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 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?
***ANSWER IS C.24+/-0.029 for apex***

24 ± 0.224

Step-by-step explanation:

Confidence interval = mean ± critical value × standard error

CI = μ ± CV × SE

The critical value is the t score or z score.  Since n > 30, we can approximate using the z score.  For 95% confidence, z = 1.96.

The standard error is:

SE = σ / √n

SE = 0.8 / √49

SE = 0.114

So the confidence interval is:

CI = 24 ± (1.96) (0.114)

CI = 24 ± 0.224

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step-by-step explanation:

answer: 230 miles ( vote ! )

step-by-step explanation:

180.39 minus 17.99 equals 162.4

then divide 162.4 by .80 to get 230