Fifty two of a test panel of 175 consumers say that they would buy cereal A if it is presented on the market and 35 of another test panel of 150 consumers say that they would buy cereal B. Construct a 99% confidence interval for the difference between the two cereals. Give the lower limit.

Answer 1

To construct a confidence interval for the difference between the proportions, we can use the formula:

\[ \text{Confidence interval} = (\hat{p}_1 - \hat{p}_2) \pm Z \cdot \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \]

where:
- \(\hat{p}_1\) is the proportion of consumers who would buy cereal A
- \(\hat{p}_2\) is the proportion of consumers who would buy cereal B
- \(n_1\) is the sample size for cereal A
- \(n_2\) is the sample size for cereal B
- \(Z\) is the z-score corresponding to the desired confidence level (in this case, 99%)

Given information:
- \(\hat{p}_1 = \frac{52}{175}\)
- \(\hat{p}_2 = \frac{35}{150}\)
- \(n_1 = 175\)
- \(n_2 = 150\)
- Confidence level = 99%
- Using a z-table or calculator, the z-score corresponding to a 99% confidence level is approximately 2.58

Plugging in the values, the confidence interval becomes:

\[ \text{Confidence interval} = \left(\frac{52}{175} - \frac{35}{150}\right) \pm 2.58 \cdot \sqrt{\frac{\frac{52}{175}(1-\frac{52}{175})}{175} + \frac{\frac{35}{150}(1-\frac{35}{150})}{150}} \]

Simplifying the expression, we get:

\[ \text{Confidence interval} = 0.091 \pm 2.58 \cdot 0.047 \]

Calculating the lower limit of the confidence interval:

\[ \text{Lower limit} = 0.091 - 2.58 \cdot 0.047 \]

\[ \text{Lower limit} \approx 0.091 - 0.121 \]

\[ \text{Lower limit} \approx -0.030 \]

Therefore, the lower limit of the 99% confidence interval for the difference between the proportions is approximately -0.030.

(Note: Since confidence intervals cannot be negative, we can consider the lower limit as 0 when interpreting the result.)