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.

Ground vs. Helicopter for Serious Injuries A study investigated rates of fatalities among patients with serious traumatic injuries. Among 61,909 patients transported by helicopter, 7813 died. Among 161,566 patients transported by ground services, 17,775 died (based on data from “Association Between Helicopter vs Ground Emergency Medical Services and Survival for Adults With Major Trauma,” by Galvagno et al., Journal of the American Medical Association, Vol. 307, No. 15). Use a 0.01 significance level to test the claim that the rate of fatalities is higher for patients transported by helicopter.

a. Test the claim using a hypothesis test.

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

c. Considering the test results and the actual sample rates, is one mode of transportation better than the other? Are there other important factors to consider?

In a sample of 61,909 patients transported by helicopter, 7813 died.

In another sample of 161,566 patients transported by ground services, 17,775 died.

It is claimed that the proportion of fatalities is higher for the patients transported by helicopter as compared to the patients transported by ground services.

Null hypothesis:The rate of fatalities is the same for the patients transported by helicopter and patients transported by ground services.

\({H_0}:{p_1} = {p_2}\)

Alternate hypothesis: The rate of fatalities is higher for the patients transported by helicopter than for the patients transported by ground services.

\({\rm{ }}{H_1}:{p_1} > {p_2}\)

Let\({n_1}\)be the sample size of patients transported by helicopter.

\({n_1} = 61909\)

Let\({n_2}\)be the sample size of patients transported by ground services.

\({n_2} = 161566\)

Assume that,\({x_1}\)and\({x_2}\)are the number of patients who died and were transported by helicopter and ground services respectively.

Let \({\hat p_1}\)be the samplerate offatalities for patients transported by helicopter.

Thus,

\(\begin{array}{c}{{\hat p}_1} = \frac{{{x_1}}}{{{n_1}}}{\rm{ }}\\ = \frac{{7813}}{{61909}}\\ = 0.1262\end{array}\)

\(\begin{array}{c}{{\hat q}_1} = 1 - {{\hat p}_1}\\ = 0.8738\end{array}\)

Let \({\hat p_2}\)be the sample rate offatalities for patients transported by ground services.

Thus,

\(\begin{array}{c}{{\hat p}_2} = \frac{{{x_2}}}{{{n_2}}}{\rm{ }}\\ = \frac{{17775}}{{161566}}\\ = 0.11002\end{array}\)

\(\begin{array}{c}{{\hat q}_2} = 1 - {{\hat p}_2}\\ = 0.88998\end{array}\)

The value of the pooled sample proportion is equal to:

\(\begin{array}{c}\bar p = \frac{{{x_1} + {x_2}}}{{{n_1} + {n_2}}}\\ = \frac{{7813 + 17775}}{{61909 + 161566}}\\ = 0.1145\end{array}\)

Hence,

\(\begin{array}{c}\bar q = 1 - \bar p\\ = 1 - 0.1145\\ = 0.8855\end{array}\)

The test statistic is equal to:

\(\begin{array}{c}z = \frac{{\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - \left( {{p_1} - {p_2}} \right)}}{{\sqrt {\frac{{\bar p\bar q}}{{{n_1}}} + \frac{{\bar p\bar q}}{{{n_2}}}} }}\\ = \frac{{\left( {0.1262 - 0.11002} \right) - 0}}{{\sqrt {\frac{{\left( {0.1145} \right)\left( {0.8855} \right)}}{{61909}} + \frac{{\left( {0.1145} \right)\left( {0.8855} \right)}}{{161566}}} }}\\ = 10.7532\end{array}\)

Referring to the standard normal distribution table, the critical value of z corresponding to\(\alpha = 0.01\)for a right-tailed test is equal to 2.33.

The p-value of the z-score equal to 10.7532 is equal to 0.000.

a.

Since the p-value is less than 0.01 and the test statistic value is greater than the critical value, the null hypothesis is rejected.

There is sufficient evidence to support the claimthat the rate of fatalities is higher for patients transported by helicopter than for patients transported by ground services.

If the level of significance for a one-tailed test is equal to 0.01, then the corresponding confidence level to construct the confidence interval is equal to 98%.

The expression of the confidence interval is as follows:

\(\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E < {p_1} - {p_2} < \left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E\)

The value of \({z_{\frac{\alpha }{2}}}\) when \(\alpha = 0.02\) is equal to 2.33.

E is margin of error and has the following formula:

\(\begin{array}{c}E = {z_{\frac{\alpha }{2}}}\sqrt {\frac{{{{\hat p}_1}{{\hat q}_1}}}{{{n_1}}} + \frac{{{{\hat p}_2}{{\hat q}_2}}}{{{n_2}}}} \\ = 2.33 \times \sqrt {\frac{{\left( {0.1262} \right)\left( {0.8738} \right)}}{{61909}} + \frac{{\left( {0.11} \right)\left( {0.89} \right)}}{{161566}}} \\ = 0.0036\end{array}\)

b.

Substituting the required values, the following confidence interval is obtained:

\(\begin{array}{c}\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E < {p_1} - {p_2} < \left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E\\(0.1262 - 0.11002) - 0.0036 < {p_1} - {p_2} < (0.1262 - 0.11002) + 0.0036\\0.0126 < {p_1} - {p_2} < 0.0198\end{array}\)

Thus, the 98% confidence interval is equal to (0.0126, 0.0198).

This confidence interval does not contain zero. This implies means there is a significant difference between the two population proportions.

Therefore, there is sufficient evidence to support the claimthatthe rate of fatalities is higher for patients transported by helicopter than for patients transported by ground services.

c.

The sample proportion of fatalities for patients transported by helicopter is approximately equal to 12.62%, and the sample proportion of fatalities for patients transported by ground services is approximately equal to 11.00%.

Hence, considering the test results and actual sample rates, it appears that ground services are a better mode of transportation than a helicopter.

Also, there are many important factors to consider, like the distance between health facilities and where the injury has happened.