Question

Please solve the questions using R language and attach the script with it!!!

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What is the probability that a penny I have will land "heads" when tossed? To analyze this question I randomly toss the coin, n=3901 times. On x = 2500 of these tosses, the penny landed "heads." Based upon this sample, I wish to estimate p, the true probability of the penny landing "heads" when tossed. Let p be the sample proportion of "heads" in our sample. Answer the following using R code. a)As a function of p and n, what is the variance of p? (use R code) Op*(1-P)/n (P^n)/(1-P) n*p*(1-P) sqrt(n*p*(1-0)) p/(1-P) sqrt(p*(1-P)/n) b) as a function of p and n, what is the standard deviation of p? (use R code) Op/(1-p) p*(1-P)/n (p^n)/(1-P) sqrt(p*(1-P)/n) sqrt(n*p*(1-P)) On*p*(1-P) c) Calculate p. d) Let ptot be the random variable representing the total number of "heads" in a sample of n tosses. As a function of p or of p and n, what is the standard deviation of ptot? (p^n)/(1-P) sqrt(n*p*(1-P)) sqrt(p*(1-P)/n) p/(1-P) p*(1-P)/n n*p*(1-P) e) What is the critical value for an approximate classical 96% confidence interval for p? f) Calculate a classical 96% confidence interval for p. ( g) What is the length of the above classical 96% confidence interval for p? h) Assuming the same p value, what sample size would have made the 96% confidence interval for p have a length of .015 or less?

Please solve the questions using R language and attach the script with it!!!

Please solve the questions using R language and attach the script with it!!!

Answer 1

(a) The correct answer is p*(1-p)/n

(b)The correct answer is sqrt(p*(1-p)/n)

(c)$\hat p=\frac{x}{n}=\frac{2500}{3901}=0.64086$

(d)V(x)=npq if x represents the number of heads in n tosses and it follows the Binomial distribution with parameter n(sample size=3901) and p(probability of success, where success refers getting head in a toss).

So standard deviation of X is sqrt(n*p*(1-p)) where is ptot.

(e) we know critical value is 2.053749

$P(|\frac{x-np}{\sqrt{np(1-p)}}|<z_{0.02})=0.96\\ z_{0.02 } \,$can be found from standard normal table$\\ P(|\frac{x/n-p}{\sqrt{p(1-p)/n}}|<z_{0.02})=0.96\\ P(|\frac{\hat p -p}{\sqrt{p(1-p)/n}}|<z_{0.02})=0.96$

From this we can find confidence interval and others answers.
#x-number of successes in n tosses of #n-number of tosses x-2500; n-3901 #(a) here p represents phat p-x/n Coin Run 4 Source vphat-p* (1-p)/n vphat #(b) sdphat-sqrt(p* (1-p)/n) sdphat #(c) phat #(d) ptot-sqrt (n p (1-p)) ptot #(e)to find the critical value we need z1 such that P(z<z1)96 (from standard normal distributio z1-qnorm(0. 02,0,1,lower. tail FALSE) #cv=crittcal value p(Iz|<z1) =0.04 CV=z1 CV #(f) ul-phat+z1 sdphat 11-phat-z1 sdphat ci-c(11,ul) I ci #(g) length of the confidence interval 1g-ul-11 1g #(h)to find the sample size n1-4 z1 2/(0.15 2) n1

> #X-number of successes in n tosses of a coin > #n=number of tosses > X-2500; n-=3901 > #(a) here p represemts phat > p-x/n > p [1] 0.6408613 > vphat-p* (1-p)/n > vphat [1] 5.899977e-05 > #(b) sdphat-sqrt (p* (1-p)/n) > sdphat [1] 0.0076811 31 > #(c) > phat [1] 0.6408613 #(d) >ptot-sqrt(n*p* (1 -p) ) ptot [1] 29.96409

#(e)to find the criti cal value we need z1 such that' P(lz]kz1)=.96 (from standard normal distribution) 1-gnorm(0.02 0.1.1ower tail FALSE) #cv=critical value pelz<z1)=0.04 CV=Z1 CV [1] 2.053749 u phat zsdphat 1-phat zsdphat ci-c(11,ul) ci [1] 0.6250862 0. 6566364 M(g)length of the confidence interval 1g-u1 11 [1] 0.03155023 Ch)to find the sample size 4 21A2/00. 15 2) ini [1] 749.8461
​​​​​​(f)96% ci (.6250862,.6566364)

(g)length of ci is 0.03155023

(H)sample size should be 750 approximately.in R code picture we have found the value of n solving this equation.

+ n Pa PZ002 16 CI ven P- O'02 Noo, t sh mld be o'IS Langtn 2 2002 2- eus 0 '152 So L 2 0'02 TO