Question 1108899: Suppose that you wish to obtain a 95% confidence interval for a population mean. The population is normally distributed, the sample size is 20, and the population standard deviation is unknown. The correct procedure to use is the t-interval procedure.
(a)If you mistakenly use the z-interval procedure, will the resulting confidence interval be too wide or too narrow? Why?
(b)Will the true confidence level associated with this interval be greater than or less than 95%?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! sample size is 20.
z-score for 95% confidence interval would be plus or minus 1.959963986.
t-score for 95% confidence interval would be plus or minus 2.093024022, using 19 degrees of freedom (sample size of 20 - 1 = 19 degrees of freedom).
if you mis-understood the z-score to be the t-score, the resulting confidence interval would be too narrow.
if the z-score was interpreted as the t-score, then a t-score of plus or minus 1.959963986 with 19 degrees of freedom would generate a confidence level of .9351665248 = 93.5%.
that would be less than the desired 95%.
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