SOLUTION: A random sample of 16 pharmacy customers showed the waiting times below (in minutes). Find a 90 percent confidence interval for μ, assuming that the sample is from a normal po

Algebra ->  Probability-and-statistics -> SOLUTION: A random sample of 16 pharmacy customers showed the waiting times below (in minutes). Find a 90 percent confidence interval for μ, assuming that the sample is from a normal po      Log On


   

Question 207138: A random sample of 16 pharmacy customers showed the waiting times below (in minutes). Find a 90 percent confidence interval for μ, assuming that the sample is from a normal population.
21 22 22 17 21 17 23 20
20 24 9 22 16 21 22 21


Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
values are shown below:
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21
22
22
17
21
17
23
20
20
24
9
22
16
21
22
21
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sum and average of all values is shown below:
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318 = sum of all numbers
16 = number of occurrences
19.875 = sum of all numbers divided by number of occurrences = average = mean = mu
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square of all values minus the mean is shown below:
example:
1.265625 = %2821-19.875%29%5E2 = %28x+-+mu%29%5E2
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1.265625
4.515625
4.515625
8.265625
1.265625
8.265625
9.765625
0.015625
0.015625
17.015625
118.265625
4.515625
15.015625
1.265625
4.515625
1.265625
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199.75 = sum of all deviations squared.
16 = number of occurrences
3.533323506 = standard deviation of population = sigma+=+sqrt%28sum%28%28x-mu%29%5E2%29%2Fn%29%29
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to find the 90% confidence interval for the population (2 tailed confidence interval assumed), do the following:
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go to the following website:
http://davidmlane.com/hyperstat/z_table.html
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got to the bottom graph.
enter a mean of 19.875
enter a sd of 3.5333
enter a shaded area of .9
select between
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your answer is that at a 90% confidence level, the waiting times can be between 14.0632 minutes and 25.6868 minutes.
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to see how this is represented on a z score, do the following:
go to the bottom graph.
enter a mean of 0
enter a sd of 1
enter a shaded area of.9
select between
----- your answer is that at a 90% confidence level, the waiting times can be between 1.6449 standard deviations below the mean and 1.6449 standard deviations above the mean.
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1.6449 standard deviations is 1.6449 * the standard deviation of 3.5333 = 5.8119
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the mean of 19.875 minus 5.8119 = 14.063 which is close enough to 14.0632 to have the discrepancy be explained by rounding error. similarly, the mean of 19.875 plus 5.8119 = 25.6869 is close enough to also have the discrepancy be explained by rounding error.
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answer to your question:
mean = 19.875
standard deviation equal 3.5333
at 90% confidence interval waiting can be between 14.0632 and 25.6869 minutes
this is between minus 1.6449 and plus 1.6449 standard deviations from the mean.
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