A Wall Street Journal article, titled "Hypertension Drug Linked to Cancer," reported on a study of several types of high-blood-pressure drugs and links to cancer. For one type, called calcium-channel blockers, $27$of $202$elderly patients taking the drug developed cancer. For another type, called beta-blockers, $28$of $424$other elderly patients developed cancer. Find a $90\%$confidence interval for the difference between the cancer rates of elderly people taking calcium-channel blockers and those taking beta-blockers. Note: The results of this study were challenged and questioned by several sources that claimed, for example, that the study was flawed and that several other studies have suggested that calcium-channel blockers are safe.

Given in the question that, A Wall Street Journal article, titled "Hypertension Drug Linked to Cancer," reported on a study of several types of high-blood-pressure drugs and links to cancer. For one type, called calcium-channel blockers,$27$of $202$elderly patients taking the drug developed cancer. For another type, called beta-blockers,$28$of $424$other elderly patients developed cancer. we need to find a $90\%$confidence interval for the difference between the cancer rates of elderly people taking calcium-channel blockers and those taking beta-blockers.

The given values are, $x_1=27,n_1=202,x_2=28,n_2=424$, and $90\%$confidence interval.

The formula for ${\tilde{p}}_1$is given by,

${\tilde{p}}_1=\frac{x_1+1}{n_1+2}$

Substitute $x_1=27,n_1=202$

${\tilde{p}}_1=\frac{28+1}{424+2}$

$=0.068$

The formula for ${\tilde{p}}_2$is given by,

${\tilde{p}}_2=\frac{x_2+1}{n_2+2}$

${\tilde{p}}_2=\frac{366+1}{872+2}$

$=0.420$

Therefore, the sample proportions are $0.137$and $0.068.$

When both sample sizes are 5 or higher, the two proportions plus-four $z$confidence interval technique can be utilized.

Foe a confidence level of $(1-\alpha )$the plus-four $z$confidence interval for $P_1-P_2$is Calculate the value of $\alpha$,

$90=100(1-\alpha )$

$\alpha =0.1$

The value of$z$at $\alpha /2$from the $z$-score table is $1.645$.

For the difference between the two-population proportion, the needed confidence interval is determined as,

$\left({\tilde{p}}_1-{\tilde{p}}_2\right)\pm z_{\alpha /2}·\sqrt{\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1-{\tilde{p}}_2\right)}{n_2+2}}=(0.137-0.068)\pm 1.945$

role="math" localid="1651336957327" $.\sqrt{\frac{0.137(1-0.137)}{204}+\frac{0.068(1-0.068)}{426}}$

$=0.069\pm 0.566$

$=-0.497$to $0.635$

As a result, the difference between two proportions ranges from $-0.497$ to $0.635$.