Assume that population mean is to be estimated from the sample size n =100, Sample mean x = 76.0cm, standard deviation s = 4.0cm. Use the results to approximate the margin of error and 95% confidence interval.
Solution:
Given that,
Point estimate = sample mean = = 76.0
sample standard deviation = s = 4.0
sample size = n = 100
Degrees of freedom = df = n - 1 = 100-1 = 99
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,99 = 1.984
Margin of error = E = t/2,df * (s /n)
= 1.984 * (4.0 / 100)
E = 0.8
The margin of error and 95% confidence interval is 0.8
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