Question

The alternating current AC breakdo?n voltage of an insulating liqud indicates its delectric strength An article ave the accompanying sample obser ations cn breakdown voltage k c a particular circut under certain conditions 62 50 53 5B 42 54 55 1 59 64 51 53 64 62 50 6B 55 56 58 S0 55 50 56 55 46 55 54 54 52 4B 48 55 57 4 63 58 53 56 54 59 53 52 51 56 EO 50 57 5 (a) Construct a boxplct cf the data. o40 55 60 650 50 65 70 40 60 70 40 60 65 (b) Calculate and interpret à gS% CI for tre average breaedawn valtigeu. (Round nur answ rs to one decimal place.) Interpret the resulting interval we are 95% corndent that the true populanon mean lies above tris iteval. we re 05% confident that the true populatian mean lias balow this intarval. We are 95% confident that this interval does not contain the oue population mean. We are 95% confident that this interval contains the true population mean. Does it appear that u has been precisely estimated? Explain. ? Thisinerval is quite wide relave to the scale cf the “ta values themselves, so it could be sgJed that the mean has not been presely est otec. This inerval is quita wide relative to th@ scale e( the data values themselves, so it could be ar gled that the rnaan has baan precise,estimated. This iterval is quite narrow relatve to the scale of the data values themselves, so it couid be argued that the mean has been precisely estimated ) This interval is quite narrow r-lative to the scale of the dats valuws tnem solves, so it could b?.rguec that the mean has not been precisely "stint c) tha investi atar beliavas that vitually all valu“ c breakdown voltage are bat eaan 40 a d 70 Y hat sample siza would be app~priata orthe 95%CI to have a wicth o L kV so that ? is estim ted to within 0.5 ki ith 9 confidence 7 Round your ans ar up to the neare t whole S pp0 number) You may need to use the appropriate table in the Appendix of Tables to anser this questien.

Answer 1

a)

option B is right

b)$\overline{x}=55.125,S=5.118,n=48$

for 95% confidence and degree of freedom = n-1=47

$T_{\frac{\alpha }{2}}=2.0117$

$CI=\overline{x}\pm T_{\frac{\alpha }{2}}(\frac{S}{\sqrt{n}})=55.125\pm 2.0117(\frac{5.118}{\sqrt{48}})$

$CI=55.125\pm 1.486$

C1 = (53.6.56.6)

Interpret the interval:

option D is right

does it appear ...................?

option D is right