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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

A medical researcher tested a new treatment for poison ivy against the traditional ointment. He concluded that the new treatment is more effective. Explain what the P-value of 0.047 means in this context.

The manufacturer of a metal stand for home TV sets must be sure that its product will not fail under the weight of the TV. Since some larger sets weigh nearly 300 pounds, the company's safety inspectors have set a standard of ensuring that the stands can support an average of over 500 pounds. Their inspectors regularly subject a random sample of the stands to increasing weight until they fail. They test the hypothesis \(\mathrm{H}_{0}: \mu=500\) against \(\mathrm{H}_{\mathrm{A}}: \mu>500,\) using the level of significance \(\alpha=0.01\). If the sample of stands fails to pass this safety test, the inspectors will not certify the product for sale to the general public. a. Is this an upper-tail or lower-tail test? In the context of the problem, why do you think this is important? b. Explain what will happen if the inspectors commit a Type I error. c. Explain what will happen if the inspectors commit a Type II error.

For each of the following situations, find thecritical value(s) for \(z\) or \(t\). a. \(\mathrm{H}_{0}: p=0.5\) vs. \(\mathrm{H}_{\mathrm{A}}: p \neq 0.5\) at \(\alpha=0.05\). b. \(\mathrm{H}_{0}: p=0.4 \mathrm{vs} . \mathrm{H}_{\mathrm{A}}: p>0.4\) at \(\alpha=0.05\). c. \(\mathrm{H}_{0}: \mu=10\) vs. \(\mathrm{H}_{\mathrm{A}}: \mu \neq 10\) at \(\alpha=0.05 ; n=36\). d. \(\mathrm{H}_{0}: p=0.5\) vs. \(\mathrm{H}_{\mathrm{A}}: p>0.5\) at \(\alpha=0.01 ; n=345\) e. \(\mathrm{H}_{0}: \mu=20\) vs. \(\mathrm{H}_{\mathrm{A}}: \mu<20\) at \(\alpha=0.01 ; n=1000\)

Spam filters try to sort your e-mails, deciding which are real messages and which are unwanted. One method used is a point system. The filter reads each incoming e-mail and assigns points to the sender, the subject, key words in the message, and so on. The higher the point total, the more likely it is that the message is unwanted. The filter has a cutoff value for the point total; any message rated lower than that cutoff passes through to your inbox, and the rest, suspected to be spam, are diverted to the junk mailbox. We can think of the filter's decision as a hypothesis test. The null hypothesis is that the e-mail is a real message and should go to your inbox. A higher point total provides evidence that the message may be spam; when there's sufficient evidence, the filter rejects the null, classifying the message as junk. This usually works pretty well, but, of course, sometimes the filter makes a mistake. a. When the filter allows spam to slip through into your inbox, which kind of error is that? b. Which kind of error is it when a real message gets classified as junk? c. Some filters allow the user (that's you) to adjust the cutoff. Suppose your filter has a default cutoff of 50 points, but you reset it to 60 . Is that analogous to choosing a higher or lower value of \(\alpha\) for a hypothesis test? Explain. d. What impact does this change in the cutoff value have on the chance of each type of error?

For each of the following situations, state whether a Type I, a Type II, or neither error has been made. a. A test of \(\mathrm{H}_{0}: \mu=25\) vs. \(\mathrm{H}_{\mathrm{A}}: \mu>25\) rejects the null hypothesis. Later it is discovered that \(\mu=24.9\). b. A test of \(\mathrm{H}_{0}: p=0.8\) vs. \(\mathrm{H}_{\mathrm{A}}: p<0.8\) fails to reject the null hypothesis. Later it is discovered that \(p=0.9\). c. A test of \(\mathrm{H}_{0}: p=0.5\) vs. \(\mathrm{H}_{\mathrm{A}}: p \neq 0.5\) rejects the null hypothesis. Later it is discovered that \(p=0.65\). d. A test of \(\mathrm{H}_{0}: p=0.7\) vs. \(\mathrm{H}_{\mathrm{A}}: p<0.7\) fails to reject the null hypothesis. Later it is discovered that \(p=0.6\).

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