Quiz: Week 2 Quiz this resut cons is ent with the tumou or k this resul unikely

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Quiz: Week 2 Quiz this resut cons is ent with the tumou or k this resul unikely lo cur with a tun out of 69%, why or why nor? c ased on these eesults dee appear that acurate voling es its can be btained by asking volenb they =(Round to one decimal place as needed ) A. This result is unusual because 638 is greater than the maximum usual value B. This result is unusual because 638 is within the range of usual values c. This result is unusual because 638 is below the minimum usual value D. This result is usual because 638 is within the range of usual values c. Does it appear that accurate voting results can be obtained by asking voters how they acted? A. Yes, because the results indicate that 69% is a possible turnout O B. No, because it appears that O C. No, because it appears that substantially more people say that they voted than the proportion of people who actualty did vote ly fewer people say that they voted than the proportion of people who achually did vete. Click to select your

Explanation / Answer

Let X = number of subjects who responded that they voted. Then, X ~ B(n, p) where n = number of subjects and p = probability of a subject responding having voted. p is taken to be 0.69, given that in the last election the election turn-out was 69%.

Part (a)

Mean = E(X) = np = 1002 x 0.69 = 691.4 = µ ANSWER 1

Variance = np(1 – p) = 1002 x 0.69 x 0.31= 214.3278

Standard Deviation = 14.6 = ANSWER 2

Part (b)   

We want to see if getting 638 subjects responding as having voted is consistent with the perceived voter turn-out of 69%

Null hypothesis - H0: p = 0.69   Vs Alternative – HA: p 0.69

To test this we will develop 95% Confidence Interval for p, which is given by:

pcap ± Z0.025sqrt{pcap(1- pcap)/n}, where

pcap = sample estimate of p = 638/1002 = 0.6367 and Z0.025 = upper 2.5% point of N(0, 1) = 1.96 [obtained using Excel Function on Standard Normal Probability]

Thus, 95% Confidence Interval for p = 0.6367 ± 1.96sqrt(0.6367 x 0.3603/1002)

= 0.6367 ± 1.96sqrt(0.0002308514)

= 0.6367 ± 0.0298

= (0.6069, 0.6665)

Since 0.69 is not within the above range, there is enough evidence to suggest that the survey result is unusual. Option A ANSWER

Part (c)

Option (c) ANSWER

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