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.

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.)

To construct a 99% confidence interval for the difference between the two cereals, we can use the formula:

CI = (p1 - p2) ± Z * sqrt( (p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2) )

Where:
p1 = proportion of consumers who would buy cereal A
p2 = proportion of consumers who would buy cereal B
n1 = sample size of the test panel for cereal A
n2 = sample size of the test panel for cereal B
Z = the z-score corresponding to the desired confidence level (99% in this case)

First, calculate the proportions of consumers who would buy each cereal:

p1 = 52/175 = 0.2971 (rounded to 4 decimal places)
p2 = 35/150 = 0.2333 (rounded to 4 decimal places)

Next, calculate the standard error:

SE = sqrt( (p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2) )
= sqrt( (0.2971 * (1 - 0.2971) / 175) + (0.2333 * (1 - 0.2333) / 150) )

Now, we need to find the z-score corresponding to the 99% confidence level. From a standard normal distribution table, the z-score for a 99% confidence level is approximately 2.576 (rounded to 3 decimal places).

Now we can calculate the margin of error:

ME = Z * SE
= 2.576 * sqrt( (0.2971 * (1 - 0.2971) / 175) + (0.2333 * (1 - 0.2333) / 150) )

Finally, we can construct the confidence interval:

CI = (p1 - p2) ± ME

Lower Limit = (0.2971 - 0.2333) - (2.576 * sqrt( (0.2971 * (1 - 0.2971) / 175) + (0.2333 * (1 - 0.2333) / 150) ))

Calculating the lower limit:

Lower Limit = 0.0638 - (2.576 * sqrt( (0.2971 * (1 - 0.2971) / 175) + (0.2333 * (1 - 0.2333) / 150) ))

Lower Limit = 0.0638 (rounded to 4 decimal places)

Therefore, the lower limit of the 99% confidence interval for the difference between the two cereals is 0.0638.

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

CI = (p1 - p2) ± Z * sqrt((p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2))

Where:
CI is the confidence interval
p1 and p2 are the proportions for cereal A and cereal B, respectively
Z is the critical value based on the desired confidence level (99% in this case)
n1 and n2 are the sample sizes for cereal A and cereal B, respectively

First, we calculate the proportions for each cereal:

p1 = 52 / 175 = 0.2971
p2 = 35 / 150 = 0.2333

Next, we determine the critical value, Z, for a 99% confidence level. Since it's a two-tailed test, we divide the desired confidence level (99%) by 2 to find the corresponding critical value from the standard normal distribution. In this case, Z is approximately 2.576.

Plugging in the values into the formula:

CI = (0.2971 - 0.2333) ± 2.576 * sqrt((0.2971 * (1 - 0.2971) / 175) + (0.2333 * (1 - 0.2333) / 150))

Simplifying the equation:

CI = 0.0638 ± 2.576 * sqrt(0.0018 + 0.0017)
= 0.0638 ± 2.576 * sqrt(0.0035)
= 0.0638 ± 2.576 * 0.0592
= 0.0638 ± 0.1525

Finally, we can calculate the lower limit of the confidence interval by subtracting the value obtained above from the difference between the proportions:

lower limit = (0.2971 - 0.2333) - 0.1525
= 0.0638 - 0.1525
= -0.0887

Therefore, the lower limit of the 99% confidence interval for the difference between the two cereals is approximately -0.0887.