In: Statistics and Probability
The employees of Electronics Inc. would like to have a dental plan as part of their benefits package. The question is: How much does a typical employee and his or her family spend per year on dental expenses? A sample of 59 employees reveals the mean amount spent last year was $2010, with a standard deviation of $600.
a. Construct a 95% confidence interval for the population mean. (Round the final answers to the nearest whole dollar.)
Population mean $ $
b. The information from part (a) was given to the president of Electronics Inc. He indicated he could afford $1780 of dental expenses per employee. Is it possible that the population mean could be $1780?
(Click to select) No. Yes. The population mean (Click to select) could not be could be $1780 because it is (Click to select) not in the in the interval constructed above.
(a)
n = 59
x-bar = 2010
s = 600
% = 95
Standard Error, SE = s/√n = 600/√59 = 78.11334659
Degrees of freedom = n - 1 = 59 -1 = 58
t- score = 2.001717468
Margin of error = t * SE = 2.00171746800345 * 78.1133465884943 = 156.3608504
Lower Limit of the confidence interval = x-bar - width = 2010 - 156.360850350396 = 1853.63915
Upper Limit of the confidence interval = x-bar + width = 2010 + 156.360850350396 = 2166.36085
The confidence interval is [$1854, $2166]
(b)
No. The population mean could not be $1780 because it is not in the interval constructed above.
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