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 57 42 54 55 61 59 64 51 53 64 62 50 68 55 56 58 51 55 50 56 55 46 56 53 54 52 47 48 55 58 48 63 57 57 55 53 59 54 53 50 56 60 50 57 59 (a) Construct a boxplot of the data. . . . I . 40 45 50 55 60 65 7040 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 . (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 above this interval. We are 95% confident that the true population mean lies below this interval.

Answer 1

#R version 3.3.2 x=c(62,55,58,50,56,48,54,58,63,57,51,57,4 2,55,57,54,50,55,55,56,53,61,55,59,59,46, 54,64,56,53,51,53,50,53,54,56,64,52,60,62 ,47,50,50,48,57,68,55,59) length(x) mean(x) sd(x) boxplot(x, horizontal=TRUE) Save it Put on a wall Run it (Shift+Enter) [ + ] Show input Absolute running time: 0.45 sec, cpu time: 0.5 sec, memory peak: 32 Mb, absolute service time: 0,46 sec [1] 48 [1] 55.04167 [1] 5.13212

50 55 60 65

from given data no 48 x = 55.042 S= 5.1321 ca) for boxplot option cc) is correct (b) gst. confidence interval is À £t* sin 58.042 +2.012 & n 11.49 5.1321 148 St*=2.012, at 1/ = 0.025 df-n-1=48-1=47 S5.042 The gst. confidence interval is (53.5 sos) we are g5% confident that this interval contains mean the true population cc) This internal is quite narrow relative to the in scale of the data values themselves so it could be argued that the mean has been precisely estimated. (d) Given: E= 1 oz5. 1321 2*31.96 2- table) N: 2*7 n:(1-96 XS:1321)? ne 101:18 n = 102