Bkstr vs Amazon

(10 pts)

For this problem, you’re going to take a small sample of textbooks that are in use here at UNCA this semester and compare their bkstr.com price to their Amazon price. You’ll then run a hypothesis test of the form

where \mu_A represents the average Amazon price and \mu_B represents the average Bookstore price. This is a lot like an example from our last class presentation.

First, you’ll get your list of books (using the menu below) with Bkstr prices already listed and pointers to the corresponding Amazon pages. Then, you’ll follow the link to look up the Amazon price so you can wrap it in a table that looks something like so:

Finally, you’ll compute the pairwise differences between the prices and run the hypothesis test. Of course, you’ll use the t-distribution, due to the small sample size.

Note: I’ll be aggregating all this data into one much larger sample so we can run a better hypothesis test on Friday. As a result, it will be of central importance for you to format your table correctly so that my program can read it. It’s really not that hard! To format your table like the following:

Simply type the following:

|Book|Format|Bkstr Price|Amazon Price|
|---|---|---|---|
|Book 1|Buy/Rent New/Used|$00.00|$00.00|
|Book 2|Buy/Rent New/Used|$00.00|$00.00|
|Book 3|Buy/Rent New/Used|$00.00|$00.00|
|Book 4|Buy/Rent New/Used|$00.00|$00.00|
|Book 5|Buy/Rent New/Used|$00.00|$00.00|

You can simply copy and paste that to get started.

H0 : μ A = μ B
H A : μ A < μ B
mean of difference = 30.907
s = 71.93 using 95% confidence level. t* = 2
t - value = (30.907 - 0) / 71.93/sqrt(5) = 0.96
= P(T<−0.96) = 0.195 thus we fail to reject the null at 95% confidence level
using DOF 4

ME = 2.776 X 71.93/(sqrt(5)) =
XBAR + ME, XBAR - ME
T* = 2.776

H_0:\mu_A=\mu_B
H_A:\mu_A<\mu_B

mean of difference = 39.41
std dev = 29.9416
My test statistic =2.943
My P value is =.021
Therefore I fail to reject the Null Hypothesis.

The sample size, mean, and standard deviation are:

The formula to calculate the test statistic is:

That yields a p-value of .0045, thus I reject the null.

Here are my books:

The mean and standard deviation are:

mean = 39.08
std dev = 55.5

Thus, my test statistic is 39.08/(55.5/sqrt(5)) = 1.57.
That yields a p-value of 0.095, thus I fail to reject the null.

H0 : μ A = μ B
H A : μ A < μ B

mean of difference: 5.788
Standard Deviation of difference: 35.47668638
Thus, my test statistic is: 5.788/(35.47668638/sqrt(5)) = 0.3641829
That yields a p-value of 0.388829, so we fail to reject the null hypothesis

H_0: \mu_A = \mu_B\
H_A: \mu_A \lt \mu_B

Mean of Difference: 4.49
Standard Deviation of Difference: 3.3524
Confidence Level: 95%
t*: 2.7764

ME:

Confidence Interval:

Test Statistic:

P-Value:

Thus, we reject the null hypothesis at a 95% confidence level.

H0 : μ A = μ B
H A : μ A < μ B
mean of difference = 65.63
standard deviation =116.517

Test statistic = 1.259
P-value = 0.214
I fail to reject the null

H0: μ A = μ B
H A: μ A < μ B
Mean of Difference: 4.268
Standard Deviation: σ: 2.548
Test Statistic: 3.7455
P Value: P(T<3.7455)=0.91695
1-0.91695= 0.08305
Thus, we fail to reject the null hypothesis at a 95% confidence level.

The mean and standard deviation are:

Mean = 4.584
Std dev = 4.23246

Thus, my test statistic is (4.584 - 0)/ (4.23246/sqrt(5)) = -2.42179
That yields a p-value of 0.036, thus I can reject the null.

The mean and standard deviation are as follows:

Mean = 13.316
Std Dev = 19.440317

Thus, my test statistic is 13.316/(19.440317/sqrt(5)) = 1.53.
That yields a p-value of 1-0.89962 = 0.10038, thus I fail to reject the null hypothesis when operating at a 95% confidence level.

Ho: ma = mb
Ha: ma < mb

For the difference between Bookstore and Amazon:
mean = 22.901999999999997
std dev=25.95146662522178
Hypothesis Test:
Ho: μ=0
Ha: μ>0
My test statistic is:
T=(22.90199-0)/(25.95146/2.23606)=1.973315
yields a p-value of: 0.059857
thus we fail to reject the null in this case.

mean of difference = 39.193999999999996
standard deviation of difference = 55.014761019202844

My test statistic is: (39.193999999999996-0/(55.014761019202844/sqrt(5)) = 1.5930351543204184

p-Value = 1 - 0.9068139127473118 = 0.09318608725268818

Thus, we fail to reject the null hypothesis at 95% confidence.

mean of difference- 11.214
s=19.79 using 95% confidence interval t=2
t value- (11.214-0)/23.92/sqrt(5)=1.27
p(t<-1.27)=.8635,1-.8635= 0.1365, fail to reject the null

My hypothesis test will be run in the form of:
H0:μA=μB
HA:μA<μB

The mean of differences is: 6.6899
The standard deviation is: 10.3365
The test statistic should look like the following: (6.6899-0)/(10.3365/sqrt(5))= 1.4472
My p-value is: 0.1107

Because my p-value is greater than my significance level of 0.05, I will fail to reject the null

My mean is 3.86 and standard deviation is 3.127746792820672
And so my test statistic is 2.1375 which gives me a p-value of 0.0831
I fail to reject the null hypothesis