A simple random sample of 60 items resulted in a sample mean of 80. The population standard deviation is σ=15.
a. Compute the 95% confidence interval for the population mean. Round your answers to one decimal place
b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean. Round your answers to two decimal places
c. What is the effect of a larger sample size on the interval estimate?
(a)
n = 60
x-bar = 80
s = 15
% = 95
Standard Error, SE = σ/√n = 15 /√60 = 1.936491673
z- score = 1.959963985
Width of the confidence interval = z * SE = 1.95996398454005 * 1.93649167310371 = 3.795453936
Lower Limit of the confidence interval = x-bar - width = 80 - 3.79545393564498 = 76.20454606
Upper Limit of the confidence interval = x-bar + width = 80 + 3.79545393564498 = 83.79545394
The confidence interval is [76.2, 83.8]
(b)
n = 120
x-bar = 80
s = 15
% = 95
Standard Error, SE = σ/√n = 15 /√120 = 1.369306394
z- score = 1.959963985
Width of the confidence interval = z * SE = 1.95996398454005 * 1.36930639376292 = 2.683791216
Lower Limit of the confidence interval = x-bar - width = 80 - 2.68379121557574 = 77.31620878
Upper Limit of the confidence interval = x-bar + width = 80 + 2.68379121557574 = 82.68379122
The confidence interval is [77.32, 82.68]
(c)
As n increases, the confidence interval narrows down. Larger sample provides a smaller margin of error.
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