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 51 54 64 62 51 68 54 56 58 51 56 50 56 56 46 55 54 54 53 48 48 55 57 49 63 58 57 55 53 59 54 52 51 56 60 50 57 59 (a) Construct a boxplot of the data. 40 45 50 55 60 65 70 40 45 50 55 60 65 70 40 45 50 55 60 65 70 40 45 50 55 60 65 70 (b) Calculate and interpret a 95% CI for true average breakdown voltage u. (Round your answers to one decimal place.) Interpret the resulting interval. We are 95% confident that this interval contains the true population mean. We are 95% confident that this interval does not contain the true population mean. We are 95% confident that the true population mean lies below this interval. We are 95% confident that the true population mean lies above this interval. Does it appear that u has been precisely estimated? Explain. 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. 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 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. (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 u is estimated to within 1 kV with 95% confidence)? (Round your answer up to the nearest whole number.) circuits

Answer 1

Using Minitab software ,

1) Open worksheet that contains the data.

2) a) to construct a Boxplot, graph-> Boxplot-> double click-> simple-> ok-> Enter graph variable as Breakdown voltage-> ok we get

s Boxplot of Break Volt Boxplot of Break. Volt. 70 65 60 C10 Break. Volt. C11 C12 C13 55 50 45 40

3) For getting 95% confidence interval for mean,

E Session Confidence interval Two-sided Results Margin of Error 2 Sample Size 57 I Workshee 01 02 CB 04 C5 C6 CY C8 C9 C10 Break. Volt. Mean1 SEMeani StDev1 Lower lim Upper lim 56.3969 56.4208 62 55.2083 0.722583 5.00620 2 51 3 54 4 57

4) For sample size estimation,

Stat-> Power and sample size-> sample size for estimation.

Boxplot of Break. Volt. MTB > Intplot Break. Volt. SUBC> Intbar, SUBC> Mean. Interval Plot of Break. Volt. MTB > Name c2 "Meani" C3 "SEMeanl" c4 "St Dev1" MTB > Statistics Break. Volt.'; SUBC> Mean Meani; MTB > InvCDE 0.95 kl: SUBC> Normal 0.0 1.0 MTB > print kl Data Display T K1 1.64485 MTB > c5=c2- (k1*c4/sqrt (48)) * ERROR * Unknown Minitab command: c MTB > let c5=c2 - (kl 047 sqrt (48)) MTB > let c5=c2+ (k1*c4/sqrt (48)) MTB > SSCI;

Tools StatGraph Editor CE MTB > let c5=c2- (k1 c4/sqrt (48)) MTB > let c5=c2+(k1*c4/sqrt (48)) MTB > SSCI, SUBC NMean: SUBO> Sigma 7.5, SUBO> Confidence 95.0 ; SUBOS IType O SUBC> MError 2 Sample Size for Estimation Method Parameter Distribution Standard deviation Confidence level confidence interval Mean Normal 75 (estimate) 95% Iwo-sided Results Margin Sample of Error Size 57 Current Worksheet Worksheet : Cybe here to search

Tools StatGraph Editor CE MTB > let c5=c2- (k1 c4/sqrt (48)) MTB > let c5=c2+(k1*c4/sqrt (48)) MTB > SSCI, SUBC NMean: SUBO> Sigma 7.5, SUBO> Confidence 95.0 ; SUBOS IType O SUBC> MError 2 Sample Size for Estimation Method Parameter Distribution Standard deviation Confidence level confidence interval Mean Normal 75 (estimate) 95% Iwo-sided Results Margin Sample of Error Size 57 Current Worksheet Worksheet : Cybe here to search