Respuesta :
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:
[tex]CI=\bar x\pm z_{\alpha/2}\ \frac{\sigma}{\sqrt{n}}[/tex]
The information provided is:
[tex]\bar x=24\\\sigma=0.8\\n=49\\\text{Confidence Level}=95\%[/tex]
The critical value of z for 95% confidence level is:
[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
*Use a z-table.
Compute the 95% confidence interval for the mean number of ounces of ketchup per bottle as follows:
[tex]CI=\bar x\pm z_{\alpha/2}\ \frac{\sigma}{\sqrt{n}}[/tex]
[tex]=24\pm1.96\cdot \frac{0.80}{\sqrt{49}}\\\\=24\pm 0.224\\\\=(23.776, 24.224)\\\\\approx (23.8, 24.2)[/tex]
Thus, the 95% confidence interval for the mean number of ounces of ketchup bottle is (23.8, 24.2).
