Question

Gourmet Cooking Two chefs, \(A\) and \(B\), rated 22 meals on a scale of \(1-10\). The data are shown in the table. Do the data provide sufficient evidence to indicate that one of the chefs tends to give higher ratings than the other? Test by using the sign test with a value of \(\alpha\) near. \(05 .\) $$ \begin{array}{ccc|ccc} \hline \text { Meal } & \text { A } & \text { B } & \text { Meal } & \text { A } & \text { B } \\ \hline 1 & 6 & 8 & 12 & 8 & 5 \\ 2 & 4 & 5 & 13 & 4 & 2 \\ 3 & 7 & 4 & 14 & 3 & 3 \\ 4 & 8 & 7 & 15 & 6 & 8 \\ 5 & 2 & 3 & 16 & 9 & 10 \\ 6 & 7 & 4 & 17 & 9 & 8 \\ 7 & 9 & 9 & 18 & 4 & 6 \\ 8 & 7 & 8 & 19 & 4 & 3 \\ 9 & 2 & 5 & 20 & 5 & 4 \\ 10 & 4 & 3 & 21 & 3 & 2 \\ 11 & 6 & 9 & 22 & 5 & 3 \\ \hline \end{array} $$ a. Use the binomial tables in Appendix I to find the exact rejection region for the test. b. Use the large-sample \(z\) statistic. (NOTE: Although the large-sample approximation is suggested for \(n \geq 25\), it works fairly well for values of \(n\) as small as \(15 .)\) c. Compare the results of parts a and b.

Short answer

Answer: Steps 4 and 6 involve comparing the calculated values to the critical values to test the hypothesis. In Step 4, the number of positive signs (p) is compared to the critical value (c) obtained from the binomial distribution. In Step 6, the computed z-statistic is compared to the critical z-value corresponding to the chosen significance level.

Step-by-step solution

Compute the signs of A's ratings minus B's ratings

For each meal, calculate the difference between A's rating and B's rating. If the difference is positive, record a "+"; if it is negative, record a "−"; and if it is equal, discard the meal from our analysis.

Calculate the number of positive and negative signs

Count the number of positive signs (A's rating higher than B's) and the number of negative signs (B's rating higher than A's). Let's use "n" to denote the count of total meals excluding equal ratings and "p" for the count of positive signs.

Determine the rejection region using the binomial distribution

Now, we'll use Appendix I - Binomial Tables with our given significance level \(\alpha \approx 0.05\) and "n" (total meals). Find the critical value "c" using the binomial distribution such that if the number of positive signs (p) is greater than or equal to "c", we reject the null hypothesis that both chefs rate the meals equally.

Conduct the hypothesis test using the binomial distribution

Compare our calculated value of "p" (positive signs) with the critical value "c". If "p" is greater than or equal to "c", then we reject the null hypothesis i.e., one of the chefs tends to give higher ratings than the other.

Calculate the large-sample z-statistic

Although our total meals count is less than 25 (22 meals), we can still use the large-sample z-statistic to approximate our test. The z-statistic is given by the formula: $$ z = \frac{p - \frac{n}{2}}{\sqrt{\frac{n}{4}}} $$ Compute the z-statistic using the value of "p" and "n".

Compare the z-statistic with the critical z-value

We will compare the computed z-statistic with the critical z-value corresponding to our chosen significance level. If the z-statistic is greater than or equal to the critical z-value, we reject the null hypothesis i.e., one of the chefs tends to give higher ratings than the other.

Compare the results from the binomial distribution and z-statistic tests

Finally, compare the results obtained in steps 4 and 6. If both tests lead to the same conclusion, it implies that the large-sample z-statistic test provides a reasonable approximation even though the sample size is less than 25.

Key concepts

Understanding the Sign Test

Let's start by understanding the sign test, which is a non-parametric method used for hypothesis testing. Upon comparing two related samples, the sign test determines if the median of the differences is zero. In the context of our chefs' meal ratings, we use this test to see if there is a consistent preference by one chef over the other.

For the sign test, we calculate the difference in ratings for each meal. If Chef A rates higher, we mark a plus sign (+), if Chef B rates higher, we assign a minus sign (−), and if they rate the same, we ignore that meal. The sign test doesn't consider the magnitude of difference, only the direction. Next, we count the number of positive and negative signs to see if there's a statistically significant difference in the ratings.

The beauty of the sign test lies in its simplicity. It doesn't require the data to be of a particular distribution, making it robust and versatile. Managing the test merely involves counts of signs and a straightforward hypothesis test which, in this exercise, pits the null hypothesis (no difference) against the alternative hypothesis (there is a difference). To evaluate the test, we refer to a binomial distribution, the details of which we'll explore next.

Exploring Binomial Distribution

Binomial distribution is central to the sign test and shines when dealing with dichotomous outcomes—situations with two possible outcomes, such as 'success' or 'failure.' In our chefs’ example, a 'success' can be viewing each meal where Chef A rated higher than Chef B.

The binomial distribution will help us calculate the probability of observing a certain number of successes (positive ratings by Chef A over Chef B, in our case) given the total number of trials (meals rated) and the probability of success on a single trial (which is 0.5 assuming there's no inherent rating bias between chefs).

When dealing with a binomial distribution, we often have to decide whether the observed result is so extreme that it falls into a rejection region—a range of values for which we would reject the null hypothesis. We find this rejection region using binomial tables or computational tools, setting our significance level (in this case near 0.05). This leads us to either embrace or discard our assumption that the chefs rate meals similarly.

The Large-Sample Z-Statistic

Now, onto the concept of the large-sample z-statistic. Typically recommended for sample sizes of 25 or larger, this approximation method is still quite robust for slightly smaller samples, like our case with 22 meals.

The z-statistic is a way to measure the number of standard deviations an observation is from the mean. For our chefs, the z-statistic helps us to understand whether the difference in the number of times Chef A rated a meal higher than Chef B is significant. Essentially, we're checking if our observed success rate (Chef A’s higher rating) is statistically significantly different from what we’d expect by random chance.

To calculate the z-statistic, we use a formula that incorporates our count of positive signs and the expected number under the null hypothesis. If our calculated z-value is larger than the critical z-value from the z-distribution table at our chosen significance level, we must consider the possibility that Chef A really does tend to give higher ratings than Chef B, leading to a rejection of the null hypothesis.

In essence, while the sign test uses direct observational data to come to a conclusion, the large-sample z-statistic introduces the concept of standard deviation to measure how far our observed data deviates from what’s expected, giving us another perspective on hypothesis testing.