(12 pts) An education researcher would like to test whether 2 nd graders retain
ID: 3217426 • Letter: #
Question
(12 pts) An education researcher would like to test whether 2nd graders retain or lose knowledge during the summer when they are presumably not in school. She asks nine 2nd graders to take a comprehension exam at the end of the school year (May), and then asks those same students to come back after the summer (late August) to retake a different but equivalent exam, to see if their level of comprehension has changed. Using the data below, test this hypothesis using an alpha level of .05.
May
August
90
100
65
80
78
92
50
60
89
90
92
98
75
70
90
96
65
87
a.(2 pts) What is the appropriate test?
b.(1 pt) State the null hypothesis:
c.(1 pt) State the alternative hypothesis:
d.(2 pts) Find the critical value:
e.(2 pts) Calculate the obtained statistic:
f.(1 pts) Make a decision:
g.(1 pt) What does your decision mean?
h.(2 pts) Use SPSS to perform this same analysis and confirm the same result. Paste your output below:
May
August
90
100
65
80
78
92
50
60
89
90
92
98
75
70
90
96
65
87
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 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).
s = sqrt [ ((di - d)2 / (n - 1) ]
s = 7.98
SE = s / sqrt(n)
S.E = 2.66
DF = n - 1 = 22 -1 = 21
t = [ (x1 - x2) - D ] / SE
t = - 3.3
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 two-tailed test, the P-value is the probability that a t statistic having 8 degrees of freedom is more extreme than 3.33; that is, less than - 3.33 or greater than - 3.33
Thus, the P-value = 0.0104
Interpret results. Since the P-value (0.0104) is less than the significance level (0.05), we have to reject the null hypothesis.
From the test we have sufficient evidence that the score of students havce changed.
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