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A coin is flipped 25 times. Let \(x\) be the number of flips that result in heads (H). Consider the following rule for deciding whether or not the coin is fair: Judge the coin fair if \(8 \leq x \leq 17\). Judge the coin biased if either \(x \leq 7\) or \(x \geq 18\). a. What is the probability of judging the coin biased when it is actually fair? b. Suppose that a coin is not fair and that \(P(\mathrm{H})=0.9\) What is the probability that this coin would be judged fair? What is the probability of judging a coin fair if \(P(\mathrm{H})=0.1 ?\) c. What is the probability of judging a coin fair if \(P(\mathrm{H})=0.6 ?\) if \(P(\mathrm{H})=0.4 ?\) Why are these probabilities large compared to the probabilities in Part (b)? d. What happens to the "error probabilities" of Parts (a) and \((b)\) if the decision rule is changed so that the coin is judged fair if \(7 \leq x \leq 18\) and unfair otherwise? Is this a better rule than the one first proposed? Explain.

Short Answer

Expert verified
The probability of judging a coin biased when it is actually fair is $P(\text{judging biased when fair}) = P(x \leq 7) + P(x \geq 18)$. For \(P(H) = 0.9\) and \(P(H) = 0.1\), the probabilities of judging a coin fair are calculated using the given range of \(8 \leq x \leq 17\). The probabilities are larger in part (c) because the probability of heads is closer to 0.5. Changing the decision rule to \(7 \leq x \leq 18\), we need to recalculate the error probabilities and compare with the original ones to determine if the new rule is better.

Step by step solution

01

Determine the probability of judging the coin biased when it is actually fair

When the coin is fair, the probability of getting heads in each flip is \(P(H)=0.5\). According to the given conditions, we judge a coin to be biased if \(x\le7\) or \(x\ge18\). To find the probability of judging the coin biased, we need to calculate the probability that we will get number of heads between these ranges using the binomial probability formula.
02

Calculate the probability for x≤7 and x≥18

Using the binomial probability formula for \(x\le7\) and \(x\ge18\), we get: \[P(x \leq 7) = \sum_{x=0}^{7} \binom{25}{x} (0.5)^x (0.5)^{25-x}\] \[P(x \geq 18) = \sum_{x=18}^{25} \binom{25}{x} (0.5)^x (0.5)^{25-x}\]
03

Calculate the total probability of judging a coin biased

Summing up the above probabilities gives us the total probability of judging a coin biased when it is actually fair. Thus: \[P(\text{judging biased when fair}) = P(x \leq 7) + P(x \geq 18)\] #b. Suppose that a coin is not fair and that \(P(H)=0.9\) What is the probability that this coin would be judged fair? What is the probability of judging a coin fair if \(P(H)=0.1 ?\) #
04

Determine the probability of judging as fair when P(H)=0.9

According to the given conditions, we judge a coin to be fair if \(8\le x\le 17\). Let's calculate the probability of judging a coin fair when \(P(H)=0.9\): \[P(8 \leq x \leq 17) = \sum_{x=8}^{17} \binom{25}{x} (0.9)^x (0.1)^{25-x}\]
05

Determine the probability of judging as fair when P(H)=0.1

Similarly, let's calculate the probability of judging a coin fair when \(P(H)=0.1\): \[P(8 \leq x \leq 17) = \sum_{x=8}^{17} \binom{25}{x} (0.1)^x (0.9)^{25-x}\] #c. What is the probability of judging a coin fair if \(P(H)=0.6 ?\) if \(P(H)=0.4 ?\) Why are these probabilities large compared to the probabilities in Part (b)? #
06

Determine the probability of judging fair when P(H)=0.6

Calculate the probability of judging a coin fair when \(P(H)=0.6\): \[P(8 \leq x \leq 17) = \sum_{x=8}^{17} \binom{25}{x} (0.6)^x (0.4)^{25-x}\]
07

Determine the probability of judging fair when P(H)=0.4

Calculate the probability of judging a coin fair when \(P(H)=0.4\): \[P(8 \leq x \leq 17) = \sum_{x=8}^{17} \binom{25}{x} (0.4)^x (0.6)^{25-x}\]
08

Explain why these probabilities are large compared to Part (b)

The probabilities calculated in this part are larger compared to Part (b) because the probability of heads is closer to 0.5 (fair coin), and thus the results tend to be less biased, making it more likely to be judged as fair in the given range. #d. What happens to the "error probabilities" of Parts (a) and (b) if the decision rule is changed so that the coin is judged fair if \(7 \leq x \leq 18\) and unfair otherwise? Is this a better rule than the one first proposed? Explain. #
09

Calculate the error probabilities

Redo the steps in Parts (a) and (b) by changing the range from \(8 \leq x \leq 17\) to \(7 \leq x \leq 18\). Calculate the error probabilities of judging a coin biased when it is actually fair and judging a coin fair when it is actually biased.
10

Compare the new error probabilities with the original ones

Compare the newly calculated error probabilities with the ones calculated in Parts (a) and (b). If the new error probabilities are lower, the new decision rule is better than the first proposed one, as it reduces the chance of making incorrect judgments.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability of Judging a Coin Biased
When flipping a fair coin, which we assume has an equal chance of landing on heads or tails, we often want to decide if the coin might be biased based on the outcomes. For example, if we flip a coin multiple times and it lands on heads much more often than tails, we may suspect the coin is not fair.

