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.

Bednets to Reduce Malaria In a randomized controlled trial in Kenya, insecticide-treated bednets were tested as a way to reduce malaria. Among 343 infants using bednets, 15 developed malaria. Among 294 infants not using bednets, 27 developed malaria (based on data from “Sustainability of Reductions in Malaria Transmission and Infant Mortality in Western Kenya with Use of Insecticide-Treated Bednets,” by Lindblade et al., Journal of the American Medical Association, Vol. 291, No. 21). We want to use a 0.01 significance level to test the claim that the incidence of malaria is lower for infants using bednets.

a. Test the claim using a hypothesis test.

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

c. Based on the results, do the bednets appear to be effective?

The trial is conducted to reduce malaria; the infants who developed malaria in 343 infants using bednets are 15 and the infants who developed malaria in 294 infants who did not use bednets are 27.

The significance level is \(\alpha = 0.01\) .

Let\({p_1},{p_2}\)be the actual proportion of incidences of malaria in infants using bednets and without using bednets, respectively.

Test the claim that theincidence of malaria is lower for infants using bednets using the following hypotheses.

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

From the given information,

\(\begin{array}{l}{n_1} = 343\,\\{x_1} = 15\\{n_2} = 294\,\\{x_2} = 27\end{array}\)

The sample proportions are

\(\begin{array}{c}{{\hat p}_1} = \frac{{{x_1}}}{{{n_1}}}\\ = \frac{{15}}{{343}}\\ = 0.0437\end{array}\]

and

\(\begin{array}{c}{{\hat p}_2} = \frac{{{x_2}}}{{{n_2}}}\\ = \frac{{27}}{{294}}\\ = 0.0918\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( {15 + 27} \right)}}{{\left( {343 + 294} \right)}}\\ = 0.0659\end{array}\]

and

\(\begin{array}{c}\bar q = 1 - \bar p\\ = 1 - 0.0659\\ = 0.9340\end{array}\).

To conduct a hypothesis test of two proportions, the test statistics 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( {0.0437 - 0.0918} \right) - 0}}{{\sqrt {\left( {\frac{{0.0659 \times 0.9340}}{{343}} + \frac{{0.0659 \times 0.9340}}{{294}}} \right)} }}\\ = - 2.439\end{array}\].

The value of the test statistic is -2.44.

Referring to the standard normal table for the negative z-score of 0.0073, the cumulative probability of -2.44 is obtained from the cell intersection for rows -2.4 and the column value 0.04.

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

which is

\(P\left( {z < - 2.44} \right) = 0.0073\].

Thus, the p-value is 0.0073.

As the p-value is less than 0.01, the null hypothesis is rejected. Thus, there is sufficient evidence to support the claim that the incidences of malaria are lower in infants using bednets.

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)} \]

For the one-tailed test at a 0.01 significance level, the associated confidence interval is 98%.

For the critical value\({z_{\frac{\alpha }{2}}}\], the cumulative area to its left is\(1 - \frac{\alpha }{2}\).

Mathematically,

\(\begin{array}{c}P\left( {Z < {z_{\frac{\alpha }{2}}}} \right) = 1 - \frac{\alpha }{2}\\P\left( {Z < {z_{\frac{{0.02}}{2}}}} \right) = 0.99\end{array}\)

Refer to the standard normal table for the critical value. The area of 0.99 corresponds to row 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{{0.0437 \times 0.9563}}{{343}} + \frac{{0.0918 \times 0.9082}}{{294}}} \right)} \\ = 0.0469\end{array}\].

Substitute the value of E in the equation for the confidence interval. So,

\(\begin{array}{c}{\rm{Confidence}}\,\,{\rm{Interval}} = \left( {\left( {{{{\rm{\hat p}}}_1} - {{{\rm{\hat p}}}_2}} \right) - {\rm{E}}\,\,{\rm{,}}\,\,\left( {{{{\rm{\hat p}}}_1} - {{{\rm{\hat p}}}_2}} \right){\rm{ + E}}} \right)\\ = \left( {\left( {0.0437 - 0.0918} \right) - 0.04685\,,\,\,\left( {0.0437 - 0.0918} \right) + 0.04685} \right)\\ = \left( { - 0.0013\,\,,\, - 0.0950} \right)\end{array}\].

Thus, the 98% confidence interval for two proportions is\( - 0.0950 < \left( {{{\rm{p}}_1} - {{\rm{p}}_2}} \right) < - 0.0012\)

As 0 does not belong to the interval, the null hypothesis is rejected. Thus, there is sufficient evidence to support the claim that the incidences of malaria are lower in infants using bednets.

c.

There is sufficient evidence to support the claim that the incidences of malaria are lower in infants using bednets.

From the results, it can be concluded that using bednets reduces the incidences of developing malaria significantly among infants as the interval is negative.