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

Questions 1 (Did the program increase store traffic? Use a pooled t-test) and Qu

ID: 3282789 • Letter: Q

Question

Questions 1 (Did the program increase store traffic? Use a pooled t-test) and Question 2 (Did the program increase store traffic? Use a paired difference t-test. ask for two different approaches to the same question. The results differ. Write a paragraph describing which approach is the most appropriate. You should end with a clear (yes-no) conclusion to the question.

Abstract: A supermarket chain wants to know if their “buy one, get one free” campaign increases customer traffic enough to justify the cost of the program. For each of 10 stores, they select two days at random to run the test. For one of those days (selected by a coin flip), the program will be in effect. They want to determine whether the program increases the mean traffic. The results in a number of customer visits to the 10 stores are in the dataset.

Supermarket study

Store #

With Program

Without Program

1

140

136

2

233

235

3

110

108

4

42

35

5

332

328

6

135

135

7

151

144

8

33

39

9

178

170

10

147

141

Questions 1 (Did the program increase store traffic? Use a pooled t-test) and Question 2 (Did the program increase store traffic? Use a paired difference t-test. ask for two different approaches to the same question. The results differ. Write a paragraph describing which approach is the most appropriate. You should end with a clear (yes-no) conclusion to the question.

Supermarket study

Store #

With Program

Without Program

1

140

136

2

233

235

3

110

108

4

42

35

5

332

328

6

135

135

7

151

144

8

33

39

9

178

170

10

147

141

Explanation / Answer

ANSWER:

Did the plan add to store traffic? Use a shared t-test.

The null premise - The denote travel through the plan and with the curriculum is identical, that is 145

The interchange suggestin - The stand for traffic by way of the list is superior than devoid of the plan, that is superior than 145

The test marker select (including which test) - We will manner 2-sample t-test using mutual variances

Let n1 be the illustration size on or after inhabitants 1 (with program), s1 be the trial usual departure of populace 1.

Let n2 be the trial size from people 2 (withou program), s2 be the model usual divergence of populace 2.

n1 = n2 = 10

mean of sample 1 (with program) = 150.1

mean of sample 2 (without program) = 147.1

sd of sample 1 (with program) = 86.98

sd of sample 2 (without program) = 86.33

Then the universal standard departure can be probable by the pooled typical departure:

2sampl_1.gif

= sqrt(((10-1)*86.98*86.98 + (10-1)*86.33*86.33)/(10+10-2) = 86.65

The test statistic is:

2sampl_2.gif

t = 150.1-147.1 / 86.65/sqrt(5) = 0.0774

with degrees of freedom equal to df = n1 + n2 - 2 = 10 + 10 -2 = 18

The critical value and decision rule - For significance level of 0.05, critical value of t at df = 18 is 1.734

The P-value - p-value for t = 0.0774 at df = 18 is 0.4696

Your answer - As the p-value is greater than 0.05, we fail to reject the null premise and bring to a close that in attendance is no arithmetic confirmation that the mean traffic with the plan is better than without the plan, that is superior than 145

Did the line up increase store up traffic? Use a matching diversity t-test.

The unfounded supposition - The divergence amid signify traffic in the company of the course and exclusive of the curriculum is nought.

The broken guess - The discrepancy involving imply traffic by way of the agenda and lacking the course is greater than 0.

The test sign chosen (as well s which test) - We will ways a matched-pairs t-test of the untrue proposition.

The differentiation among the sample = 4,-2,2,7,4,0,7,-6,8,6

denote of the differences = 3

Sd of the diffrences = 4.52

average error of the disparity, SE = 4.52/sqrt(10) = 1.43

quantity of autonomy = 10 -1 = 9

t = mean difference/SE = 3/1.43 = 2.098

The serious cost and conclusion rule - For implication level of 0.05, dangerous price of t at df = 9 is 1.833

The P-value - p-value for t = 2.098 at df = 9 is 0.0326

Your answer - As the p-value is lower than 0.05, reject the null hypothesis and bring to a close that the distinction sandwiched between signify traffic with the plan and devoid of the series is bigger than zero.

initial premise testing determine whether the plan increases the signify interchange superior than 145 in which we fail to reject the unsound premise, so there is no numerical confirmation that plan increase the transfer better than 145. Second premise taxing determine whether is there any disparity in the mean travel with or devoid of the list. We rejected the null supposition and close that the signify traffic with plan is greater than wothout plan. In the difficulty, the purpose is to establish whether the plan increase the mean interchange. So we ought to use the subsequent suggestion easy which use paired disparity t-test

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Chat Now And Get Quote