Question

4) An educator believes that new reading activities in the classroom will help elementary school pupils improve their reading ability. She arranges for a third grade class of 10 students to follow these activities for an 8-week period. A control classroom of 12 third graders follows the same curriculum without the activities. At the end of the 8 weeks, all students are given the Degree of Reading Power (DRP) test, which measures the aspects of reading ability that the treatment is designed to improve. The table below represents the results of this test (assume the distributions of the treatment and control groups are normally distributed). See Data Set for values. 43 71 52 57 33 46 43 Treatment 21 Control 42 58 55 57 55 43 26 10 53 42 37 42 28 48 Based on the evidence provided, does it appear that these new reading activities actually improve scoring on the DRP at the 5% significance level? YES NO Please show the four steps of your analysis in the space provided below...

Answer 1

Let denote the reading scores of treatment and control stores respectively.

111, 112

Hypotheses : To test null hypothesis Ho :M1 - U2 = 0 against alternative hypothesis H:M1-M2 > 0 This is a one tailed right sided test as the alternative hypothesis contains '>' sign. Here, 1 = 48.10, si = 14.20, ni = 10,2 = 40.08, S2 = 13.32, n2 = 12 The test statistic can be written as: (21 2.2) - 0 t= + ni n2 which under Ho follows a t distribution with 18 df. where +9) + ni n2 df 14.202 10 14.202 10 10-1 13.322 -)2 12 13.322 12 12-1 ~18 + ni-1 n2-1 and standard error for the difference between means 14.202 13.322 SE = + = 5.912730 ni n2 10 12

Decision rule / Rejection region : We reject Ho at 0.05 level of significance if P-value < 0.05 or if t stat > to.05,df 0.4 density 0.3- 0.2- 0.1 - 0.0- -2 0 2 4 t Now, Value of the test statistic : The value of the test statistic is (48.10 – 40.08) tstat = = 1.355832 1.356 14.202 10 + 13.322 12 associated degrees of freedom = 18 The critical value = tcritical t0.05,18 =1.734064 1.734 P - value = P(+18 > tstat) = P(t18 > 1.355832) = 1 - P(t18 < 1.355832) 1 – 0.904040 = 0.095960 ~ 0.0960 Conclusion: Since p-value > 0.05 and tstat * tcritical = 1.734, so We fail to reject Ho at 0.05 level of significance Hence, we can conclude that mean for first population is not significantly greater than the mean for second population.

There is not sufficient evidence to support the claim that new reading activities significantly improve scoring on the DRP.