Question

Suppose that a restaurant chain claims that its bottles of ketchup contain 24 ounces of ketchup on average, with a standard deviation of 0.4 ounces. If you took a sample of the 49 bottles of ketchup what would be the approximate 95% confidence interval for a mean number of ounces of ketchup per bottle in the sample?

Answer 1

Explanation:

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

Confidence Interval = Sample Mean ± (Z-score * Standard Error)

Where:

Sample Mean = 24 ounces (average number of ounces of ketchup)

Standard Error = Standard Deviation / √(sample size) = 0.4 / √(49) ≈ 0.057

The Z-score for a 95% confidence level is approximately 1.96.

Now, let's calculate the confidence interval:

24 ± (1.96 * 0.057)

This gives us the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample as 24 ± 0.112.

Therefore, the correct option is:

B. 24 ± 0.114

⊂⁠(⁠(⁠・⁠▽⁠・⁠)⁠)⁠⊃

Answer 2

The confidence interval is (23.888, 24.112)

Explanation:

The restaurant chain claims that its bottles of ketchup contain 24 ounces of ketchup on average. So, the mean is 24 ounces.

Standard Deviation = 0.4

Number of bottles used for sample = 49, so n = 49

Confidence level = 95% or 0.95

z=1.96

Hence, the confidence interval is given by the expression:

`24+1.96*(0.4)/(√(49) )` and

`24-1.96*(0.4)/(√(49) )`

Hence, Confidence interval = (23.888, 24.112)