Test statistic - expected vs. observed

Question

5. In the past few years, attention has been focused on hybrid learning versus traditional learning. Administrators in a large urban district take a random sample of 50 ninth graders and compare the algebra achievement levels of those who took pre-algebra in a hybrid learning format and those who did not. If achievement is independent of the hybrid learning format, then the proportions at each level should be equal. You need to use the observed counts in the frequency table below to determine if there is an association between learning via the hybrid format and algebra achievement at the 0.05 significance level. What is the value of the test statistic?

Below Grade
Level At Grade
Level Advanced
Hybrid learning 8 6 6
Traditional learning 6 15 9

Answer 1

Let's start with the appropriate matrix of values and the null hypothesis

 

Null hypothesis: Ho: There is no association between learning via the hybrid format and algebra achievement 

Alternative hypothesis: Ha: There is an association between learning via the hybrid format and algebra achievement 

 

                          Below Grade  Level At Grade    Level Advanced

Hybrid Learning         8                       6                        6

Traditional Learning   6                       15                      9

 

Using these values as the observed counts, we need to develop the expected values. They are determined by summing each row and column for the marginal counts. The new matrix becomes

 

                                  Below Grade      Level At Grade        Level Advanced  Sum
Hybrid Learning                    8                 6                               6                20
Traditional Learning              6                  15                             9                30

Sum                                   14                 21                            15               50

 

The expected value for Below Grade  and  Hybrid is 14 (the column total) times 20 (the row total) divided by 50 (the grand total) which yields 5.6

 

This is a chi-squared distribution and the test statistic is 

 

chi-squared test statistic = ∑ (observed_i - expected_i)²÷ expected_i =

 

(8-5.6)²÷ 5.6 + (6-8.4)²÷ 8.4 +(6-8.4)²÷ 8.4+ (15-12.6)²÷ 12.6 + (6-6)²÷ 6+ (9-9)²÷9

 

chi-squared test statistic = 2.857 with 2 degrees of freedom, df where df = number of rows of original matrix less 1 times the number of columns less 1. Since we have two rows and three columns, the df = 1*2 or 2

 

A chi-squared table reveals that the p-value of this test is 0.2397 

Since the p-value is greater than the significance level of .05, we cannot reject the null.

Therefore we conclude that there is no association between learning via the hybrid format and algebra achievement.