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 risk start of the year, the kids a standardized reading test for their age group. After 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 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? 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. f) What is the effect size in terms of proportion of variance accounted for (r^2)? Do you characterize it as being small medium, or large? g.) Calculate a 95% confidence interval.

Explanation / Answer

Solution:-

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

Null hypothesis: d< 0

Alternative hypothesis: d > 0

Note that these hypotheses constitute a one-tailed t pair 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 [ ((di - d)2 / (n - 1) ] =

s2 = 36 (Given)

s = 6

SE = s / sqrt(n)

S.E = 1.2

DF = n - 1 = 25 -1 = 24

t = [ (x1 - x2) - D ] / SE

t = 2.5

tcritical = 1.711

where di 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 one-tailed test, the P-value is the probability that a t statistic having 24 degrees of freedom is more than 2.5

Thus, the P-value = 0.0098

Interpret results. Since the P-value (0.0098) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we conclude that we have sufficient evidence in the favor of the claim that there is improvements in the their scores.

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