Question

The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. An article gave the accompanying sample observations on breakdown voltage (kV) of a particular circuit under certain conditions.

62	50	54	58	42	54	56	61	59	64	50	54	64	62	51	68
54	56	57	51	55	50	56	56	46	55	53	55	52	48	47	56
58	48	63	57	57	56	54	59	53	53	50	56	60	50	56	59

(a)

Construct a boxplot of the data.

B-) Calculate and interpret a 95% CI for true average breakdown voltage μ. (Round your answers to one decimal place.)

( , )

*Interpret the resulting interval.

We are 95% confident that the true population mean lies below this interval.

We are 95% confident that this interval does not contain the true population mean.    

We are 95% confident that the true population mean lies above this interval.

We are 95% confident that this interval contains the true population mean.

*Does it appear that μ has been precisely estimated? Explain.

This interval is quite narrow relative to the scale of the data values themselves, so it could be argued that the mean has been precisely estimated.

This interval is quite wide relative to the scale of the data values themselves, so it could be argued that the mean has been precisely estimated.    

This interval is quite narrow relative to the scale of the data values themselves, so it could be argued that the mean has not been precisely estimated.

This interval is quite wide relative to the scale of the data values themselves, so it could be argued that the mean has not been precisely estimated.

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 1 kV (so that μ is estimated to within 0.5 kV with 95% confidence)? (Round your answer up to the nearest whole number.)

circuits

Answer 1

Solution : Given that 62;50;54;58;42;54;56;61;59;64;50;54;64;62;51;68;54;56;57;51;55;50;56;56;46;55;53;55;52;48;47;56;58;48;63;57;57;56;54;59;53;53;50;56;60;50;56;59

X = 55.1042, σ = 5.0713, n = 48, 95% Confidence interval for Z = 1.96

(a) Box plot :

Box-Plot Graph: * 68 Outlier 64 Maximum Third Quartile 58 55 Median First Quartile 43 Minimum

(b) 95% CI for true average breakdown voltage μ : X +/- Z*σ/sqrt(n)
= 55.1042 +/- 1.96*5.0713/sqrt(48)
= (53.6695 , 56.5389)
= (53.7 , 56.5) rounded

=> option D. we are 95% confident that this interval conatins the true population mean.

=> option A. This interval is quite wide relative to the scale of the data values themselves, so it could be argued that the mean has been precisely estimated.

c) s = range/4 = (70-40)/4 = 7.5, 95% Confidence interval for Z = 1.96, E = 0.5

n = (Z*s/E)^2 = (1.96*7.5/0.5)^2 = 864.36 = 865

n = 865