You\'ve decided to use an intervention program for reading skills for at risk ch

Question

You've decided to use an intervention program for reading skills for at risk children (n = 25). At the start of the year, the kids take a standardized reading test for their age group. After a year of intervention, they take a follow-up standardized test. On average there was a 3 point difference between time one and time two scores across the students. The variance for these differences is 36. Your primary objective is to determine whether your intervention improved reading test scores. a.) State your null and alternative hypotheses: b.) What type of research design is this? Circle one: Matched or Repeated Measures c). What are your degrees of freedom and critical value(s) given the alpha = .05 level? d.) Calculate your test statistic. e.) What would you conclude about the efficacy of the intervention? Write a conclusionary statement and be sure to report your results as it would appear in an APA compliant manuscript (e.g. t(df) = t_obs, p

Explanation / Answer

Q.5 Mean difference = 3

variance of difference = 36

(a) Null Hypothesis : H0 : There is no significant difference between average of time one scores and time 2 scores.

ALternative Hypothesis : Ha : There is significant increase in average time two scores with respect to average time two scores.

(b) It is Matched research design.

(c) degree of freedom dF = n-1 = 25 -1 = 24

(d) Test Statistic

t = Mean difference / sqrt (variance/n) = 3/ sqrt (36/25) = 3/ 1.2 = 2.5

(e) tcritical for dF = 24 and alpha = 0.05 level =>  tcritical = 1.711

so t > tcritical , so we cannot reject the null hypothesis and can conclude that there is statistically evidence that there is significant increase in test score after intervention.

p - value = The P-Value is .009827.

The result is significant at p < .05.

(f) Correlation COefficient r = t/ sqrt ( n - 2 + t2 ) = (2.5)/ sqrt ( 25 -2 + 2.52) = 2.5/ sqrt (23 + 6.25) = 0.4622

so Effect size =  COefficient of determination r2 in paired data

so Effect size = 0.46222  = 0.2136

so effect size is small

(g) calculate 95 % confidence interval of difference of means

95 % confidence interval of difference of means = ( x1- bar - x2-bar) +- t0.975, 24 s/(sqrt (n))

= 3 +- 2.064 (6/5) = (0.5232 , 5.4768)

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