In: Statistics and Probability
x̄ = 55 s = 15 n = 101
Answer)
Given info
Mean = 55
S.d = 15
N = 101
1)
As the population s.d is not given and we are given with sample s.d as the best estimate we will use t distribution to estimate the interval
Degrees of freedom is = n-1 = 100
For 100 dof and 99% confidence level critical value t from t table is = 2.63
Margin of error (MOE) = t*s.d/√n = 2.63*15/√101 = 3.92
Interval is given by
(Mean - MOE, Mean + MOE)
[51.08, 58.92].
You can be 99% confident that the population mean (μ) falls between 51.08 and 58.92.
B)
Critical value t for 95% confidence level with 100 dof = 1.98
Moe = 1.98*15/√101 = 2.96
[52.04, 57.96].
You can be 95% confident that the population mean (μ) falls between 52.04 and 57.96.
C)
Critical value t for 90% confidence level with 100 dof = 1.66
Moe = t*s.d/√n = 1.66*15/√101 = 2.48
[52.52, 57.48].
You can be 90% confident that the population mean (μ) falls between 52.52 and 57.48.
Similarity is that they all give us the range in which population parameter may lie with certain confidence level
Differences is that
We can see that as the confidence level decreases, width of the interval also decreases.