Question

Suppose that a restaurant chain claims that its bottles of ketchup contain 24

ounces of ketchup on average, with a standard deviation of 0.2 ounces. If you
took a sample of 49 bottles of ketchup, what would be the approximate 95%
confidence interval for the mean number of ounces of ketchup per bottle in
the sample?
A. 24 ± 0.229
B. 24±0.114
C. 24 +0.029
D. 24 ± 0.057

Answer 1

To calculate the 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample, use the formula CI = x ± (z * (s / sqrt(n))). Plugging in the values, the interval is 24 ± 0.057.

Step-by-step explanation:

To calculate the 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample, we can use the formula:

CI = x ± (z * (s / sqrt(n)))

Where:

• CI is the confidence interval
• x is the sample mean
• z is the z-value for the desired confidence level (1.96 for 95% confidence)
• s is the sample standard deviation
• n is the sample size

Plugging in the values from the question, we get:

CI = 24 ± (1.96 * (0.2 / sqrt(49))) = 24 ± 0.057

Therefore, the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample is 24 ± 0.057.