Question

Suppose you work for a survey research company. In a typical survey, you mail questionnaires to 150 companies. Of course, some of these companies might decide not to respond. Assume that the nonresponse rate is 45%; that is, each company's probability of not responding, independently of the others, is 0.45. Round your answers to three decimal places and if your answer is zero, enter "0". a. If your company requires at least 90 responses for a valid survey, find the probability that it will get this many. Use a data table to see how your answer varies as a function of the nonresponse rate (for a reasonable range of response rates surrounding 45%).

I cannot get a table to work with the probabilities for

25%
30%
35%
40%
45%
50%
55%
60%
65%

Answer 1

(a)

For non-response rate of 25%, number of responses from 150 companies X will follow Binomial distribution with parameters n = 150 and, p = 1 - 0.25 = 0.75

Using Normal approximation of Binomial theorem, X will follow Normal distribution with mean = np = 150*0.75 = 112.5 and standard deviation = sqrt(150*0.75*(1-0.75)) = 5.303

Probability to get at least 90 responses for a valid survey = P(X 90) = P(X > 89.5) (Using Continuity Correction)

= P[Z > (89.5 - 112.5)/5.303] = P[Z > -4.38] = 1

For non-response rate of 30%, number of responses from 150 companies X will follow Binomial distribution with parameters n = 150 and, p = 1 - 0.3 = 0.7

Using Normal approximation of Binomial theorem, X will follow Normal distribution with mean = np = 150*0.7 = 105 and standard deviation = sqrt(150*0.7*(1-0.7))  = 5.612

Probability to get at least 90 responses for a valid survey = P(X 90) = P(X > 89.5) (Using Continuity Correction)

= P[Z > (89.5 - 105)/5.612] = P[Z > -2.76] = 0.997

For non-response rate of 35%, number of responses from 150 companies X will follow Binomial distribution with parameters n = 150 and, p = 1 - 0.35 = 0.65

Using Normal approximation of Binomial theorem, X will follow Normal distribution with mean = np = 150*0.65 = 97.5 and standard deviation = sqrt(150*0.65*(1-0.65)) = 5.842

Probability to get at least 90 responses for a valid survey = P(X 90) = P(X > 89.5) (Using Continuity Correction)

= P[Z > (89.5 - 97.5)/5.842] = P[Z > -1.37] = 0.915

For non-response rate of 40%, number of responses from 150 companies X will follow Binomial distribution with parameters n = 150 and, p = 1-0.40 = 0.6

Using Normal approximation of Binomial theorem, X will follow Normal distribution with mean = np = 150*0.6 = 90 and standard deviation = sqrt(150*0.6*(1-0.6)) = 6

Probability to get at least 90 responses for a valid survey = P(X 90) = P(X > 89.5) (Using Continuity Correction)

= P[Z > (89.5 - 90)/6] = P[Z > -0.08] = 0.533

For non-response rate of 45%, number of responses from 150 companies X will follow Binomial distribution with parameters n = 150 and, p = 1 - 0.45 = 0.55

Using Normal approximation of Binomial theorem, X will follow Normal distribution with mean = np = 150*0.55 = 82.5 and standard deviation = sqrt(150*0.55*(1-0.55)) = 6.093

Probability to get at least 90 responses for a valid survey = P(X 90) = P(X > 89.5) (Using Continuity Correction)

= P[Z > (89.5 - 82.5)/6.093] = P[Z > 1.15] = 0.125

For non-response rate of 50%, number of responses from 150 companies X will follow Binomial distribution with parameters n = 150 and, p = 1 - 0.5 = 0.5

Using Normal approximation of Binomial theorem, X will follow Normal distribution with mean = np = 150*0.5 = 75 and standard deviation = sqrt(150*0.5*(1-0.5)) = 6.124

Probability to get at least 90 responses for a valid survey = P(X 90) = P(X > 89.5) (Using Continuity Correction)

= P[Z > (89.5 - 75)/6.124] = P[Z > 2.368] = 0.009

For non-response rate of 55%, number of responses from 150 companies X will follow Binomial distribution with parameters n = 150 and, p = 1 - 0.55 = 0.45

Using Normal approximation of Binomial theorem, X will follow Normal distribution with mean = np = 150*0.45 = 67.5 and standard deviation = sqrt(150*0.45*(1-0.45)) = 6.093

Probability to get at least 90 responses for a valid survey = P(X 90) = P(X > 89.5) (Using Continuity Correction)

= P[Z > (89.5 - 67.5)/6.093] = P[Z > 3.61] = 0

For non-response rate of 60%, number of responses from 150 companies X will follow Binomial distribution with parameters n = 150 and, p = 1 - 0.60 = 0.4

Using Normal approximation of Binomial theorem, X will follow Normal distribution with mean = np = 150*0.4 = 60 and standard deviation = sqrt(150*0.4*(1-0.4)) = 6

Probability to get at least 90 responses for a valid survey = P(X 90) = P(X > 89.5) (Using Continuity Correction)

= P[Z > (89.5 - 60)/6] = P[Z > 4.92] = 0

For non-response rate of 65%, number of responses from 150 companies X will follow Binomial distribution with parameters n = 150 and, p = 1 - 0.65 = 0.35

Using Normal approximation of Binomial theorem, X will follow Normal distribution with mean = np = 150*0.35 = 52.5 and standard deviation = sqrt(150*0.35*(1-0.35)) = 5.842

Probability to get at least 90 responses for a valid survey = P(X 90) = P(X > 89.5) (Using Continuity Correction)

= P[Z > (89.5 - 52.5)/5.842] = P[Z > 6.33] = 0