Question

Question B: A student doing her statistics project wants to test the theory that students who sit in the front of a class tend to have higher GPA's than those who choose to sit in the back Below are the GPA's, sorted by class seating position for a random sample. Use a 0.05 significance level to perform the appropriate test to determine if there is a difference in the mean GPA's for the three positions. Does the student's theory appear correct? Why or why not? Position GPA Front 3.06 2.89 3.58 4.0 3.52 3.33 3.89 3.56 Middle 2.86 2.63 3.56 2.11 2.93 3.39 3.65 2.52 Back 2.58 2.65 3.06 2.46 2.87 2.92 3.22 3.64 Questions 5-8 will address this test. 2.5 pts Question 5 Question B, Part 1 of 4 What is the value of the test statistic (rounded to two decimal places)? Stop sharing Hide Il app honorock.com is sharing your screen Jose Luis Resillo Need 925 665-0480 FAX #

Answer 1

We need to perform a one-way ANOVA to test this hypothesis.

Detailed solution given below.

We are given Front: #1 = 3.47875 Si = 0.1433 n1 = 8 Middle: X, = 2.95625 S = 0.2939 n2 = 8 Back: 3 = 2.925 S; = 0.1472 n2 = 8 Step 1 - State the hypotheses H:H1 = H2 = 143 H : At least one mean is different form the other. Step 2 Find the critical value No. of group = K = 3 Total no. of observation = N = 8+8+8 = 24 dfN = K -1 = 2 dfD = N-k= 21 a = 0.05 Using a and dfN = 2, dfD = 21, we find the critical value Feritical = 0.0514 Step 3 Compute the test value a Grand mean AGM (X1X91)+(x,xn)+(Xgxng) N (3.47875x8)+(2.95625x8)+(2.325x8) KGM 3.12 b. Sum of squares within and between group SSbetween = n1(X - XGM)? + n2(X, - XGM) +ng( g - cm)? SSbetween = 8(3.47875 – 3.12)” +8(2.95625 – 3.12)2 +8(2.925 – 3.12)2 = 1.5483 S Swithin = (n1 - 1)S + (n2 - 1)S; + (n3 – 1)S3 S Swithin (8 - 1)0.1433+ (8 - 1)0.2939 +(8 – 1)0.1472 = 4.0908

c. Variance within and between group Sbetween = SSteen k-1 1.5483 3-1 = 0.7742 SS mitiven Swithin 4.0908 (8-1)+(8-1)+(8-1) 0.1948 (n1-1)+(12-1)+(13-1) d. Fstatistics Fstat Stress Sihvers 0.7742 0.1948 -3.9743 e. Pvalue Using Fstat and dfN = 2, dfD = 21, we find the pvalue Pvalue = 0.0344 Step 4 Since the pvalue is less than the 0.05, we Reject Step 5 There is enough evidence to reject the null hypothesis and conclude that at least one mean is different from the others. Summary Table Sum of squares d.f. Mean square F Pvalue Between 1.5483 2 0.7742 3.9743 0.0344 Within 4.0908 21 0.1948 Total 5.6391 23

We have sufficient evidence to claim that the mean GPA is differs for the different seating positions.