Let's solve the problem step by step: --- ## **Problem Restatement** - **Objective:** Judge the effectiveness of an ad campaign for a new product. - **Method:** Compare customer purchase proportions **before** and **after** the campaign. - **Sample Sizes:** - Before: 96 customers, 78 purchased - After: 78 customers, all 78 purchased --- ## **Step 1: State the Proportions** Let: \( n_1 = \) sample size before = **96** \( x_1 = \) number who purchased before = **78** \( n_2 = \) sample size after = **78** \( x_2 = \) number who purchased after = **78** Calculate sample proportions: \[ \hat{p}_1 = \frac{x_1}{n_1} = \frac{78}{96} = .8125 \] \[ \hat{p}_2 = \frac{x_2}{n_2} = \frac{78}{78} = 1.00 \] --- ## **Step 2: State the Difference in Proportions** \[ \hat{p}_2 - \hat{p}_1 = 1.00 - .8125 = .1875 \] --- ## **Step 3: Find the Standard Error (SE) of the Difference** The standard error for the difference in two proportions is: \[ SE = \sqrt{ \frac{\hat{p}_1 (1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2 (1 - \hat{p}_2)}{n_2} } \] Plug in the numbers: - For before: \(\hat{p}_1 = .8125\), \(n_1 = 96\) - For after: \(\hat{p}_2 = 1.00\), \(n_2 = 78\) \[ SE = \sqrt{ \frac{.8125 \times (1 - .8125)}{96} + \frac{1.00 \times (1 - 1.00)}{78} } \] \[ = \sqrt{ \frac{.8125 \times .1875}{96} + } \] \[ = \sqrt{ \frac{.15234375}{96} } \] \[ = \sqrt{.001587} \] \[ = .03987 \] --- ## **Step 4: Find the Critical Value (z\*) for 95% Confidence** For 95% confidence, \(z^* = 1.96\) --- ## **Step 5: Calculate the Confidence Interval** \[ \text{CI} = (\hat{p}_2 - \hat{p}_1) \pm z^* \times SE \] \[ = .1875 \pm 1.96 \times .03987 \] \[ = .1875 \pm .0782 \] So, - **Lower limit:** \(.1875 - .0782 = .1093\) - **Upper limit:** \(.1875 + .0782 = .2657\) --- ## **Step 6: Write the Final Answer** A 95% confidence interval for the difference in proportions is: \[ (.11, .27) \] (rounded to two decimal places) --- ## **Final Answer** **The 95% confidence interval for the difference in proportions between before and after the ad campaign is:** \[ \boxed{(.11, .27)} \] --- **Explanation:** - This interval does NOT contain , suggesting a statistically significant increase in the proportion of customers purchasing the product after the advertising campaign.

# Solution Verification and Explanation ## **Problem Restatement** - **Objective:** Judge the effectiveness of an ad campaign for a new product. - **Method:** Compare customer purchase proportions **before** and **after** the campaign. - **Sample Sizes:** - Before: 96 customers, 78 purchased - After: 78 customers, all 78 purchased --- ## **Step 1: State the Proportions** Let: - \( n_1 = \) sample size before = **96** - \( x_1 = \) number who purchased before = **78** - \( n_2 = \) sample size after = **78** - \( x_2 = \) number who purchased after = **78** ### **Calculate Sample Proportions** 1. **Before the Campaign:** \[ \hat{p}_1 = \frac{x_1}{n_1} = \frac{78}{96} \approx 0.8125 \] **Explanation:** This indicates that approximately 81.25% of customers purchased the product before the campaign. 2. **After the Campaign:** \[ \hat{p}_2 = \frac{x_2}{n_2} = \frac{78}{78} = 1.00 \] **Explanation:** This indicates that 100% of customers purchased the product after the campaign. --- ## **Step 2: State the Difference in Proportions** ### **Calculate the Difference** \[ \hat{p}_2 - \hat{p}_1 = 1.00 - 0.8125 = 0.1875 \] **Explanation:** The difference of 0.1875 suggests a significant increase in purchase proportion after the campaign. --- ## **Step 3: Find the Standard Error (SE) of the Difference** ### **Standard Error Formula** \[ SE = \sqrt{ \frac{\hat{p}_1 (1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2 (1 - \hat{p}_2)}{n_2} } \] ### **Plug in the Numbers** 1. For before: \(\hat{p}_1 = 0.8125\), \(n_1 = 96\) 2. For after: \(\hat{p}_2 = 1.00\), \(n_2 = 78\) ### **Calculate Each Component** \[ SE = \sqrt{ \frac{0.8125 \times (1 - 0.8125)}{96} + \frac{1.00 \times (1 - 1.00)}{78} } \] \[ = \sqrt{ \frac{0.8125 \times 0.1875}{96} + \frac{0}{78} } \] \[ = \sqrt{ \frac{0.15234375}{96} } \] \[ = \sqrt{0.001587} \approx 0.03987 \] **Explanation:** The standard error measures the variability of the difference in proportions, which is approximately 0.03987. --- ## **Step 4: Find the Critical Value (z\*) for 95% Confidence** ### **Critical Value** For a 95% confidence level: \[ z^* = 1.96 \] **Explanation:** This value is derived from the standard normal distribution for a two-tailed test. --- ## **Step 5: Calculate the Confidence Interval** ### **Confidence Interval Formula** \[ \text{CI} = (\hat{p}_2 - \hat{p}_1) \pm z^* \times SE \] ### **Calculate the Interval** \[ = 0.1875 \pm 1.96 \times 0.03987 \] \[ = 0.1875 \pm 0.0782 \] #### **Lower and Upper Limits** - **Lower limit:** \(0.1875 - 0.0782 \approx 0.1093\) - **Upper limit:** \(0.1875 + 0.0782 \approx 0.2657\) **Explanation:** This confidence interval indicates where the true difference in proportions is likely to fall. --- ## **Step 6: Write the Final Answer** ### **Final Confidence Interval** \[ \boxed{(0.11, 0.27)} \] **Explanation:** The 95% confidence interval for the difference in proportions suggests a statistically significant increase in purchase likelihood after the ad campaign, as the interval does not include zero. --- ## **Conclusion** The calculations have been verified, and the steps clearly outline the reasoning and methods used. The increase in purchase proportion after the ad campaign can be considered statistically significant based on the confidence interval obtained.