Problem

30. The following data gives the retail price (per square-foot) and durability score (assigned. by an independent firm) for randomly selected types of laminate flooring available in a particular hardware store. Test the claim that there is a monotonic correlation between a laminate floor's retail price and durability score. Use the 0.05 significance level and Spearman's rank correlation test. $\begin{array}{cc}\text { Price (dollars) } & \text { Durability Score } \\ 0.69 & 8.2 \\ 0.85 & 9.5 \\ 0.97 & 9.1 \\ 0.99 & 8.9 \\ 1.09 & 11.0 \\ 1.29 & 11.5\end{array}$ What is the value of the TEST STATISTIC used in this hypothesis test?

Solution

Step 1 :Given the retail price and durability score for different types of laminate flooring, we are asked to test the claim that there is a monotonic correlation between a laminate floor's retail price and durability score. We will use the 0.05 significance level and Spearman's rank correlation test.

Step 2 :The given data is as follows: Price (in dollars) = [0.69, 0.85, 0.97, 0.99, 1.09, 1.29] and Durability Score = [8.2, 9.5, 9.1, 8.9, 11.0, 11.5].

Step 3 :We first calculate the ranks of the two variables. The ranks for price are [1, 2, 3, 4, 5, 6] and for durability score are [1, 4, 3, 2, 5, 6].

Step 4 :Next, we calculate the differences between the ranks of corresponding variables. The differences are [0, -2, 0, 2, 0, 0].

Step 5 :We then square these differences to get [0, 4, 0, 4, 0, 0].

Step 6 :We sum up these squared differences to get a total of 8.0.

Step 7 :Finally, we calculate the Spearman's rank correlation coefficient using the formula \(1 - \frac{6 \times \text{sum of squared differences}}{n \times (n^2 - 1)}\), where n is the number of observations. Substituting the values, we get \(1 - \frac{6 \times 8.0}{6 \times (6^2 - 1)}\), which simplifies to approximately 0.771.

Step 8 :Thus, the value of the test statistic used in this hypothesis test, which is the Spearman's rank correlation coefficient, is approximately \(\boxed{0.771}\).

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