Exhibit 20.11.A wine magazine wants to know if chefs can tell the difference between duck liver pate and wet dog food.Eighteen chefs were asked to rate both the pate 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. Refer to Exhibit 20.11.Assuming has the standard normal distribution,the value of the test statistic for the sign test is:
A)3.767
B)-3.767
C)0
D)None of the above

Question

Exhibit 20.11.A wine magazine wants to know if chefs can tell the difference between duck liver pate and wet dog food.Eighteen chefs were asked to rate both the pate 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.  Refer to Exhibit 20.11.Assuming  has the standard normal distribution,the value of the test statistic for the sign test is:
A)3.767 
B)-3.767 
C)0 
D)None of the above

Quizplus · Accepted Answer

To calculate the sign test statistic, we need to find the number of positive and negative differences between the ratings of the pate and dog food. We have 15 positive differences (where pate rating is higher than dog food rating) and 3 negative differences. Since we have an even number of observations, we will ignore the one observation that has the same rating for both pate and dog food. 
The sign test assumes a binomial distribution for the number of positive differences, where the probability of success (i.e., pate being rated higher) is 0.5 under the null hypothesis (i.e., chefs cannot tell the difference between pate and dog food). Using the binomial formula with n=15 and p=0.5, we get a mean of 7.5 and a standard deviation of 1.94. 
The test statistic is calculated as (observed number of positive differences - expected number of positive differences) divided by the standard deviation. In this case, the observed number of positive differences is 15 and the expected number of positive differences is 7.5. Therefore, the test statistic is (15-7.5)/1.94 = 3.87 (rounded to three decimal places). Since the sign test is a one-tailed test, we look up the one-tailed p-value for a normal distribution with mean=0 and standard deviation=1. The p-value is 0.0000707 (from a standard normal distribution table). Since the test statistic is positive, we reject the null hypothesis at the 5% significance level if the p-value is less than 0.05. Therefore, we can conclude that the chefs can distinguish between duck liver pate and wet dog food based on their ratings, and the test statistic is significant at the 5% level. The correct choice is B.