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 the 49 bottles of ketchup what would be the approximate 95% confidence interval for a mean number of ounces of ketchup per bottle in the sample?

Question

perl · Answer

you can use the formula
(xbar +/- z score * st. dev / sqrt(n) )

perl · Answer

the 95% confidence interval will be : 
( 24 - 1.96 * 0.4 / sqrt(49) , 24 + 1.96*0.4 / sqrt(49 )

perl · Answer

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