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" 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 55 60 50 56 58
a) Construct a boxplot of the data and comment on interesting features.
b) Calculate and interpret a 95% Confidence Interval for true average breakdown voltage Î¼. Does it appeas that Î¼ 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% Confidence Interval to have a width of 2kV (so that Î¼ is estimated to within 1 kV with 95% confidence?)

The black dot in the middle of the box represents the sample mean (X[bar]) as you can see it is also moved to the right side of the distribution, affected by the presence of an outlier.

Whiskers:

The upper whisker is smaller than the lower one and there is a value that can be considered an outlier.

Altogether you can say that the distribution of the data set is skewed to the left.

b)

To estimate the population mean you need the population to be at least normal. The box plot shows that the distribution is not symmetrical but, considering that the sample size is n= 48 you can apply the Central Limit Theorem and approximate the sampling distribution to normal.

X[bar]≈N(μ;σ²)

As the distribution is approximate, using the sample standard deviation (since we don't know the population value) for the distribution standard deviation is also valid ⇒ [tex]Z= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }[/tex]

X[bar] ± [tex]Z_{1-\alpha /2}[/tex] * [tex]\frac{S}{\sqrt{n} }[/tex]

∑X= 2626 ∑X²= 144950 n= 48

X[bar]= ∑X/n= 2626/48= 54.71

S²= 1/n[∑X²- (∑X)²/n]= 1/47[144950-(2626)²/48]= 27.36

S= 5.23

[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.96[/tex]

54.71 ± 1.96 * [tex]\frac{5.23}{\sqrt{48} }[/tex]

[53.22; 56.18]kV

c)

95% CI

amplitude a=2 kV i.e. semi amplitude d= 1 kV

The semi amplitude or margin of error for the CI can be calculated as:

d= [tex]Z_{1-\alpha /2}[/tex] * [tex]\frac{S}{\sqrt{n} }[/tex]

From this formula you can clear the sample size

[tex]\frac{d}{Z_{1-\alpha /2}} = \frac{S}{\sqrt{n} }[/tex]

[tex](\frac{d}{Z_{1-\alpha /2}} )*\sqrt{n} = S[/tex]

[tex]\sqrt{n} = S* (\frac{Z_{1-\alpha /2}}{d} )[/tex]

[tex]n = (S* (\frac{Z_{1-\alpha /2}}{d} ))^2[/tex]

[tex]n= (5.23*(\frac{1.96}{1} ))^2= 105.08= 106[/tex]