Question

The question below is fill-in-the-blank. Be sure that you follow the instructions since it is computer graded. Either fill in the exact value you get from StatCrunch (no rounding) OR write your answer selection exactly as it is written.
A soft-drink company is conducting research to select a new design for the can. A random sample of participants has been selected. Instead of a typical taste test with two different sodas, they gave each participant the same soda in two different cans. One can was predominantly red, and the other predominantly blue. The order was chosen randomly. Participants were asked to rate each drink on a scale of 1 to 10. The data are recorded below and α=0.05.
Rater Red Blue
1 4 2
2 7 4
3 3 9
4 8 8
5 5 2
6 9 9
7 7 3
8 5 7
9 6 6
10 9 9
11 8 7
12 3 10
13 6 3
14 8 4
15 9 5
16 7 6
Does this sample indicate that there is a difference in the ratings based on can color? Test an appropriate hypothesis. Assume conditions have been met and that µd = red - blue
Hypotheses:
H0: µd=0; The mean difference of ratings is zero.
H0: µd (<, >, not equal to) 0; The mean difference of ratings is (greater than, less than, different than) zero.
Calculations: Type each value EXACTLY as they are given in StatCrunch
Mean:
Std. Err.:
DF:
T-Stat:
P-value:
Conclusion:Since the P-value is (low, high), we (reject, fail to reject) the null hypothesis. There (is/is not) sufficient evidence to suggest that can color (increases, reduces, changes) the average rating of the soda.

Answer 1

Answer:

Step-by-step explanation:

Hello!

The random sample of n= 16 participants was asked to taste test two different sodas cans and rate them on a scale from 1 to 10. The cans contained the same type of soda but differ in color.

Since each participant tasted and ranked the two types of soda cans, you have per person a pair of data (X₁;X₂):

Where

X₁: Rank given to the red soda.

X₂: Rank given to the blue soda.

As said, the data is arranged in pairs per experimental unit, so these two variables are dependent, to study them is best to determine a third variable, the difference between X₁ and X₂:

Xd: X₁ - X₂

Assuming Xd~N(μd;σd²)

The objective is to test if there is any difference between the ratings based on the can color. The statistic hypotheses are:

H₀: μd=0

H₁: μd≠0

To calculate the descriptive statistics you have to calculate first the difference of each pair of ratings (see table in attachment)

Mean:

X[bar]d= ∑Xd/n= -6/16= -0.38

Variance

$S_d^2=\frac{1}{n-1} [sumX_d^2-\frac{(sumXd)^2}{n} ]= \frac{1}{15}[64-\frac{(-6)}{^16} ]= 4.12$

Standard deviation

Sd= √4.12= 2.03

Statistic $t= \frac{X[bar]_d-Mu_d}{\frac{Sd}{\sqrt{n} } } ~~t_{n-1}$

Df: n-1= 16-10 15

$t_{H_0}= \frac{-0.38-0}{\frac{2.03}{4} } = -0.74$

P-value:

P(t₁₅≤-0.74) + P(t₁₅≥0.74)= P(t₁₅≤-0.74) + (1 - P(t₁₅≤0.74))= 0.2354 + (1 - 0.7646)= 0.4708

The decision rule for the p-value approach is:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

α: 0.05

The p-value is greater than the level of significance, the decision is to not reject the null hypothesis.

At a 5% significance level, there is no significant evidence to reject the null hypothesis. You can conclude that there is no difference between the rankings of the red and blue cans. I.e. there is no evidence to suggest that the can color changes the average ranting of the soda.

I hope this helps!