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Quantitative Usability, Statistics & Six Sigma by Jeff Sauro



April 14, 2008 | asked by Jeff Phillips :

Question : A random sample of 25 households finds that an average of 2.3 people reside in each house (the standard deviation is 0.35). With a 95% confidence level, what is your estimation of the population average?

Answer :

Since this sample size is relatively small (25) I recommend building your confidence interval using the a t-statistic. Assuming the data is roughly normally distributed:

  1. Calculate the standard error of the mean: standard deviation / square root of the sample size = .35/5 = .07
  2. Find the critical value from the t-distribution for 24 degrees of freedom and a probability of .05. You can use MS Excel and type =TINV(.05,24) and should get 2.06.
  3. Calculate the margin of error: which is the standard error of the mean times the t-value = .07*2.06 = .1442.
  4. Generate the Lower and Upper bounds of the confidence interval by adding and subtracting the margin of error to the mean= 2.3+.1442 = 2.442 and = 2.3-.1442 =2.1558

So your 95% confidence interval around the mean is (2.1558, 2.442).


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Question Tags

Tag Name # Vote
Confidence Intervals2
Margin of Error2
Standard Error of the Mean2
t-distribution2
t-statistic2
TINV2

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