Question

What would be the approximate 95% confidence interval for the mean number of ounces of catchup bottle in the sample

Answer 1

The 95% confidence interval for the mean number of ounces of ketchup bottle is (23.8, 24.2).

Explanation:

The complete question is:

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 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?

Solution:

The (1 - α)% confidence interval for the population mean is:

`CI=\bar x\pm z_(\alpha/2)\ (\sigma)/(√(n))`

The information provided is:

`\bar x=24\\\sigma=0.8\\n=49\\\text{Confidence Level}=95\%`

The critical value of z for 95% confidence level is:

`z_(\alpha/2)=z_(0.05/2)=z_(0.025)=1.96`

*Use a z-table.

Compute the 95% confidence interval for the mean number of ounces of ketchup per bottle as follows:

`CI=\bar x\pm z_(\alpha/2)\ (\sigma)/(√(n))`

`=24\pm1.96\cdot (0.80)/(√(49))\\\\=24\pm 0.224\\\\=(23.776, 24.224)\\\\\approx (23.8, 24.2)`

Thus, the 95% confidence interval for the mean number of ounces of ketchup bottle is (23.8, 24.2).