Question

1. The research around Leona’s Tacos has created quite a buzz and business is great. Leona is planning on expanding her menu and she’s going to start with a new taco filling. She is considering her own black bean mix “Savor” and Hadey’s famous braise of seitan “Praise”. She naturally performs a blind taste test with 20 randomly selected judges from Avocado Park. Each judge is randomly served a taco filled with either “Savor” or “Praise” and asked to assess the experience using a rubric that results in a score from 0-100. Leona’s findings can be found in the data below questions
   1. Identify the populations of interest.
   2. Identify the variable of interest.
   3. What type of variable is being studied here?
   4. What would be a suitable parameter for determining if the taco filling receive different scores?
   5. Write a null hypothesis for Leona’s study.
   6. Write an alternative hypothesis for Leona’s study.
   7. Leona decides to use a significance level of 0.05. What conditions must be met to test the hypothesis in part e?
   8. Calculate a P-Value for your null hypothesis from part e and make a decision about the null hypothesis based on this P-Value and the significance level stated in part g.
   9. Summarize your decision from part h.

"Savor"BlackBeans: 71, 80, 81, 77, 82, 72, 79, 85, 78, 76

"Praise"Seitan: 83, 79, 77, 85, 80, 86, 78, 84, 83, 80

Answer 1

a. Population of interest: The judges from Avocado Park

b. Variables of interest: Scores of "Savor"BlackBeans and Scores of "Praise"Seitan.

c. Continuous variable (score lies in 0-100)

d.

g.

We assume that two samples come from two independent normal populations with equal population variances.

h.

R output:

> shapiro.test(SavorBlackBeans)

Shapiro-Wilk normality test

data: SavorBlackBeans
W = 0.97033, p-value = 0.8939

> shapiro.test(PraiseSeitan)

Shapiro-Wilk normality test

data: PraiseSeitan
W = 0.94196, p-value = 0.575
From above test we see that p-value for both samples is greater than 0.05 so assumption of normality holds.

R output:

F test to compare two variances

data: SavorBlackBeans and PraiseSeitan
F = 1.9526, num df = 9, denom df = 9, p-value = 0.3332
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.4849985 7.8611608
sample estimates:
ratio of variances
1.952601
From above test since p-value>0.05, assumption of equal variances holds.

Two Sample t-test

data: SavorBlackBeans and PraiseSeitan
t = -2.0183, df = 18, p-value = 0.05871
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-6.9391513 0.1391513
sample estimates:
mean of x mean of y
78.1 81.5
From above test, since p-value=0.05871>0.05 so we fail to reject null hypothesis at 5% level of significance.

R code:

SavorBlackBeans=c( 71, 80, 81, 77, 82, 72, 79, 85, 78, 76)
PraiseSeitan=c( 83, 79, 77, 85, 80, 86, 78, 84, 83, 80)
shapiro.test(SavorBlackBeans)
shapiro.test(PraiseSeitan)
var.test(SavorBlackBeans,PraiseSeitan,alternative = "two.sided")
t.test(SavorBlackBeans,PraiseSeitan,var.equal=TRUE)

i. From h, we conclude that there is insufficient evidence that there is insignificant diference in the mean scores of taco filling.