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.

Cardiac Arrest at Day and Night A study investigated survival rates for in hospital patients who suffered cardiac arrest. Among 58,593 patients who had cardiac arrest during the day, 11,604 survived and were discharged. Among 28,155 patients who suffered cardiac arrest at night, 4139 survived and were discharged (based on data from “Survival from In-Hospital Cardiac Arrest During Nights and Weekends,” by Puberty et al., Journal of the American Medical Association, Vol. 299, No. 7). We want to use a 0.01 significance level to test the claim that the survival rates are the same for day and night.

a. Test the claim using a hypothesis test.

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

c. Based on the results, does it appear that for in-hospital patients who suffer cardiac arrest, the survival rate is the same for day and night?

The two groups of patients are formed on the basis of the time of suffering cardiac arrest, which is day or night.

Among the 58593 who suffered during the day, 11604 survived and got discharged. Among the 28155 who suffered during the night, 4139 survived and got discharged.

The significance level is $\alpha =0.01$to test the claim that the survival rate is the same in both situations.

Let $p_1,p_2$be the population proportion of the survival rate for patients who suffered cardiac arrests during the day and night, respectively.

The following hypotheses are formulated for the claim that the survival rates are the same for the day and night:

$$\begin{gathered} H_0:p_1=p_2 \\ {H}_1:p_1\ne p_2 \end{gathered}$$

As per information for two groups,

The sample proportions for the two groups are as follows.

$$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{11604}{58593} \\ =0.1980 \end{gathered}$$

$$\begin{gathered} {\hat{p}}_2=\frac{x_2}{n_2} \\ =\frac{4139}{28155} \\ =0.1470 \end{gathered}$$

The sample pooled proportions are calculated as follows.

$$\begin{gathered} \overline{p}=\frac{\left(x_1+x_2\right)}{\left(n_1+n_2\right)} \\ =\frac{\left(11604+4139\right)}{\left(58593+28155\right)} \\ =0.1815 \end{gathered}$$

and

.$$\begin{gathered} \overline{q}=1-\overline{p} \\ =1-0.1815 \\ =0.8185 \end{gathered}$$

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

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

Here, $\overline{p}$is the pooled sample proportion, and $\overline{q}=1-\overline{p}$.

Substitute the values. So,

.$$\begin{gathered} z=\frac{\left({\hat{p}}_1-{\hat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\left(\frac{\overline{p}\overline{q}}{n_1}+\frac{\overline{p}\overline{q}}{n_2}\right)}} \\ =\frac{\left(0.1980-0.1470\right)-0}{\sqrt{\left(\frac{0.1815\times 0.8185}{58593}+\frac{0.1815\times 0.8185}{28155}\right)}} \\ =18.261 \end{gathered}$$

The value of the test statistic is 18.261.

Referring to the standard normal table for the positive z-score of 0.9999, the cumulative probability of 18.26 is obtained from the cell intersection for rows 3.50 and above and the column value 0.00.

That is,$P\left(Z<18.26\right)=0.9999$.

For the two-tailed test, the p-value is twice the area to the right of the test statistic.

$$\begin{gathered} 2P\left(Z>18.261\right)=2\times \left(1-P\left(Z<18.261\right)\right) \\ =2\times \left(1-0.9999\right) \\ =0.0002 \end{gathered}$$

Thus, the p-value is 0.0002.

As the p-value is less than the significance level of 0.01, the null hypothesis is rejected.

Hence, there is not enough evidence to support the claim that the survival rate is the same in the day and night.

b.

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

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

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

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

The confidence interval for the two-tailed test with a level of significance of 0.01 is 99%.

The critical value $z_{\frac{\alpha }{2}}$has the cumulative area to its left as $1-\frac{\alpha }{2}$.

Mathematically,

$$\begin{gathered} P\left(Z<z_{\frac{\alpha }{2}}\right)=1-\frac{\alpha }{2} \\ P\left(Z<z_{\frac{0.01}{2}}\right)=1-0.005 \\ P\left(Z<z_{0.005}\right)=0.995 \end{gathered}$$

From the standard normal table, the area 0.995 is observed corresponding to the intersection of the row value 2.5 and column values 0.07 and 0.08, which implies that the critical value is 2.576.

The margin of error is as follows.

$$\begin{gathered} E=z_{\frac{\alpha }{2}}\times \sqrt{\left(\frac{{\hat{p}}_1\times {\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2\times {\hat{q}}_2}{n_2}\right)} \\ =2.576\times \sqrt{\left(\frac{0.198\times 0.802}{58530}+\frac{0.147\times 0.853}{28155}\right)} \\ =0.0069 \end{gathered}$$

Substitute the value of E in equation (1).

$$\begin{gathered} \text{Confidence} \text{interval}=\left(\left({\hat{p}}_1-{\hat{p}}_2\right)-E , \left({\hat{p}}_1-{\hat{p}}_2\right)+E\right) \\ =\left(\left(0.198-0.147\right)-0.0069 , \left(0.198-0.147\right)+0.0069\right) \\ =\left(0.051-0.0069 , 0.051+0.0069\right) \\ =\left(0.0441 , 0.0579\right) \end{gathered}$$

Thus, the 99% confidence interval for two proportions is $0.0441<\left(p_1-p_2\right)<0.0579$.

As the value 0 is not included in the 99% confidence interval, it can be inferred that the survival rates are significantly different for day and night patients.

c.

As the confidence limits are positive, it can be inferred thatthe survival rates are high for patients who had a cardiac arrest during the day than the patients who had a cardiac arrest during the night.