Toss The Coin Four Times If It Lands All Heads

Toss The Coin Four Times If The Coin Lands Either All Heads Or All Ta

Toss the coin four times. If the coin lands either all heads or all tails, reject Ho: p=1/2 (the p denotes the chance for the coin to land on heads). Complete parts a and b. a. what is the probability of a type 1 error for this procedure? b. if p=4/5, what is the probability of a type II error for this procedure? Suppose that a class contains 15 boys and 30 girls, and that 10 students are to be selected at random for a special assignment. Find the probability that exactly 3 boys will be selected.

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The given problem involves two distinct statistical scenarios: one about a coin-tossing hypothesis test and the other about a hypergeometric probability related to student selection. The first part concerns evaluating the probabilities of Type I and Type II errors in a hypothesis test based on observing the outcomes of repeated coin tosses. The second part involves calculating the probability of selecting a specific number of boys from a mixed group through random sampling. Each scenario demonstrates fundamental concepts in probability and statistical testing, which are integral to understanding the behavior of random processes and sampling distributions.

Analysis of the Coin Toss Hypotheses

The first part of the problem concerns an experiment where a coin is tossed four times, and the rule for rejecting the null hypothesis (H0: p = 1/2) is based on observing whether all outcomes are heads or all tails. Specifically, the test is constructed as follows: if the four tosses are all heads or all tails, then H0 is rejected; otherwise, H0 is not rejected. This decision rule makes this a specific form of a binomial test, focusing on extreme outcomes that could signal a deviation from the assumed probability p=1/2.

Part a: Probability of a Type I Error

A Type I error occurs when the null hypothesis is true (p=1/2), but the test incorrectly rejects it. Under these circumstances, the probability of a Type I error, denoted as alpha, is equal to the probability of observing either all heads or all tails when p=1/2. For four independent coin tosses, the probability of all heads is (1/2)^4 = 1/16, and similarly, the probability of all tails is also 1/16. Since these two events are mutually exclusive, the total probability of either all heads or all tails, given p=1/2, is:

• α = P(all heads or all tails | p=1/2) = P(all heads) + P(all tails) = 1/16 + 1/16 = 2/16 = 1/8.

Therefore, the probability of a Type I error for this procedure is 1/8 or 0.125.

Part b: Probability of a Type II Error when p=4/5

A Type II error occurs when the alternative hypothesis (p ≠ 1/2, in this case p=4/5) is true but the test fails to reject H0. The test rejects H0 when all four tosses are uniformly heads or tails, i.e., the extreme outcomes. When p=4/5, the probability of heads in a single toss is 0.8, and tails 0.2.

Under p=4/5, the probability of observing all heads in four flips is (0.8)^4 = 0.4096, while the probability of all tails is (0.2)^4 = 0.0016. Since the test rejects H0 when all outcomes are heads or tails, the probability of not rejecting H0 (i.e., failing to observe either all heads or all tails) when p=4/5 is:

• β = P(not rejecting H0 | p=4/5) = P(not all heads and not all tails | p=4/5) = 1 - [P(all heads) + P(all tails)] = 1 - (0.4096 + 0.0016) = 1 - 0.4112 = 0.5888.

Thus, the probability of a Type II error when p=4/5 is approximately 0.5888, which indicates a substantial chance (almost 59%) that the test will fail to reject H0 even if the true p is 0.8.

Probability of Selecting Exactly 3 Boys from the Class

The second problem addresses the probability of selecting exactly 3 boys out of 10 students randomly chosen from a class with 15 boys and 30 girls, totaling 45 students. This is a hypergeometric probability problem, which calculates the chance of obtaining a specific number of successes (boys) in a sample drawn without replacement from a finite population.

Calculating the Hypergeometric Probability

The hypergeometric probability formula is:

P(X = k) = [C(K, k) * C(N - K, n - k)] / C(N, n)

Where:

• N = total population size = 45
• K = total number of boys = 15
• n = number of students selected = 10
• k = number of boys selected = 3

First, compute each combination:

• C(K, k) = C(15, 3) = 455
• C(N - K, n - k) = C(30, 7) = 2035800
• C(N, n) = C(45, 10) = 1,221,759,297

Pluging these values into the hypergeometric formula gives:

P(3 boys) = (455 * 2,035,800) / 1,221,759,297 ≈ (924,669,000) / 1,221,759,297 ≈ 0.757

Therefore, the probability that exactly 3 boys are selected out of 10 students randomly chosen from the class is approximately 0.757, or 75.7%.

Conclusion

This analysis exemplifies the application of probability theory in decision-making and sampling scenarios. The first scenario demonstrates how specific outcome-based tests can be used to evaluate hypotheses about probabilities, highlighting the importance of understanding Type I and Type II errors. The second illustrates the hypergeometric distribution's role in finite population sampling, which has widespread applications in statistics, quality control, and survey sampling. Both types of problems underscore the relevance of probability calculations in real-world decision-making processes and hypothesis testing.

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