\\ = \left( {0.00559,\,0.0123} \right)\end{array}\).

Thus, the 90% confidence interval for two proportions is \(0.00559 < \left( {{p_1} - {p_2}} \right) < 0.0123\).

The 90% confidence interval does not include the hypothesized value 0, which implies that the null hypothesis is rejected.

It can be concluded that there is sufficient evidence to support the claim that seat belts are effective as the proportion of fatalities is higher when occupants are not wearing seatbelts.

c.

From the testing of the hypothesis and using the confidence interval, the fatality rate differs in each group. The interval shows the positive range estimate, which implies that the proportion of fatalities without wearing seat belts is larger than the proportion wearing seat belts.