Questions: In a comparative study of two new drugs, A and B, 375 patients were treated with drug A, and 275 patients were treated with drug B. (The two treatment groups were randomly and independently chosen.) It was found that 266 patients were cured using drug A and 198 patients were cured using drug B. Let p1 be the proportion of the population of all patients who are cured using drug A, and let p2 be the proportion of the population of all patients who are cured using drug B. Find a 95% confidence interval for p1-p2. Then find the lower limit and upper limit of the 95% confidence interval. Carry your intermediate computations to at least three decimal places. Round your responses to at least three decimal places. (If necessary, consult a list of formulas.) Lower limit: Upper limit:

In a comparative study of two new drugs, A and B, 375 patients were treated with drug A, and 275 patients were treated with drug B. (The two treatment groups were randomly and independently chosen.) It was found that 266 patients were cured using drug A and 198 patients were cured using drug B. Let p1 be the proportion of the population of all patients who are cured using drug A, and let p2 be the proportion of the population of all patients who are cured using drug B. Find a 95% confidence interval for p1-p2. Then find the lower limit and upper limit of the 95% confidence interval.

Carry your intermediate computations to at least three decimal places. Round your responses to at least three decimal places. (If necessary, consult a list of formulas.)

Lower limit:
Upper limit:

Transcript text: In a comparative study of two new drugs, $A$ and $B, 375$ patients were treated with drug $A$, and 275 patients were treated with drug $B$. (The two treatment groups were randomly and independently chosen.) It was found that 266 patients were cured using drug A and 198 patients were cured using drug B . Let $p_{1}$ be the proportion of the population of all patients who are cured using drug A , and let $p_{2}$ be the proportion of the population of all patients who are cured using drug B. Find a $95 \%$ confidence interval for $p_{1}-p_{2}$. Then find the lower limit and upper limit of the $95 \%$ confidence interval. Carry your intermediate computations to at least three decimal places. Round your responses to at least three decimal places. (If necessary, consult a list of formulas.) \begin{tabular}{|ll|} \hline Lower limit: & $\square$ \\ Upper limit: & $\square$ \\ \hline \end{tabular}

Solution

Solution Steps

Step 1: Calculate Sample Proportions

For drug $ A $, the sample proportion of patients cured is given by:

\[ \hat{p}_1 = \frac{266}{375} \approx 0.709 \]

For drug $ B $, the sample proportion of patients cured is given by:

\[ \hat{p}_2 = \frac{198}{275} \approx 0.720 \]

Step 2: Determine the Confidence Interval for the Difference in Proportions

To find the $ 95\% $ confidence interval for the difference in proportions $ p_1 - p_2 $, we use the formula:

\[ (\hat{p}_1 - \hat{p}_2) \pm z \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}} \]

Where $ z $ is the critical value for $ 95\% $ confidence, which is approximately $ 1.96 $.

Calculating the difference in sample proportions:

\[ \hat{p}_1 - \hat{p}_2 \approx 0.709 - 0.720 = -0.011 \]

Next, we calculate the standard error:

\[ \sqrt{\frac{0.709(1 - 0.709)}{375} + \frac{0.720(1 - 0.720)}{275}} \approx \sqrt{\frac{0.709 \cdot 0.291}{375} + \frac{0.720 \cdot 0.280}{275}} \approx \sqrt{0.000610 + 0.000724} \approx \sqrt{0.001334} \approx 0.0365 \]

Now, we can compute the margin of error:

\[ 1.96 \cdot 0.0365 \approx 0.0715 \]

Thus, the confidence interval is:

\[ (-0.011) \pm 0.0715 \]

Calculating the lower and upper limits:

\[ \text{Lower limit} = -0.011 - 0.0715 \approx -0.0815 \] \[ \text{Upper limit} = -0.011 + 0.0715 \approx 0.0605 \]

Step 3: Round the Results

Rounding the limits to three decimal places gives:

\[ \text{Lower limit} \approx -0.081 \] \[ \text{Upper limit} \approx 0.061 \]