I NEED THE ANSWERS FOR PARTS A,B,C,D,E,F,G,H PLEASE USE R SCRIPT FOR PART H How

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I NEED THE ANSWERS FOR PARTS A,B,C,D,E,F,G,H PLEASE USE R SCRIPT FOR PART H
How bad is the wireless network at JJ? Out of a random sample of 121 JJ students, 75 students said they had a network connection at JJ lost in the past month. Let p be the (unknown) true proportion JJ students who have lost a network connection in the last month. Let phat be the sample proportion How bad is the wireless network at JJ? Out of a random sample of 121 JJ students, 75 students said they had a network connection at JJ lost in the past month. Let p be the (unknown) true proportion JJ students who have lost a network connection in the last month. Let phat be the sample proportion
My Noters O Ask Your Teacher 6. Question Details How bad is the wireless network at JJ? Out of a random sample of 121 JJ students, 75 students said they had a network connection at 33 lost in the past month. Let p be the (unknown) true proportion J3 students who have lost a network connection in the last month. Let phat be the sample proportion a)Obtain an unbiased point estimate of p. b) We wish to construct a 94 % classical confidence interval for p, what is the critical value multiplier zstar? c) Create a 94% classical confidence interval for p? ( ) d) How long is the 94% classical confidence interval for p? e) In terms of p and n, give the formula for the standard deviation of the distribution of the sample proportion phat. (R code) O p (1-P)/n o sqrt(n*p*(1-p)) On*p*(1-p) O sqrt(p*(1-p)/n) f) Based on this data estimate the standard deviation of the sample proportion of JJ students who lost a connection in the last month. g)Assuming the same value of the sample proportion what is the smallest sample size for which the length of the 94% confidence interval for p is less than or equal to .12? h) Copy your R script for the above into the text box here.

Explanation / Answer

A) p = 75/121 = 0.62

B) At 94% Confidence interval the critical value is 1.89

C) The Confidence interval is

p +/- z* * sqrt(p * (1 - p )/n)

= 0.62 +/- 1.89 * sqrt(0.62 * 0.38/121)

= 0.62 +/- 0.083

= 0.537, 0.703

D) The width of the confidence interval= 0.703 - 0.537 = 0.166

E) Option-D) sqrt (p * (1 - p)/n)

F) standard deviation = sqrt (P * (1 - P)/n)

= sqrt (0.62 * 0.38/121) = 0.044

G) Margin of error = 0.12

Or, z* * sqrt (p * (1 - p)/n ) = 0.12

Or, 1.89 * sqrt (0.62 * 0.38 /n) = 0.12

Or, sqrt(n) = 1.89 * sqrt (0.62 * 0.38)/0.12

Or, sqrt(n) = 7.64

Or, n = 58

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