To statistically determine if a coin is biased, we set up a decision rule based on the number of observed heads in a series of flips. Using binomial distribution, which applies to events with two possible outcomes—like flipping a coin—we can calculate the probability of declaring a coin biased under our decision rule, even when the coin is actually fair, which is our error probability.

This error probability represents the chance of a Type I error, or a false positive, in hypothesis testing wherein we incorrectly reject the null hypothesis (that the coin is fair). Recognizing the possibility of this error is crucial in statistical hypothesis testing and helps us in refining our decision rules to minimize incorrect judgments.
Calculation Using Binomial Distribution
The binomial distribution is used to determine the probability of observing a specific number of 'successes' in a fixed number of independent trials. For coin flips, 'success' could be defined as the coin landing on heads.

To calculate the probability of observing a certain number of heads using the binomial distribution, we apply the formula:
  • The number of trials (n), which is how many times we flip the coin.
  • The number of successes (x), the number of heads we're interested in.
  • The probability of success on a single trial (p), the chance of getting a head on a flip.
The binomial formula is then
\[P(X = x) = \binom{n}{x}p^x(1-p)^{n-x}\]
This formula calculates the likelihood of getting exactly x heads in n tosses. However, to find the probability of getting at least or at most a certain number of heads, we sum up the probabilities for all relevant values of x.
Error Probability
Error probability in the context of a coin toss refers to the chances of making an incorrect judgment about the coin's fairness. There are two types of errors that can occur:
  • Type I error (false positive): Concluding that the coin is biased when it is actually fair.
  • Type II error (false negative): Failing to detect that the coin is biased when it is not fair.
The error probabilities are determined based on the decision rule we set for judging a coin as biased or fair. In the original exercise, the decision rule has implicit error probabilities—which are the probabilities that a fair coin will be judged biased or that a biased coin will be judged fair. It's important to understand and manage these error probabilities when designing and evaluating statistical tests to avoid incorrect conclusions.
Fair Coin Probability
The probability of a fair coin refers to the likelihood of a coin landing on heads or tails without any bias. For a perfectly fair coin, this probability is 0.5 for each outcome. In hypothesis testing, we operate under the assumption that the coin is fair, and any deviations from the expected number of heads or tails in a series of flips would need to be analyzed to determine if they are due to random chance or if they suggest an actual bias.

When we consider a series of flips, we need to account for the natural fluctuations that can occur even with a fair coin. Even a fair coin might show an uneven distribution of heads and tails over a short series of flips purely by chance. By applying the binomial distribution, we can calculate the chances of various outcomes and decide if the observed results fall within the range we would expect for a fair coin, thereby helping us to avoid incorrectly judging a fair coin as biased.

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Most popular questions from this chapter

You are to take a multiple-choice exam consisting of 100 questions with five possible responses to each question. Suppose that you have not studied and so must guess (randomly select one of the five answers) on each question. Let \(x\) represent the number of correct responses on the test. a. What kind of probability distribution does \(x\) have? b. What is your expected score on the exam? (Hint: Your expected score is the mean value of the \(x\) distribution.) c. Calculate the variance and standard deviation of \(x\). d. Based on your answers to Parts \((\mathrm{b})\) and \((\mathrm{c}),\) is it likely that you would score over 50 on this exam? Explain the reasoning behind your answer.

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A grocery store has an express line for customers purchasing at most five items. Consider the random variable \(x=\) the number of items purchased by a randomly selected customer using this line. Make two tables that represent two different possible probability distributions for \(x\) that have the same standard deviation but different means.

Thirty percent of all automobiles undergoing an emissions inspection at a certain inspection station fail the inspection. a. Among 15 randomly selected cars, what is the probability that at most 5 fail the inspection? b. Among 15 randomly selected cars, what is the probability that between 5 and 10 (inclusive) fail the inspection? c. Among 25 randomly selected cars, what is the mean value of the number that pass inspection, and what is the standard deviation? d. What is the probability that among 25 randomly selected cars, the number that pass is within 1 standard deviation of the mean value?

Suppose that fund-raisers at a university call recent graduates to request donations for campus outreach programs. They report the following information for last year's graduates: $$\begin{array}{lllll}\text { Size of donation } & \$ 0 & \$ 10 & \$ 25 & \$\end{array}$$ 0.30 0 Proportion of calls 0.45 .20 0.05 Three attempts were made to contact each graduate. A donation of $$\$ 0$$ was recorded both for those who were contacted but declined to make a donation and for those who were not reached in three attempts. Consider the variable \(x=\) amount of donation for a person selected at random from the population of last year's graduates of this university. a. Write a few sentences describing what donation amounts you would expect to see if the value of \(x\) was observed for each of 1000 graduates. b. What is the most common value of \(x\) in this population? c. What is \(P(x \geq 25)\) ? d. What is \(P(x>0)\) ?

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