4. Interval estimation of a population proportion Aa Aa Think about the followin

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4. Interval estimation of a population proportion Aa Aa 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 1,536 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. (Data source: These data were adapted from Ben Mansour, Selima, Jouini, Elyes, Marin, Jean-Michel, Napp, Clotilde, & Robert, Christian. (2008). Are risk-averse agents more optimistic? A Bayesian estimation approach. Journal of Applied Econometrics, 23(6), 843-860.) Someone is described as "risk loving" if the maximum amount he or she is willing to pay to play is greater than $50, the game's expected value. Suppose in this 1,536-person sample, 20 people are risk loving. Let p denote the proportion of the adult population aged 22 to 55 who are risk loving and 1- p, the proportion of the same population who are not risk loving. Use the sample results to estimate the proportion p The proportion of adults in the sample who are risk loving is sample who are not risk loving is . The proportion 1- of adults in the 0.01302 0.20101 0.79899 0.98698 You because conclude that the sampling distribution of p mated by a normal distribution, The sampling distribution of standard deviation of has a mean and an estimated

Explanation / Answer

4. Interval estimation of a population proportion

a)

p^(risk loving) = x/n = 20/1536 = 0.013020833 [ANSWER, FIRST BLANK]

Hence,

1 - p^ = 0.986979167 [ANSWER, 2ND BLANK]

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b)

You can conclude that the sampling distribution p^ can be approximated by a normal distribution, because both the number of risk loving (20) and not risk loving (1516) are both >= 5. [ANSWER]

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c)

It has a mean of

p = unknown,

and the standard deviation is, as n = 1536,

s = standard deviation = sqrt(p(1-p)/n) =    0.002892529 [ANSWER]

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d)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.013020833          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.002892529          
              
Now, for the critical z,              
alpha/2 =   0.005          
Thus, z(alpha/2) =    2.575829304          
Thus,              
Margin of error = z(alpha/2)*sp =    0.007450662          
lower bound = p^ - z(alpha/2) * sp =   0.005570171          
upper bound = p^ + z(alpha/2) * sp =    0.020471495          
              
Thus, the confidence interval is              
              
(   0.005570171   ,   0.020471495   ) [ANSWER, CONFIDENCE INETRVAL]

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