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

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

(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 μis estimated to within 0.5 kV with 95% confidence)? (Round your answer up to the nearest whole number.)

circuits

Answer 1

(B) Copy and paste the data values into excel sheet, use excel function AVERAGE and STDEV.S to find the mean and sample standard deviation, we get

mean (x bar) = 55.06 and sample standard deviation (s) = 5.088 with sample size n = 48

degree of freedom = n-1= 48-1 = 47

t critical = T.INV.2T(alpha,df) = T.INV.2T(0.05,47) = 2.012

CI =

(C) Given that margin of error (E) = 0.5

by range rule of thumb, standard deviation =(max-min)/4 = (70-40)/4 = 30/4 = 7.5

z score for 95% confidence level is 1.96 (using z table)

sample size n = ((z*sd)/E)^2

= ((1.96*7.5)/0.5)^2

= 29.4^2

= 864.36

= 864 (rounded to nearest whole number) or 865 (rounded to next whole number)