Suppose you randomly sample 25 cans and find the mean amount

Suppose you randomly sample 25 cans and find the mean amount of Coke in this sample is 11.85 ounces and the standard deviation is 0.30 ounces.

Construct a 95% confidence interval estimate for the mean amount of Coke in the all such cans. Make a statement about this confidence interval in context of the problem

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Solution

As per question

n = 25

Mean = x^- = 11.85

Standard Deviation = S_x= 0.30

We know Confidence Interval = x^- +- (z.S_x)/sqrt(n)

where x^- is mean , z is the z value (Seen from table), S_x is the standard deviation , n is th number of samples

For 95% C.I. , z = 1.96

=> 95% C.I =   x^- +- (1.96.S_x)/sqrt(n)

=> 95% C.I = 11.85 +- (1.96*0.30/sqrt(25)

=> 95% C.I = 11.85 +- (1.96*0.30/sqrt(25)

=> 95% C.I = 11.85 +- (0.588/5)

=> 95% C.I = 11.85 +- 0.1176

=> 95% C.I = (11.85 - 0.1176, 11.85 + 0.1176)

=> 95% C.I = (11.7324, 11.9676)

This defines a range of values that can be 95% certain contains the mean amount of coke.