Question

A teacher gives a third grade class of n = 16 a vocabulary test at the beginning of the school year. To evaluate the changes that occur during the year, students are tested again at the end of the year. Their test scores revealed an average improvement of MD = 5.7 points with s2 = 144. Does this constitute a significant improvement? Use a one tail, related-samples t test with an alpha level of .05.

a. Calculate Cohen's D (Round 2 decimal places)

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d< 0

Alternative hypothesis: u_d > 0

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

s = sqrt [ (\sum (d_i - d)² / (n - 1) ]

s = 12

SE = s / sqrt(n)

S.E = 3.0

DF = n - 1 = 16 -1

D.F = 15

t = [ (x₁ - x₂) - D ] / SE

t = 1.90

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 15 degrees of freedom is greater than 1.90.

Thus, the P-value = 0.038

Interpret results. Since the P-value (0.038) is less than the significance level (0.05), we have to reject the null hypothesis.

Reject H₀, this constitute a significant improvement.

a) The Cohen's D is 0.48

d = 0.475