Question

A Californian Chenin Blanc was evaluated on a 9-pt scale representing varietal nature (with 1=very dissimilar and 9=very similar in varietal nature). Permission was sought for a Jamaican wine made from Chenin Blanc grapes grown in Jamaica to also be called “Chenin Blanc.” It was decided that if the “varietal nature” scores of the Jamaican wine (as assigned by 8 experts) were not significantly different from the California Chenin Blanc scores, permission will be given. (8 pts)

Expert	CA Chenin Blanc	Jamaican Chenin Blanc
1	8.4	6.1
2	7.8	7.9
3	8.2	7.8
4	9.7	9.2
5	9.1	8.5
6	8.5	7.5
7	7.4	7
8	8.0	8.4

Question 1: What is the value of S?

Question 2: What is the critical value of t? This is the value from the table.

Question 3: What is the calculated t-value?

Question 4: Are there significant differences between the wines? State this in terms of the relationship between calculated t-value and critical t-value. (2 pts)

Question 5: Can this Jamaican wine be called “Chenin Blanc?” Look back to the question above and state the p-value in the response. (2 pts)

Question 6: Would using more experts provide a more robust conclusion? Why or why not?

Answer 1

Question 1: What is the value of S?

To calculate the value of the S, we have to calculate the both means and variances.

Expert	CA Chenin Blanc	Diff (X - M)	Sq. Diff (X - M)2		Jamaican Chenin Blanc	Diff (X - M)	Sq. Diff (X - M)2
1	8.4	0.01	0		6.1	-1.7	2.89
2	7.8	-0.59	0.35		7.9	0.1	0.01
3	8.2	-0.19	0.04		7.8	0	0
4	9.7	1.31	1.72		9.2	1.4	1.96
5	9.1	0.71	0.51		8.5	0.7	0.49
6	8.5	0.11	0.01		7.5	-0.3	0.09
7	7.4	-0.99	0.98		7	-0.8	0.64
8	8	-0.39	0.15		8.4	0.6	0.36
	Mean: 8.39		sum=SS1: 3.75		Mean: 7.80		sum=SS2: 6.44

CA Chenin Blanc calculation(sample 1)

N₁: 8
df₁ = N - 1 = 8 - 1 = 7
M₁: 8.39
SS₁: 3.75
s²₁ = SS₁/(N - 1) = 3.75/(8-1) = 0.54

Jamaican Chenin Blanc (Sample 2)

N₂: 8
df₂ = N - 1 = 8 - 1 = 7
M₂: 7.8
SS₂: 6.44
s²₂ = SS₂/(N - 1) = 6.44/(8-1) = 0.92

s²_p = ((7/14) * 0.54) + ((7/14) * 0.92) = 0.73

S = 0.8544

Question 2: What is the critical value of t? This is the value from the table.

The Critical Value in the table for the t-test as,

DF=n1+n2-2=8+8-2=14, alpha=0.05 , t=2.145

Question 3: What is the calculated t-value?

²_M1 = s²_p/N₁ = 0.73/8 = 0.09 (calculated above)
s²_M2 = s²_p/N₂ = 0.73/8 = 0.09

t = (M₁ - M₂)/√(s²_M1 + s²_M2) = 0.59/√0.18 = 1.38

t=1.38

Question 4: Are there significant differences between the wines? State this in terms of the relationship between calculated t-value and critical t-value. (2 pts)

Here tcal<t_critical, We fail to reject the null hypothesis and conclude that there is no difference.

Question 5: Can this Jamaican wine be called “Chenin Blanc?” Look back to the question above and state the p-value in the response. (2 pts)

yes, there is no difference in it, so cal call. p-value in excel as, =T.DIST.2T(1.37734,14) = 0.19003

p-value(0.1900) > 0.05, we fail to reject the null hypothesis.

Question 6: Would using more experts provide a more robust conclusion? Why or why not?

Yes, if we add the more experts then the more robust conclusion. because this is the very small sample size.

Thanks