Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

1. Suppose we decide to use the t procedures to analyze the data. calculate a 95

ID: 3062798 • Letter: 1

Question

1. Suppose we decide to use the t procedures to analyze the data. calculate a 95% confidence interval for the population mean percentage (use R software if you can, show the code too please)

2. Give an appropriate interpretation of the 95% confidence interval given by R, in the context of the problem. To what population do your conclusions apply? Comment on any biases that might be present

3. If you feel there is an appropriate hypothesis test to carry out on this data (for just the data in the data set, and not any other data from the paper), then carry it out and properly interpret the results. If you do not feel there is a natural hypothesis test of interest in this situation, then say so and justify your position.

percincrease 23 -13 20 20 20 20 21 23 24 24 24 24 24 24 25 25 26 26 29 30 31 31 31 20 -9 4 -6 -6 24 20 10 10 10 10 35 36 37 38 39 39 40 45 48 49 50 73 19 12 12 13 13 19 20 20 20 4 total 133 data 5 (continue next rw) (continue next rw)

Explanation / Answer

Here, the population mean percentage inc is not given. Assuming it to be 10. The following is the r code to be implemented for t test and confidence interval.

y <- c(-23,-13,-11,-11,-9,-8,-8,-7,-7,-6,-6,-6,-5,-5,-4,-4,-3,-3,-3,-3,-2,-2,1,1,2,2,2,3,3,3,3,3,4,4,5,6,6,6,6,6,6,6,7,7,8,8,8,8,8,9,9,9,9,9,10,10,10,10,11,11,12,12,13,13,19,20,20,20,20,20,20,20,21,23,24,24,24,24,24,24,25,25,26,26,29,30,31,31,31,33,35,36,37,38,39,39,40,45,48,49,50,73,20,-11,1,8,4,5,8,-3,-2,6,6,9,2,11,24,-6,-6,8,-3,20,2,6,19,3,-8,6,11,6,3,5,5)
t.test(y,mu=10) # Ho: mu=10

Output looks like this: -

data: y
t = 0.74526, df = 132, p-value = 0.4574
alternative hypothesis: true mean is not equal to 10
95 percent confidence interval:
8.358215 13.626747
sample estimates:
mean of x
10.99248

1)

95 percent confidence interval:
8.358215 13.626747

2)

We are 95% confident that the mean percentage increase lies between 8.358215 and 13.626747

3)

data: y
t = 0.74526, df = 132, p-value = 0.4574
alternative hypothesis: true mean is not equal to 10

Since p-value is greater than a (i.e. 0.05), we cannot reject the null hypothesis.