Interval Estimation of Population Portion Think about the following game: A fair

Question

Interval Estimation of Population Portion

Think about the following game: A fair coin is tossed 10 times. Each time the toss results in heads, you receive $10; for tails, you get nothing. What is the maximum amount you would pay to play the game?

As a statistics student, you are aware that the game is a 10-trial binomial experiment. A toss that lands on heads is defined as a success, and because the probability of a success is 0.5, the expected number of successes is 10(0.5) = 5. Since each success pays $10, the expected value of the game is 5(10) = $50.

Suppose each person in a random sample of 1536 adults between the ages of 22 and 55 is invited to play this game.Each person is asked the maximum amount they are willing to pay to play.

Someone is described as "risk neutral" if the maximum amount he or she is willing to pay to play is equal to $50, the games expected value. Suppose in this 1536-person sample, 34 people are risk neutral. Let p denote the proportion of the adult population aged 22 to 55 who are risk neutral and 1-p bar (p with a bar over top), the proportion of the same population who are not risk neutral. Use these sample results to estimate the proportion on p.

The proportion of p bar (p with bar over top)of adults in the sample who are risk neutral is _______. The proportion 1 - p bar of adults in the sample who are not risk neutral is _______.

You _________ (can or cannot) conclude that the sampling distribution of p bar can be approximately by a normal distribution, because ___________.

The sampling distribution of p bar has a mean _________ and an estimated standard deviation of _______.

You can be 99% confident that interval estimate _______ to _______ includes the population proportion p, the proportion of adults aged 22 to 55 who are risk neutral.

Explanation / Answer

The proportion, pbar=x/n, where, x denote number of events, and n denotes number of trials.

=34/1536=0.0221 (ans)

The proportion, 1-pbar=1-0.0221=0.9779 (ans)

One can (ans) conclude, because, both n=1536 and n-x=1536-34=1502 are greater than 5 (ans).

Mean: pbar=0.0221 (ans) standard deviation=sqrt[pbar(1-pbar)/n]=sqrt[0.0221(1-0.0221)/1536]=0.0038 (ans)

The 99% c.i for true population proportion is as follows:

pbar+-zalpha/2 sqrt[pbar(1-pbar)/n], where, pbar is sample proportion, z denotes z critical at alpha/2, n is sample size.

=0.0221+-2.576 sqrt[0.0221(1-0.0221)/1536]

=0.0124, 0.0318 (ans)

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.