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A basketball player with a poor foul-shot record practices intensively during the off-season. He tells the coach that he has raised his proficiency from \(60 \%\) to \(80 \%\). Dubious, the coach asks him to take 10 shots, and is surprised when the player hits 9 out of 10. Did the player prove that he has improved? a. Suppose the player really is no better than before-still a \(60 \%\) shooter. What's the probability he can hit at least 9 of 10 shots anyway? (Hint: Use a Binomial model.) b. If that is what happened, now the coach thinks the player has improved when he has not. Which type of error is that? c. If the player really can hit \(80 \%\) now, and it takes at least 9 out of 10 successful shots to convince the coach, what's the power of the test? d. List two ways the coach and player could increase the power to detect any improvement.

Short Answer

Expert verified
a. The probability that an unimproved player (still shooting at 60%) could score at least 9 out of 10 shots can be calculated with the binomial model. b. The coach's belief that the player has improved when he actually hasn't constitutes a Type I error. c. The power of the test can be calculated as \( 1 - \beta \), where \( \beta \) is the probability of making a Type II error when the player has in fact improved to an 80% shooting rate. d. The power of the test could be increased by having the player perform more shots and/or by lowering the benchmark for evidence of improvement.

Step by step solution

01

Calculate binomial probability

The probability that the player can hit at least 9 of 10 shots if he's still a 60% shooter is calculated using the binomial model. The binomial probability formula is \[P(X=k) = C(n,k) * p^{k} * (1-p)^{n-k}\]. We need to calculate the probability for 9 and 10 successful shots, and add those together. \[ P(X=9) = C(10,9) * (0.6)^{9} * (0.4)^{1}\] \[ P(X=10) = C(10,10) * (0.6)^{10} * (0.4)^{0}\] Total probability, \(P(X \geq 9) = P(X=9) + P(X=10)\)
02

Identifying the statistical error

If the coach believes the player has improved but he actually hasn't, this is a type I error. This type of error occurs when a true null hypothesis is incorrectly rejected.
03

Calculating the power of the test

Test power is the probability of correctly rejecting the null hypothesis when it is false. If the player is now an 80% shooter, the coach will be convinced with at least 9 successful shots. The power of this test can be calculated as \[ Power = 1 - \beta \], where \( \beta \) is the probability of making a type II error (not rejecting the null hypothesis when it should be). So, \( \beta = P(X \leq 8) \), when success probability \( p = 0.8 \). Calculate \( P(X \leq 8) \) using the binomial probability formula, summing from 0 to 8, and then subtract from 1 to get the power.
04

Increasing the power of the test

To increase the power of the test, two strategies the coach and player could use are: 1) Increase the number of trials: If the coach asks the player to take more than 10 shots, the power of the test will increase because of a larger sample size. 2) Set a different benchmark: If the coach decides he will be convinced with 8 out of 10 successful shots rather than 9, this lowers the standard for evidence of improvement and thus increases the power to detect an improvement.

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Key Concepts

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

Type I and II Errors
Understanding Type I and II errors is crucial when interpreting the results of hypothesis testing in statistics. A Type I error, also known as a false positive, occurs when a true null hypothesis is incorrectly rejected. This is akin to a court wrongly convicting an innocent person. In the basketball player's case, if the coach concludes the player has improved based on the 9 successful shots, but the player's ability is still at 60%, the coach has made a Type I error.

A Type II error, also called a false negative, happens when the null hypothesis is not rejected when it actually is false. It’s similar to a guilty person being acquitted. So, if the player has indeed improved to an 80% shooting rate and the coach's test fails to detect this improvement, a Type II error has occurred.

To manage these errors, one must understand their relationship with the significance level and the power of a test. Typically, researchers choose a significance level (commonly 5%), which directly affects the probability of making a Type I error. Lowering this level makes committing a Type I error less likely but increases the risk of a Type II error.
Statistical Power
Statistical power is the measure of a test's ability to detect an effect, if there is one to be detected. It is the probability of correctly rejecting the null hypothesis when it is, in fact, false. In simpler terms, it’s the test's ability to identify a true improvement or change. In our basketball player’s scenario, if the player really has improved his foul-shot record to 80%, the statistical power reflects the likelihood that the coach's test will confirm this improvement.

The formula used to calculate power is \( Power = 1 - \beta \), where \( \beta \) represents the probability of committing a Type II error. To enhance the power of a test, one could increase the sample size or adjust the test's criteria, such as the number of shots needed to demonstrate improvement. An increased sample size provides more data, making it simpler to detect true effects. Altering the necessary criteria, however, can be a balancing act between the risk of Type I and Type II errors.
Hypothesis Testing
Hypothesis testing is a formal method used to test whether a hypothesis regarding a parameter in a population can be supported or not. The process usually begins with a null hypothesis (\( H_0 \)), which represents a skeptic's stance or a default position that there is no effect or no difference. The basketball player’s percentage remaining at 60% serves as our null hypothesis in the example.

An alternative hypothesis (\( H_1 \)) is what the researcher wants to prove, such as the player's improvement in shot proficiency to 80%. After determining the null and alternative hypotheses, a test statistic is chosen, and based on the data, a decision is made as to whether to reject the null hypothesis. This decision hinges on whether the test statistic falls within a predetermined 'rejection region', which is related to the level of significance chosen for the test.

It's crucial for a coach or a researcher to clearly define the criteria for judging improvements (like the 9 out of 10 shots) and to understand the implications of Type I and II errors as well as the statistical power in interpreting the results of the hypothesis test. This clarity ensures a more reliable assessment of performance or research findings.

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Most popular questions from this chapter

Which of the following are true? If false, explain briefly. a. A very high P-value is strong evidence that the null hypothesis is false. b. A very low P-value proves that the null hypothesis is false. c. A high P-value shows that the null hypothesis is true. d. A P-value below 0.05 is always considered sufficient evidence to reject a null hypothesis.

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In a drawer are two coins. They look the same, but one coin produces heads \(90 \%\) of the time when spun while the other one produces heads only \(30 \%\) of the time. You select one of the coins. You are allowed to spin it once and then must decide whether the coin is the \(90 \%\) - or the \(30 \%\) -head coin. Your null hypothesis is that your coin produces \(90 \%\) heads. a. What is the alternative hypothesis? b. Given that the outcome of your spin is tails, what would you decide? What if it were heads? c. How large is \(\alpha\) in this case? d. How large is the power of this test? (Hint: How many possibilities are in the alternative hypothesis?) e. How could you lower the probability of a Type I error and increase the power of the test at the same time?

A new reading program may reduce the number of elementary school students who read below grade level. The company that developed this program supplied materials and teacher training for a large-scale test involving nearly 8500 children in several different school districts. Statistical analysis of the results showed that the percentage of students who did not meet the grade- level goal was reduced from \(15.9 \%\) to \(15.1 \%\). The hypothesis that the new reading program produced no improvement was rejected with a P-value of 0.023 . a. Explain what the P-value means in this context. b. Even though this reading method has been shown to be significantly better, why might you not recommend that your local school adopt it?

Which of the following are true? If false, explain briefly. a. A very low P-value provides evidence against the null hypothesis. b. A high P-value is strong evidence in favor of the null hypothesis. c. AP-value above 0.10 shows that the null hypothesis is true. d. If the null hypothesis is true, you can't get a P-value below 0.01

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