SOLUTION: A supermarket manager wishes to make an estimate of the average time in minutes a customer spends at the checkout counter. It is known from previous studies that the variance is 10
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Question 1084647: A supermarket manager wishes to make an estimate of the average time in minutes a customer spends at the checkout counter. It is known from previous studies that the variance is 10.8. How large a sample is required if she wants a 95% confidence interval for the population mean that extends no further than 0.7 minutes from the sample mean. Round up your answer to a whole number.
I gound the answer to be 8468 but it is incorrect. Answer by jim_thompson5910(35256) (Show Source):
where,
n = min sample size needed
z = critical value drawn from the standard normal (Z) distribution
sigma = population standard deviation
E = margin of error desired
At 95% confidence, we can find the z critical value to be approximately z = 1.96
Look at the statistics table.
Locate "95%" at the bottom where the confidence levels are. The value just above "95%" is "1.960" which is the same as 1.96
A table like the one I have linked is often found in the back of any statistics textbook.
So this is why z = 1.96, which is approximate.
The variance is given to be 10.8. Take the square root of the variance to get the standard deviation (sigma)
sigma = sqrt(variance)
sigma = sqrt(10.8)
sigma = 3.286335345031
which is approximate
The error desired is E = 0.7 because we want the upper or lower end of the confidence interval to be at most 0.7 units away from the center xbar.
Plug those three values mentioned above into the formula below
n = ( (z*sigma)/E )^2
n = ( (1.96*3.286335345031)/0.7 )^2
n = ( (6.44121727626076)/0.7 )^2
n = ( 9.2017389660868 )^2
n = 84.6720000000002
n = 85 Round up to the nearest whole number
Min Sample Size: n = 85
On the last step, we ALWAYS round up regardless of the decimal portion.
This rounding up is to ensure that we clear the hurdle.
If we rounded down, then E > 0.7, which is too large.
The goal is to make
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