A political pollster is conducting an analysis of sample results in order to mak
ID: 3236893 • Letter: A
Question
A political pollster is conducting an analysis of sample results in order to make predictions on election night. Assuming a two-candidate election, if a specific candidate receives at least 53% of the vote in the sample, that candidate will be forecast as the winner of the election. You select a random sample of 100 voters. Complete parts (a) through (c) below.
A polic in ictions ming a two-candi lection 8pecific candidate rec es al least 53% Complete part What probabity thi didato wil the entage of her wot The probability is that a candidate will be forecast as the winner when the yuvulation v centage of her vue is 50.1% what is the probahlity that candidata will ha torecast as the winnar whan tha panulatinn ntage ot har watAiR The probability is that a candidate wil beforecast as the vi when the popuation ercentage of her vote is S7% Round to four deci laces WMidl is lhe probs that a candidste wil be rorecast as the winriter wtleri Ilie puvulativni per verksae vule is 40% (srnd she will tually lose the el he probability isL that a candidata will ha toracaat as the winnar whan 1ha pnnulatinn narrantage or har vata ia 4ERi Round to four at the ng thi size. Com the differenc idele will be of to four decl The probability is that a candidate will beforecast as the winner when the population percentage of her vote is 57% o four decim laces The probability is that a candidata wll be forecast as he Anner when the Population parantage of har vote is 49% Round to four deci O A. Increasing the sample size by a iau ut of 4 decreases the standard error by a iactur of 2 Charging the standard error doubles the itude of the standardized Z value. ing the sample size by R tactor et by factor 2 Changing to helt of origi ing the sample size ctor of 4 by a factor of 2. Changi doubles gnitude of the standardz val O DA Inciessingure sample size by a laulur of 4 decreases the standurd error by of 2 C the standurd error decreases the standardized Z- ofits vrignal vsue. the vote the nal kandidele the elect You select H mple of 100 volereExplanation / Answer
For All questions Z = (X-Mean)/SD
Mean = n*p
Standard deviation = Sqrt[p*(1-p)n]
For a,b, c n = 100 and X = 53
(a) Mean = 100*0.501 = 50.1
SD = Sqrt[0.501*0.495*100] = 5
P(X>53) = 1-P(X<53)
Z = (53-50.1)/5 = 0.98. The p value at Z = 0.58 is 0.7190
Therefore the required probability = 1-0.7190 = 0.2810
(b) Mean = 100*0.57 = 57
SD = Sqrt[0.57*0.43*100] = 4.95
P(X>53) = 1-P(X<53)
Z = (53-57)/4.95 = 0.81. The p value at Z = 0.81 is 0.79103
Therefore the required probability = 1-0.7903 = 0.2090
(c) Mean = 100*0.49 = 49
SD = Sqrt[0.49*0.51*100] = 5
P(X>53) = 1-P(X<53)
Z = (53-49)/5 = 0.8. The p value at Z = 0.8 is 0.78814
Therefore the required probability = 1-0.78814 = 0.2119
For question d, e and f, n = 400
Als0 X = 0.53*400 = 212
(d) Mean = 400*0.501 = 200.4
SD = Sqrt[0.501*0.495*400] = 9.96
P(X>212) = 1-P(X<212)
Z = (212-200.4)/9.96 = 1.16. The p value at Z = 1.16 is 0.87698
Therefore the required probability = 1-0.87698 = 0.1230
(e) Mean = 400*0.57 = 228
SD = Sqrt[0.57*0.43*400] = 9.90
P(X>212) = 1-P(X<212)
Z = (212-228)/9.9 = 1.62. The p value at Z = 1.62 is 0.94738
Therefore the required probability = 1-0.94738 = 0.0526
(f) Mean = 400*0.49 = 196
SD = Sqrt[0.49*0.51*400] = 10
P(X>212) = 1-P(X<212)
Z = (212-196)/10 = 1.6. The p value at Z = 1.6 is 0.9452
Therefore the required probability = 1-0.9452 = 0.0548
In choose the correct answer, OPTION C is the correct answer. You can see from above, when was increased to 400(4 times) the SD(standard error), increased by 2, and hence the Z value also doubled by 2.