invest_3ed.pdf

# B state the appropriate null and alternative

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(b) State the appropriate null and alternative hypotheses that the manager wants to test. [ Hint : Which is the “by chance” hypothesis and which hy pothesis is the manager trying to gather evidence for?] H 0 : H a : (c) Suppose the manager decides to give the player a sample of 20 at-bats. How likely do you think it is that this 0.333 hitter will be able to convince the manager he has improved from the 0.250 hitter he used to be? Explain. So the question is, how likely it is that someone who is now genuinely a 0.333 player will be able to demonstrate his improvement to the manager (will convince the manager to reject the null hypothesis)? We will investigate this through a two-step process. First, we need to determine how many hits the player would need to get to convince the manager that he’s better than a 0.250 hitter. Then we need to determine how likely a 0.333 hitter is to perform that well. (d) Use the One Proportion Inference applet to see the null distribution for the number of hits obtained in 20 at-bats for a 0.250 hitter. (Check the Exact Binomial box instead of Draw Samples.) Based on your graph, (roughly) how many hits would the hitter need to get to convince you that he is better than a 0.250 hitter?

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Chance/Rossman, 2015 ISCAM III Investigation 1.6 55 We can analyze this scenario two different ways : simulation and binomial probabilities. Let’s start with simulation. (e) Open the Power Simulation applet. x Specify .250 as the hypothesized probability of success x Specify .333 as the alternative probability of success. x Enter 20 as the sample size x Enter 1000 as the number of repetitions. x Click the Draw Samples button. The top graph displays how many hits a 0.250 hitter got in the 20 at-bats for 1000 different sessions. The bottom graph displays how many hits a 0.333 hitter got in 20 at-bats for 1000 different sessions. Do the two distributions have much overlap? What does this say about how likely the player is to convince the manager that he really has improved in the sample of 20 at-bats? Explain. Let’s assume the manager will be convinced the player has improved if the player gets so many hits in his 20 at-bats, that there is less than a 5% chance that a 0.250 hitter would perform that well. In other words, the manager wants to test the above null hypothesis and needs a p-value of 0.05 or smaller in order to be convinced that the player’s long -run probability of getting a hit is now above 0.250. This pre- specified cut-off value for the p-value is called the level of significance . Definition: The level of significance , often denoted by D (“alpha”), is a standard that can be specified in advance for deciding when a p-value is small enough to provide convincing evidence against the null hypothesis. Then if the study’s p -value < D , we reject the null hypothesis in favor of the alternative and the result is said to be “statistically significant at the D - level.” Otherwise we “fail to reject” the null hypothesis, not finding convincing evidence to reject it. Common choices for D are 0.01 and 0.05.
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