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The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. The article "Testing Practices for the AC Breakdown Voltage Testing of Insulation Liquids" (IEEE Electrical Insulation Magazine, 1995: 21–26) gave the accompanying sample observations on breakdown voltage (kV) of a particular circuit under certain conditions.
62 50 53 57 41 53 55 61 59 64 50 53 64 62 50 68
54 55 57 50 55 50 56 55 46 55 53 54 52 47 47 55
57 48 63 57 57 55 53 59 53 52 50 65 60 50 56 68
a. Construct a boxplot of the data and comment on interesting features.
b. Calculate and interpret a 95% CI for true average breakdown voltage m. Does it appear that m has been precisely estimated? Explain.
c. Suppose the investigator believes that virtually all values of breakdown voltage are between 40 and 70. What sample size would be appropriate for the 95% CI to have a width of 2 kV (so that m is estimated to within 1 kV with 95% confidence)?

Sagot :

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Answer:

(53.50 ; 56.75)

127

Step-by-step explanation:

Given the data:

41, 46, 47, 47, 48, 50, 50, 50, 50, 50, 50, 50, 52, 52, 53, 53, 53, 53, 53, 53, 54, 54, 55, 55, 55, 55, 55, 55, 55, 56, 56, 57, 57, 57, 57, 57, 59, 59, 60, 61, 62, 62, 63, 64, 64, 65, 68, 68

Using calculator :

Standard deviation, s = 5.74

Mean, m = 55.125

Sample size, n = 48

95% confidence interval :

Zcritical at 95% = 1.96

Mean ± Zcritical* s/sqrt(n)

55.125 ± 1.96 * 5.74/sqrt(48)

Lower boundary :

55.125 - 1.96(0.8284976) = 53.50

Upper boundary :

55.125 + 1.96(0.8284976) = 56.75

(53.50 ; 56.75)

Sample size, n :

n = [(Zcritical * s) / Error margin]²

Error margin = 1

Zcritical at 95% = 1.96

s = 5.74

n = [(1.96 * 5.74) / 1]²

n = 11.2504²

n = 126.57

n = 127

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