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Question

a: Graci x 0 Cengage& x SmySAITLocx G what are fir x g Tur itin-R x- Tur 021&quiz; action- takeQuizeiquiz probGuidHONAPCOA801010000003e5260080000&ck-m; 15230 Attempts: Keep the Highest:8 S. 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 o.s, the expected number of successes is 10o.5)- 5. Since each success pays $10, the expected value of the game is 5($10), S50. 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 he or she is 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 neutral" if the maximum amount he or she is willing to pay to play is equal to $50, the game's 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, the proportion of the same population who are not risk neutral. Use the sample results to estimate the proportion p The proportion 1-p of adults in the The proportion p of adults in the sample who are risk neutral is sample who are not risk neutral is You conclude that the sampling distribution of P can be approximated by a normal distribution, The sampling distribution of P has a mean and an estimated Distributions Use the who are risk neutral, You can be 90% confident that the interval estimate LCL to UCL includes the . P, the

Explanation / Answer

proportion of risk neutral, pcap = 34/1536 = 0.0221

Proportion who are not risk neutral, 1 - pcap = 1 - 0.0221 = 0.9779

You can conclude that the sampling distribution pcap can be approximated by a normal distribution, because np >= 5 and n(1-p) >= 5

Sampling distribution,

p = 0.0221

sd = sqrt(p*(1-p)/n) = sqrt(0.0221*(1-0.0221)/1536) = 0.0038

n = 1536

p = 0.0221

z-value of 90% CI = 1.6449

SE = sqrt(p*(1-p)/n) = 0.00375

ME = z*SE = 0.00617

Lower Limit = p - ME = 0.01596

Upper Limit = p + ME = 0.02831

90% CI (0.016 , 0.0283 )

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