PLEASE ANSWER ALL PARTS For this question, you will use JMP to analyze the Defla
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PLEASE ANSWER ALL PARTS
For this question, you will use JMP to analyze the Deflategate data set. Each student is randomly assigned one of 10 versions of this dataset. Your version is 2.
Each version differs very slightly from the rest. Be careful that you do not use a different version; using any other version will result in your answers being marked as incorrect.
You can find version 2 of the Deflategate data on Canvas under "Additional Class Stuff / Data for WebAssign Homework".
It is recommended that you read the .pdf document outlining what is in this dataset before you begin. You can find this document on Canvas under "Additional Class Stuff / Data".
1. Report the values of the following:
a. Mean halftime PSI for Colts =
b. Standard deviation of mean halftime PSI for Colts =
c. Standard error of mean halftime PSI for Colts =
2. The statistics you reported in part 1. above refer to raw PSI readings at halftime, but what is really of interest is the difference between PSI before the game and PSI at halftime.
Create a new column in JMP containing the decrease in PSI, calculated as (PSI_Pregame - PSI_Halftime). You could calculate these by hand, but JMP will do it for you if you right-click on a column and choose formula. See the class notes for more details.
3. Now we want to compare the average decrease in PSI between the two teams. Use "Analyze / Fit Y by X", select the appropriate variables (see the notes if you have trouble with this), hit "OK", then select "t Test" from the drop down menu on the next screen. You'll get some information that won't make sense (for now), but you'll also get a 95% CI for the difference in mean PSI decrease for the Colts vs. Patriots.
JMP will default to subtracting the mean for Colts from the mean for Patriots.
4. There are some inferential questions for which it is not entirely clear what would be meant by a "population" mean. This is one such case (just ask yourself, what is "population mean PSI decrease"?). For now, assume that the idea of a population mean difference in PSI decrease is meaningful, and that we are drawing inference on it using our sample of footballs, which can be treated as a random sample from the population of all footballs that could have been selected for this football game.
Does this 95% CI suggest that the observed difference in mean PSI decrease is consistent with the "population" mean difference being zero?
a. Yes, because the interval contains zero
b. No, because the interval contains zero
c. Yes, because the interval does not contain zero
d. No, because the interval does not contain zero
d. No, because the interval is not centered on zero
e. Yes, because the interval is not centered on zero
5. Some people criticized the Wells report on the grounds that the data were biased by the fact that the footballs had been moved indoors at halftime, and the Patriots' footballs were measured first, giving the Colts' footballs more time to warm up and gain air pressure naturally. Can we use the 95% CI created in part 3. to assess this claim?
a. Yes, confidence intervals account for error due to both sampling variability and bias.
b. No, confidence intervals do not account for error due to either sampling variability or bias.
c. Yes, confidence intervals account for error due to bias, but not error due to sampling variability, and in this case we're only interested in error due to bias.
d. No, confidence intervals account for error due to sampling variability, but not error due to bias.
e. Yes, confidence intervals are numbers, and once you have a number you are basically holding truth in your hands.
PSl Prega PSI_Halftim Team me 2 P 3 P 4 P 5 P 6 P 7 P 8 P 9 P 10 P 11 P 12 C 13 C 14 C 15 C 12.5 12.5 12.5 12.5 12.5 12.5 12.5 12.5 12.5 12.5 12.5 13 13 13 13 11.52 10.72 11.21 10.65 11.27 11.49 11.94 11.09 10.96 10.28 12.9 12.77 12.43 12.56Explanation / Answer
1. R codes :
> data1=read.csv(file.choose(),header=T) #importing the csv file into R
> attach(data1)
> names(data1)
[1] "Team" "PSI_Pregame" "PSI_Halftime"
> PSI_Halftime[which(Team=="P")]
[1] 11.52 10.72 11.21 10.65 11.27 11.49 11.94 11.09 10.96 10.28 11.00
> PSI_Halftime[which(Team=="C")]
[1] 12.90 12.77 12.43 12.56
a) #MEAN
> mean(PSI_Halftime[which(Team=="C")])
[1] 12.665
b)#STANDARD DEVIATION
> sd(PSI_Halftime[which(Team=="C")])
[1] 0.2101587
c)#STANDARD ERROR
> sd(PSI_Halftime[which(Team=="C")])/length(PSI_Halftime[which(Team=="C")])
[1] 0.05253967
d)#95%CI
> t.test(PSI_Halftime[which(Team=="C")])
One Sample t-test
data: PSI_Halftime[which(Team == "C")]
t = 120.53, df = 3, p-value = 1.259e-06
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
12.33059 12.99941
sample estimates:
mean of x
12.665
e)#MEAN
> mean(PSI_Halftime[which(Team=="P")])
[1] 11.10273
f)#STANDARD DEVIATION
> sd(PSI_Halftime[which(Team=="P")])
[1] 0.4608707
g)#STANDARD ERROR
> sd(PSI_Halftime[which(Team=="P")])/length(PSI_Halftime[which(Team=="P")])
[1] 0.04189734
h)#95% CI
> t.test(PSI_Halftime[which(Team=="P")])
One Sample t-test
data: PSI_Halftime[which(Team == "P")]
t = 79.9, df = 10, p-value = 2.304e-15
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
10.79311 11.41234
sample estimates:
mean of x
11.10273
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