Two teaching methods (\'standard\' and \'new\') are compared based on a result o
ID: 3154134 • Letter: T
Question
Two teaching methods ('standard' and 'new') are compared based on a result of a reading test given at the end a learning period. Of a random sample of students. 10 are taught by new method and 12 are taught the standard method. The test results are given in the table Use the table data to estimate the true mean difference between the test scores for the new method and the standard method. Use a 95% confidence interval interpret the result from part A. What assumptions must hold in order for the estimate to be valid? Are they satisfied?Explanation / Answer
a.
Let u1 = mean of new method
u2 = mean of standard method
Calculating the means of each group,
X1 = 76.4
X2 = 72.58333333
Calculating the standard deviations of each group,
s1 = 5.83476173
s2 = 5.946096249
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 10
n2 = sample size of group 2 = 12
Thus, df = n1 + n2 - 2 = 20
Also, sD = 2.520075957
For the 0.95 confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.025
t(alpha/2) = 2.085963447
lower bound = [X1 - X2] - t(alpha/2) * sD = -1.440119664
upper bound = [X1 - X2] + t(alpha/2) * sD = 9.073452997
Thus, the confidence interval for u1 - u2 is
( -1.440119664 , 9.073452997 ) [ANSWER]
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B)
We are 95% confident that the true mean difference between the new method and the standard method is between -1.440 and 9.073. As 0 is inside this interval, then there is no significant difference between the means of the new methods.
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c)
Each of the samples must have come from an approximately normally distributed population, and that the measurements need to be independent. They are satisfied, yes.
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