A team of researchers matched two random samples of 6 5th graders each. To one o
ID: 3179556 • Letter: A
Question
A team of researchers matched two random samples of 6 5th graders each. To one of the samples, they provided a new educational program designed to boost reading skills; the other sample served as the control group (they did not get the reading intervention). At the end of the school year, the researchers tested the 5th graders on their reading proficiency, and found a sample mean difference of 70.1 and standard deviation of 49.4. Report your findings from a one-tailed test ( = .01) of the null hypothesis.
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 > 0
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 28.52
DF = 10
t = [ (x1 - x2) - d ] / SE
t = 2.46
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
The observed difference in sample means produced a t statistic of 2.46. We use the t Distribution Calculator to find P(t > 2.46) = 0.01684
Interpret results. Since the P-value (0.01684) is greater than the significance level (0.01), we cannot reject the null hypothesis.
From the test we do not have sufficient evidence in the favor of the claim that difference is 70.1.
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