Homework V STAT 509: Statistics for Engineers March 26, 2018 Problem #1. Hack-a-

Question

Homework V STAT 509: Statistics for Engineers March 26, 2018 Problem #1. Hack-a-Shaq is a basketball strategy initially instituted in the NBA by the Dallas Mavericks coach Don Nelson to hinder the scoring ability of the opposing team by continuously committing personal fouls against one of its opposing players, the player chosen being the one with the weakest free throw peroentage among players on the court. Some of the previous victims of this strategy were Shaquille ONeal, Dennis Rodman etc. and recently there are Dwight Howard, DeAndre Jordan, Andre Drummond, ete. In particular Mr. Drunmond's career free throw percentage is 41.8%. In 2017-2018 season. Iowever, he has shown quite improved daily stats on free throws as follows 0 6 5 1 054032 0 0 Free Throw Attempt 4 10 3 2 8 8 8 4 0 5 8 4 5 21 40 3 1 0 3 3 2 8 1 6 1 1 ree Throw Free Throw 3 8 13 3 10 4 3 2 7 4 6 12 2 8 3 2 6 4 0 0 2 0 2 3 3 2 1 2 5 3 443 Free Throw Free Throw Attempt 6 2 0 2 0 4 44 4 2 2 844 610 8 Free Throw Attempt 6 8 5 3 836 5 2 8 46 7 4 16 This data can be obtained by the following code on RStudio download.file("http://people.stat.sc.edu/taeho/teaching/lab/AD_FT.RData", destfilAD FT.RData):loadC"AD FT. RData") Given this data, please answer the following questions to formally test whether he has im- proved his free throw accuracy under 0.01 significance level. 1. Please describe the target parameter and what is the value of the point estimate? 2. Check whether we have enough sample sizes to ensure the normality assumption. 3. State null and alternative hypothesis. Calculate the test statistic, z 5. Calculate the p-value. 6. Draw your conchusion and provide the interpretation within the context of the problem

Explanation / Answer

1. Here taget parameter is " proportion of free throw made in 2017 - 18 season"

Point estimate = 201/333 = 0.6036

Here the value 201 is total free throws made and 333 is total free throws attempted in given 68 attempts.

(2) Yes, as sample size = 333 so we can see that sample size > 30 to ensure normality assumption.

Parellaly, number of successes and number of failures are greater than 5.

(3) Null Hypothesis : H0 : p = p0

where po is the standard proportion of free throws made which is = 0.418

and p is the population parameter.

(4) Here standard error of proportion sep = sqrt(p0 * ( 1- p0)/n) = sqrt [0.418 * (1 - 0.418)/333] = 0.027

test statistic

Z = (p^ - p0)/sep = (0.6036 - 0.418)/ 0.027 = 6.8669

sohere p - value = Pr(Z > 6.8669) = 0.0000

we reject the null hypothesis.

6. Here we draw the conclusion that here proportion of free throws made in 2017 -18 is greater than his standard proportion of 0.418.

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