Question

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

Answer 1

Answer:

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

Step-by-step 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:

$CI=\bar x\pm z_{\alpha/2}\ \frac{\sigma}{\sqrt{n}}$

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:

$CI=\bar x\pm z_{\alpha/2}\ \frac{\sigma}{\sqrt{n}}$

     $=24\pm1.96\cdot \frac{0.80}{\sqrt{49}}\\\\=24\pm 0.224\\\\=(23.776, 24.224)\\\\\approx (23.8, 24.2)$

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