Independent measurements of the leakage current I on a capacitor yielded the following data: 2.71, 2.66, 2.78, 2.67, 2.71, 2.69, 2.70, 2.73 mA. Assuming that the distribution of the random quantity I...

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Given Data:
2.71, 2.66, 2.78, 2.67, 2.71, 2.69, 2.7, 2.73
We calculate sample mean and std deviation from given data.
Sample Mean, Ī =
∑
(I)
n
= 21.658 = 2.70625
Sample Variance, s2 =
∑
(I − Ī)2
n− 1 =
0.009787
7 = 0.001398
Sample std dev, s =
√
s2 =
√
0.001398 ≈ 0.037393
95% CI for E(I) using t-dist
Sample Mean = i = 2.70625
Sample Standard deviation = s = 0.037393
Sample Size = n = 8
Significance level = α = 1− 0.95 = 0.05
Degrees of freedom for t-distribution, d.f . = n− 1 = 7
Critical Value = tα/2,df = t0.025,df=7 = 2.365 (from t-table, two-tails, d.f . = 7 )
Margin of Error = E = tα/2,df ×
si√
n
= 2.365×...