3. 4/9 points I Previcus Answers My Notes Ask Your Teacher We wish to determine

Question

3. 4/9 points I Previcus Answers My Notes Ask Your Teacher We wish to determine the unknown probability p, that a voter supports the controversial tulip proposal. We wish to estimate p. We take a random sample of n-3100 voters. x-1331 of these voters support the tulip proposal. Let p be the sample proportion of voters supporting the tulip proposal. Answer the following using R code. a)As a function of p, what is the variance of p? (use R code) 107.4200 x b) As a function of p, what is the standard deviation ofb? (use R code)0089 c) Calculate 0.4293 d) Let ptot be the random variable representing theto number of voters in the sample who support the tulip l. As a function of p, what is the standard deviation ofo? 27.5596 e) what is the critical value for an approximate classical 93% confidence interval for p? 61885 f) Calculate a classical 93% confidence interval for p. ( 0.4133 g) what is the length of the above classical 93% confidence interval for p? 0.0322 h) Assuming the same value, what sample size would have made the 93% confidence interval for p have a length or .08 or less? 0.0322 Copy your R script for the above into the text box here. 0.4455 This answer has not been graded yet.

Explanation / Answer

> n=3100

> x=1331

> phat=x/n;phat                                       # c)value of phat

[1] 0.4293548

> variance=(phat*(1-phat))/n                  # a)variance of phat

> variance

[1] 7.903525e-05

> sd=sqrt(variance);sd                              # b)standard deviation of phat

[1] 0.008890177

>

>

>ptot=1                                                       

> sd2=sqrt((ptot*(1-ptot))/n);sd2               # d) standard deviation of ptot

[1] 0

>

>

> alpha=0.07                                            

> a=1-(alpha/2)

>CV= qnorm(a);CV                                          # e)Critical value (CV)

[1] 1.811911

> phat

[1] 0.4293548

>

>

> CV=qnorm(a);CV                                       # Critical value (CV)

[1] 1.811911

> L=0.08                                                       # Length of P for 93% CI

> M=L/2                                                       # Margine of error

> n=((CV^2)/M^2)*phat*(1-phat);n                       # h)Sample size n

[1] 502.7315

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