Question

Unknown to an experimenter, when a certain unfair coin is tossed, the probability of obtaining a head is p = 0.60. The experimenter tosses the coin 300 times in order to estimate p. What is the probability that the experimenter's point estimate of p will be within the interval (0.58, 0.63)? (Note: Do NOT use normal approximation to binomial; rather, use the sampling distribution of p.) (a) What are the expected number of successes and failures? (Provide answers as integers.) Expected successes Expected failures: (b) What is the probability that the experimenter's point estimate of p will be within the interval (0.58, 0.63)? (Round your answer to four decimal places.) 0.6467

Answer 1

The expected number of successes is 180 and the expected number of failures is 120. The probability that the experimenter's point estimate of p will be within the interval (0.58, 0.63) is approximately 0.6467 or 64.67%.

Step-by-step explanation:

To find the expected number of successes and failures, we can multiply the probability of success (p = 0.60) by the number of trials (300). So the expected number of successes is 0.60 x 300 = 180, and the expected number of failures is 300 - 180 = 120.

To find the probability that the experimenter's point estimate of p will be within the interval (0.58, 0.63), we can calculate the sampling distribution of p. The standard error of the proportion (SE) is given by

SE = sqrt(p(1-p)/n),

where p is the probability of success and n is the number of trials. In this case, SE = sqrt(0.60(1-0.60)/300) = 0.025.

Next, we can calculate the z-scores for the lower and upper bounds of the interval. The z-score is given by z = (p - p0)/SE, where p0 is the point estimate of p. For the lower bound, z = (0.58 - 0.60)/0.025 = -0.8, and for the upper bound, z = (0.63 - 0.60)/0.025 = 1.2.

Using a z-table or calculator, the probability that a standard normal random variable falls between -0.8 and 1.2 is approximately 0.6467, or 64.67%.