Question

Check my answer please


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
I think it's d

Answer 1

@SolomonZelman

Answer 2

@jim_thompson5910

Answer 3

@perl

Answer 4

yeah you should have

ME = t*s/sqrt(n)

ME = 2.0106*0.8/sqrt(49)

ME = 0.22978

due to rounding error, the margin of error is a bit off, but it's close enough to 0.229

Answer 5

Thank you