已知n=16,
.
X=2.125,s=0.017.
(1)因为σ=0.01已知,故
.
X?μ
σ
n=
2.125?μ
0.01
16=400(2.125-μ)~N(0,1),
从而P(|400(2.125-μ)|<1.645)=1-2×(1-0.95)=0.90.
由|400(2.125-μ)|<1.645 可得,
2.1208875<μ<2.1291125,
故均值的置信度为90%的置信区间为:(2.1208875,2.1291125).
(2)因为σ未知,故
.
X?μ
s
n~t(n-1),
即:
2.125?μ
0.017
16=
4(2.125?μ)
0.017~t(15).
由已知条件,t0.05(15)=1.7531,
故由|
4(2.125?μ)
0.017|<1.7531可得,
2.05049325<μ<2.19950675,
故均值的置信度为90%的置信区间为:(2.05049325,2.19950675).