Problem

Construct a confidence interval for $p_{1}-p_{2}$ at the given level of confidence. \[ x_{1}=371, n_{1}=515, x_{2}=439, n_{2}=587,90 \% \text { confidence } \] The researchers are $\square \%$ confident the difference between the two population proportions, $p_{1}-p_{2}$, is between $\square$ and $\square$. (Use ascending order. Type an integer or decimal rounded to three decimal places as needed.) on Viewer le Get more help . Clear all

Solution

Step 1 :Given values are \(x_{1} = 371\), \(n_{1} = 515\), \(x_{2} = 439\), \(n_{2} = 587\) and the confidence level is 90%.

Step 2 :Calculate the sample proportions \(p_{1}\) and \(p_{2}\) using the formulas \(p_{1} = \frac{x_{1}}{n_{1}}\) and \(p_{2} = \frac{x_{2}}{n_{2}}\). This gives \(p_{1} = 0.720\) and \(p_{2} = 0.748\).

Step 3 :Calculate the standard error (se) using the formula \(se = \sqrt{\frac{p_{1}(1 - p_{1})}{n_{1}} + \frac{p_{2}(1 - p_{2})}{n_{2}}}\). This gives \(se = 0.027\).

Step 4 :Find the z-score for the given confidence level. The z-score for a 90% confidence level is 1.645.

Step 5 :Calculate the confidence interval using the formulas \(ci_{lower} = (p_{1} - p_{2}) - z \times se\) and \(ci_{upper} = (p_{1} - p_{2}) + z \times se\). This gives \(ci_{lower} = -0.071\) and \(ci_{upper} = 0.016\).

Step 6 :\(\boxed{\text{Final Answer: The researchers are 90% confident the difference between the two population proportions, } p_{1}-p_{2}, \text{ is between } -0.071 \text{ and } 0.016.}\)

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