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Verizon [17]
3 years ago
7

Which of the following best represents the highest potential for nonresponse bias in a sampling strategy? Describe why this opti

on should be considered nonresponse a. Surveying a population on Sunday mornings for a new needs assessment b. Submitting a post online advertising the need for participants in a new study c. Asking people leaving a local election to take part in an exit poll d. Posting a leaflet in the elevator of a university asking for students to take part in a paid study
Mathematics
1 answer:
natta225 [31]3 years ago
7 0

Answer:

c. Asking people leaving a local election to take part in an exit poll

Step-by-step explanation:

Asking people leaving a local election to take part in an exit poll best represents the highest potential for nonresponse bias in a sampling strategy because of the importance of the local election compared to the exit polls.

It is worthy of note that nonresponse bias occurs when some respondents included in the sample do not respond to the survey. The major difference here is that the error comes from an absence of respondents not the collection of erroneous data. ...

Oftentimes, this form of bias is created by refusals to participate for one reason or another or the inability to reach some respondents.

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a) 0.0025 = 0.25% probability that none of the appeals will be successful.

b) 0.0207 = 2.07% probability that exactly one of the appeals will be successful.

c) 0.9768 = 97.68% probability that at least two of the appeals will be successful

Step-by-step explanation:

For each appeal, there are only two possible outcomes. Either it is succesful, or it is not. The probability of an appeal being succesful is independent of other appeals, so we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

45% of first-round appeals were successful.

This means that p = 0.45

Suppose 10 first-round appeals have just been received by a Medicare appeals office.

This means that n = 10

(a) Compute the probability that none of the appeals will be successful.

This is P(X = 0).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.45)^{0}.(0.55)^{10} = 0.0025

0.0025 = 0.25% probability that none of the appeals will be successful.

(b) Compute the probability that exactly one of the appeals will be successful.

This is P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{10,1}.(0.45)^{1}.(0.55)^{9} = 0.0207

0.0207 = 2.07% probability that exactly one of the appeals will be successful.

(c) What is the probability that at least two of the appeals will be successful

This is P(X \geq 2)

Either less than two appeals are succesful, or at least two are. The sum of the probabilities of these events is decimal 1. So

P(X < 2) + P(X \geq 2) = 1

P(X < 2) = P(X = 0) + P(X = 1) = 0.0025 + 0.0207 = 0.0232

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.0232 = 0.9768

0.9768 = 97.68% probability that at least two of the appeals will be successful

6 0
3 years ago
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