SOLUTION: The Population mean time a customer spends at a drive-thru window at a local fast food store is 48.6 secs with a population standard deviation of 12.2 secs. The manager institutes

Question 854731: The Population mean time a customer spends at a drive-thru window at a local fast food store is 48.6 secs with a population standard deviation of 12.2 secs. The manager institutes a change in the drive thru process in an effort to reduce the time customers spend in the drive thru. After the change, a random sample of 41 cars results in a sample mean of 45.4 secs. What is the probability that a sample of 41 cars with a sample mean of 45.4 or less came from the original population?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
population mean equals 48.6 seconds
population standard deviation equals 12.2 seconds.
sample mean equals 45.4 seconds.
sample size equals 41.

you need to calculate the standard error of the mean.
that is equal to standard deviation of the population divided by the square root of the sample size.

in this problem that becomes:

se = 12.2 / sqrt(41) = 1.9053

accuracy to 4 decimal places is more than you would need.

you now use the following formula to find the z-score.

z = (x-m)/se

x is the sample mean.
m is the population mean.
se is the standard error that you just calculated.

this formula becomes:

z = (45.4 - 48.6) / 1.9053

you get:

z = -1.68 rounded to the nearest 2 decimal places.

2 decimal places is all that you need because most z-score tables don't provide any further detail than that.

you then look in the z-score table to find the area under the normal distribution curve that is to the left of that z-score.

i used the following table:

http://lilt.ilstu.edu/dasacke/eco148/ztable.htm

the area to the left of the z-score of -1.68 in that table is shown to be:

p(z < -1.68) = .0465

this means that the probability of a reduction in the customer's mean time being due just to chance is .0465.

since this is less than the normally stated alpha of .05, the results are considered to be statistically significant.

a one sided alpha was used because the study was whether the customer wait time was reduced and not just changed plus or minus. the direction of the test was therefore one sided resulting in a one sided alpha being used. if the test was two sided because you were looking for a change in either direction, then the alpha of .05 would have been split between the high end of the distribution curve and the low end of the distribution curve and the results would not have been statistically significant because the alpha on the low end would have .025 rather than .05. the resulting probability of .0465 would then not have been less than the alpha.

so bottom line:

the probability that the results were not due to chance is .0465 which is statistically significant.

this is similar to saying that the probability that the results came from the original population is .0465.

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