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 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+-0.029
B. 24+-0.057
C. 24+-0.229
D. 24+-0.114

Answer 1

D. 24+-0.114

Explanation:

The question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle given the following statistics;

Sample mean = 24

standard deviation, σ = 0.4

Sample size , n = 49

The confidence interval for a population mean is calculated using the formula;

Sample mean ± z-score*(σ/sqrt(n))

The z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

Substituting these values into the formula;

24 ± 1.96*
`(0.4)/(√(49) )`

= 24 ± 0.112

From the choices given, choice D is closest to the above expression;