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 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 ± 0.114
D. 24 ± 0.229

Answer 1

a

Answer 2

Elaborate @cram ?

Answer 3

okay if you take 0.8 and times it by 49 /95% there is the answer

Answer 4

The confidence interval for a \((1-\alpha)100\%\) confidence level is given by\[\left(\bar{x}-Z_{\alpha/2}\frac{\sigma}{\sqrt n},~\bar{x}+Z_{\alpha/2}\frac{\sigma}{\sqrt n}\right)\]
with \(\bar{x}\) being the sample mean (24), \(Z_{\alpha/2}\) the critical value (1.96 for 95% confidence), \(\sigma\) the standard deviation (0.8), and \(n\) the sample size (49).
\[\left(24-1.96\frac{0.8}{\sqrt {49}},~24+1.96\frac{0.8}{\sqrt {49}}\right)\]
Basically it's a matter of simplifying the margin of error term (after the \(\pm\)).
\[1.96\frac{0.8}{7}\approx0.224\]
which makes D the closest answer.