iz . Retesting, A professor wanted to know if retesting students from the same m
ID: 3355957 • Letter: I
Question
iz . Retesting, A professor wanted to know if retesting students from the same material prove test scores. He gave an exam to the students in his STP226 class and n ,two weeks later, he gave them the second exam from the same material. The data below gives scores from both tests for a random sample of 6 students (in percentage points): then student John Mary Peter Paul 67 91 90 79 82 Test1: X1 56 Test2:X2 157 85 72 93 Does the data present enough evidence that retesting improves scores? Test using -0.05, use matched pairs t-test, assume that differences have normal distribution. M: 2 2Explanation / Answer
Test Score of of Test 1 is 56, 79, 82, 67, 91,88
Test Score of of Test 2 is 57, 77, 85, 72, 90, 93
Difference in test scores is -1, 2, -3, -5, 1, -5
Mean of Difference in test scores (d ) is -1.83
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: Mean difference of test scores is zero. That is, d = 0
Alternative hypothesis: Mean difference of test scores is not zero. That is, d 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
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). Let 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.
s = sqrt [ ((di - d)2 / (n - 1) ] = sqrt[44.833/(6-1) ] = sqrt(8.966) = 2.994
Standard deviation of Difference in test scores (s) is 2.994
SE = s / sqrt(n) = 2.994 / [ sqrt(6) ] = 1.222
DF = n - 1 = 6 -1 = 5
t = [ (x1 - x2) - D ] / SE = (d - D)/ SE = (-1.83 - 0)/1.222 = -1.4975
Since we have a two-tailed test, the P-value is the probability that a t statistic having 5 degrees of freedom is more extreme than -1.4975; that is, less than -1.4975 or greater than 1.4975.
We use the t Distribution Calculator to find P(t < -1.4975) = 0.0973, and P(t > 1.4975) = 0.0973. Thus, the P-value = 0.0973 + 0.0973 = 0.1946.
Interpret results. Since the P-value (0.1946) is greater than the significance level (0.05), we cannot reject the null hypothesis and conclude that the mean difference of test scores of test 1 and test 2 is not 0. So, at 5% significance level, there is not sufficient evidence that retesting students from the same material will significantly improve the test scores.
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