use a normal distribution because σ is known and the data are normally distributed.
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population std dev , σ = 1.2200
Sample Size , n = 25
Sample Mean, x̅ = ΣX/n = 5.0880
Level of Significance , α =
0.01
' ' '
z value= z α/2= 2.5758 [Excel
formula =NORMSINV(α/2) ]
Standard Error , SE = σ/√n = 1.2200 /
√ 25 = 0.244000
margin of error, E=Z*SE = 2.5758
* 0.24400 = 0.628502
confidence interval is
Interval Lower Limit = x̅ - E = 5.09
- 0.628502 = 4.4595
Interval Upper Limit = x̅ + E = 5.09
- 0.628502 = 5.7165
99% confidence interval is (
4.46 < µ < 5.72
)
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interpret :
answer: option a)
Use the standard normal distribution or the t-distribution to construct a 99% confidence interval for the...
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