Question

3. Can taking chess lessons and playing chess improve memory? The online article described a study in which sixth-grade students who had not previously played chess participated in a program in which they took chess lessons and played chess daily for 9 months. Each student took a memory test (the Test of Cognitive Skills) before starting the chess program and again at the end of the 9-month period. The table below shows the results for 12 randomly selected students.

Student	1	2	3	4	5	6	7	8	9	10	11	12
Pre-chess Memory Score	410	610	640	675	500	780	610	625	450	720	675	670
Post-chess Memory Score	550	670	850	650	710	795	700	795	690	710	785	675
Memory Score Difference

Test the claim at the α = 0.05 level of significance that students who participated in the chess program achieve higher memory scores after completion of the program.

1. Fill in the last row of the table with the differences in memory.

2. Choose the correct Hypothesis for this test. (circle or highlight the correct answer)

A. H₀: µ_d = 0

H_A: µ_d > 0

B. H₀: µ₁ = µ₂

H_A: µ₁ > µ₂

C. H₀: µ_d = 0

H_A: µ_d < 0

D. H₀: µ₁ = µ₂

H_A: µ₁ ≠ µ₂

3. Find the P-value and use it to make a decision about the null hypothesis. Show work.

4. Summarize the final conclusion in the context of the original claim.

Answer 1

Let us denote the difference

d = Post-chess Memory Score - Pre-chess Memory Score

Let ud denotes the population mean of differnce. Hypotheses : To test null hypothesis Hould = 0 against alternative hypothesis H1 :Nd > 0 This is a one sided right tailed test as alternative hypothesis H contains only > sign. Let à denotes the sample mean of difference, sd denotes the sample standard deviation of difference and n denotes the sample size. Here, d = 101.2500, så = 8732.3864, sd = 93.4472, n= 12 The test statistic can be written as: Vnā -0.0) t= sd which under H, follows at distribution with n - 1 df. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tobs > t0.05,n–1 = 1.795885 0.4 02 0.1 0.0 - Now, Value of test statistic: The value of the test statistic tobs V12(101.2500 - 0) 93.4472 3.753351 3.75 associated degrees of freedom = 11 The critical value = critical = 10.05,11 = 1.795885 ~ 1.796 P-value = P(t11 > 3.753351) = 1 - P(t11 < 3.753351) =1-0.998404 = 0.001596 -0.0016 Conclusion : Since p-value < 0.05 and tobs > tcritical = 1.795885, so we reject H, at 0.05 level of significance Hence, we can conclude that mean difference is significantly greater than 0

There is sufficient evidence to support the claim that students who participated in the chess program achieve higher memory scores after completion of the program.