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. Assuming the sampling distribution of can be approximated by the standard normal distribution, the value of the test statistic is ________.
A) 3.767
B) -3.767
C) 0
D) None of these choices is correct.

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.  Assuming the sampling distribution of  can be approximated by the standard normal distribution, the value of the test statistic is ________.
A) 3.767 
B) -3.767 
C) 0 
D) None of these choices is correct.

Quizplus · Accepted Answer

To test the hypothesis that the chefs cannot tell the difference between duck liver pâté and wet dog food, we can use a two-sample t-test. However, since the sample size is small (n=18), and the population standard deviation is unknown, we must use a t-distribution. 
The null hypothesis is that the mean rating of duck liver pâté is equal to the mean rating of wet dog food. The alternative hypothesis is that they are not equal.

H0: μpâté - μdogfood = 0 
Ha: μpâté - μdogfood ≠ 0

We calculate the test statistic using the formula: 
t = (x̄pâté - x̄dogfood) / (sp√2/n)

where x̄pâté and x̄dogfood are the sample means for the ratings of pâté and dog food, sp is the pooled standard deviation, and n is the sample size.

From the table, we can calculate: 
x̄pâté = 3.944, x̄dogfood = 1.333 
sp = √[((n1-1)s1^2 + (n2-1)s2^2) / (n1+n2-2)] = 1.344 
t = (3.944 - 1.333) / (1.344√2/18) ≈ 3.767

Since the alternative hypothesis is two-sided, we will use a significance level of α = 0.05/2 = 0.025 (for a two-tailed test).

The degrees of freedom for the t-distribution are (n1 + n2 - 2) = 18, so the critical value for t(0.025, 18) is ±2.101. Because the calculated t-value of 3.767 is greater than the critical value of 2.101, we reject the null hypothesis.

Conclusion: The chefs can tell the difference between duck liver pâté and wet dog food, as they rated the pate significantly higher than dog food.

A Wine Magazine Wants to Know If Chefs Can Tell