Question

Suppose that a restaurant chain claims that its bottles of ketchup contain 24 ounces of ketchup on average, with a standard deviation of 0.8 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

Answer: 24+- 0.229

Explanation:

Just took the test

Answer 2

Confidence Interval = (23.776, 24.224)

Explanation:

Restaurant chain claims that its bottles of ketchup contain 24 ounces of ketchup on average.

⇒ Mean = 24 ounces.

Standard Deviation = 0.8

Number of bottles used for sample = 49

⇒ n = 49

Confidence level = 95%

Corresponding z value with 95% confidence level = 1.96

Now, confidence interval is given by the following expression :

`Mean \pm z^** (\sigma)/(√(n))\\\\=24\pm1.96* (0.8)/(√(49))\\\\=24\pm 0.224`

Hence, Confidence Interval = (23.776, 24.224)