A wine magazine wants to know if chefs can tell the difference between duck liver pâté and wet dog food.Eighteen chefs were asked to rate both the pâté and dog food on a scale from 1 to 5,with 1 corresponding to "inedible" and 5 to "very tasty." The results are shown in the following table. Using the p-value approach and α = 0.05,the appropriate conclusion is ________.
A) do not reject the null hypothesis;conclude chefs prefer pâté
B) reject the null hypothesis;conclude there is a difference in the preference between pâté and dog food
C) reject the null hypothesis;we cannot conclude there is a difference in the preference between pâté and dog food
D) do not reject the null hypothesis;conclude nothing

Question

A wine magazine wants to know if chefs can tell the difference between duck liver pâté and wet dog food.Eighteen chefs were asked to rate both the pâté and dog food on a scale from 1 to 5,with 1 corresponding to "inedible" and 5 to "very tasty." The results are shown in the following table.  Using the p-value approach and α = 0.05,the appropriate conclusion is ________.
A) do not reject the null hypothesis;conclude chefs prefer pâté 
B) reject the null hypothesis;conclude there is a difference in the preference between pâté and dog food 
C) reject the null hypothesis;we cannot conclude there is a difference in the preference between pâté and dog food 
D) do not reject the null hypothesis;conclude nothing

Quizplus · Accepted Answer

We will use a two-sample t-test assuming equal variances to test if there is a difference in the average ratings between the pâté and dog food. The null hypothesis is that there is no difference in the means of the ratings of pâté and dog food. The alternative hypothesis is that there is a difference in the means of the ratings of pâté and dog food.

We use a two-tailed test since we want to know if there is a difference in preference between the two foods.

The test statistic is:

t = ( x1 - x2 ) / [ s_p * sqrt( 1/n1 + 1/n2 ) ]

where x1 and x2 are the sample means, s_p is the pooled standard deviation, n1 and n2 are the sample sizes.

Using the data given, we have:

x1 = 3.65, x2 = 1.06, n1 = n2 = 18

s1 = 1.38, s2 = 0.35

s_p = sqrt[ ( (n1-1) * s1^2 + (n2-1) * s2^2 ) / (n1 + n2 - 2) ]
s_p = sqrt[ ( (18-1) * 1.38^2 + (18-1) * 0.35^2 ) / (18 + 18 - 2) ]
s_p = 0.96

t = ( 3.65 - 1.06 ) / ( 0.96 * sqrt( 1/18 + 1/18 ) )
t = 5.99

Using a t-table with 34 degrees of freedom and 0.05 significance level, the critical value is 2.032. Since our calculated t-value is greater than the critical value, we reject the null hypothesis.

Therefore, the appropriate conclusion is to reject the null hypothesis; conclude there is a difference in the preference between pâté and dog food.

A Wine Magazine Wants to Know If Chefs Can Tell