Question

Question 3: EatWell Inc., a large fast-food chain company of Canada, wants to test two versions of a new product before launching its full production. They select a random sample of individuals among their regular clients and ask them to participate in an experiment to rate the product according to an assessment grid worth 20 points. The 15 individuals who accepted to participate had to come to a specific company location on two occasions during a given week to test the two versions of the product. It was a randomized experiment in which the version assigned to participants during their first visit was selected at random and participants did not know which version they were rating. Scores data of each participant are included in the Minitab file EatWell_Scores.mtw and in the Excel file Assign2.xlsx under the EatWell_Scores tab.

a) Examine and comment on the distribution of the ratings for the two versions of the product and decide which statistical test is the most appropriate for this type of data. Justify your selection based on the assumptions and conditions required for the most appropriate test selected.  

b) Regardless of your answer in a) above, perform a relevant parametric test using a 10% significance level to determine if there is a difference between the two versions of the product. Make sure you show your Minitab or other software output along with an interpretation of the results (manual calculations are not required here).  

c) Now, perform a Wilcoxon signed rank test to verify if there is a significant difference at the 10% level between the two versions of the product based on these ratings. Show your manual calculations including the T+ and T- statistics and conclude based on the critical value approach.  

d) Use Minitab or other software to perform the test in c) above and comment on the computed Pvalue and your conclusion in c).  

e) Compare your results from the parametric and non-parametric tests above and state what the final conclusion regarding the two versions of the product should be.

here is the data from EatWell_Scores.mtw

Individual	Version A	Version B
1	17	16
2	17	18
3	20	17
4	11	15
5	15	12
6	16	15
7	15	14
8	16	13
9	12	12
10	16	13
11	19	18
12	14	14
13	17	15
14	18	16
15	18	17

Answer 1

(a)

Please refer graph for descriptive statistics.
Stat-Basic Statistics-Graphical Summary

MTB > GSummary 'Version A' 'Version B'.

We select paired t-test (Student's t).

Assumptions for t-test:
• Data are paired and come from the same population.(Each individual rate two versions, therefore, forming pair)
• The observations are chosen randomly and independent of one another.
• Independence:Independence of observations is reasonably assumed as the data collection process was random without replacement. In our example, it is reasonable to assume that the participating individuals are independent of one another.
• The data are measured at least on an ordinal scale (here ratings)
• The variable should be approximately normally distributed.To test the assumption of normality, a variety of methods are available, but the simplest is to inspect the data visually using a tool like a histogram.(Here,the shape for both the versions looks approximately symmetric and bell-shaped). Also,Both Versions are approximately normal as p-value > alpha=0.1.

(b)

Paired t-test
H0:Null Hypothesis, H1:Alternative Hypothesis
H0: d=difference=0 VS H1=d0.

Menu bar-Stat-Paired t-test(Students's t-test for difference of means)

MTB > Paired 'Version A' 'Version B';
SUBC> Confidence 90.0;
SUBC> Test 0.0;
SUBC> Alternative 0;
Paired T-Test and CI: Version A, Version B
Paired T for Version A - Version B
N Mean StDev SE Mean
Version A 15 16.067 2.434 0.628
Version B 15 15.000 2.000 0.516
Difference 15 1.067 1.870 0.483
90% CI for mean difference: (0.216, 1.917)
T-Test of mean difference = 0 (vs ≠ 0): T-Value = 2.21 P-Value = 0.044
Here, p-value=0.044 which is less than alpha=0.1 (10%=10/100=0.1). We Reject H0.

(c)

We use Normal approximation here(n=13)
T=min(T+,T-)=min(74.5,16.5)=16.5
Here,take n=13 (discarding values di=0, there are two such values, effective sample size=15-2=13)
E(T)=n(n+1)/4=(13*14)/4= 45.5
V(T)=n(n+1)(2n+1)/24=(13*14*27)/24=204.75
Note that :Two or more observations/differences may be equal. If so, average the ranks across the tied observations and reduce the variance by (t^ 3−t)/ 48 for each group of t tied ranks. Ex. di=1 for 6 observations. (6^3-6)/48=4.375, similarly, (2^3-2)/48=0.125, (4^3-4)/48=1.25.
We must reduce the variance by 4.375+0.125+1.25=5.75.

Thus, V(T)=204.75-5.75=199.
Square root of V(T)=Square root (199)= 14.10674
Z=(16.5-45.5)/14.10674= -2.055755

Here, Z (two-sided) test, reject H0: if |z|> tabulated z. Z~N(0,1), tab Z at 10% level of significance= 1.64.

Using R command:pnorm( -2.055755)
P(Z<-2.055755)=0.01990307. p-value=2*0.01990307=0.03980614 (Since, it is two-sided test)
p-value< alpha=0.1, We reject H0.

Rejecting H0 means concluding that d0.

Versions A and B differ.

R-commands (Complete Analysis in R-Software):

> A
[1] 17 17 20 11 15 16 15 16 12 16 19 14 17 18 18
> B
[1] 16 18 17 15 12 15 14 13 12 13 18 14 15 16 17
> t.test(A,B,paired=TRUE)
Paired t-test
data: A and B
t = 2.2097, df = 14, p-value = 0.04429
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.03134129 2.10199204
sample estimates:
mean of the differences
1.066667

> library(MASS)
> wilcox.test(A,B,paired=TRUE)
Wilcoxon signed rank test with continuity correction
data: A and B
V = 74.5, p-value = 0.04335
alternative hypothesis: true location shift is not equal to 0
Warning messages:
1: In wilcox.test.default(A, B, paired = TRUE) :
cannot compute exact p-value with ties
2: In wilcox.test.default(A, B, paired = TRUE) :
cannot compute exact p-value with zeroes

> 74.5+16.5
[1] 91
> n=13
> n*(n+1)/2
[1] 91
> n*(n+1)/4
[1] 45.5
> n*(n+1)*((2*n)+1)/24
[1] 204.75

> (16.5-45.5)/14.10674
[1] -2.055755
> pnorm( -2.055755)
[1] 0.01990307
> 2*pnorm( -2.055755)
[1] 0.03980614