Forty sixth graders were randomly selected from a school district. Then, they we
ID: 3204599 • Letter: F
Question
Forty sixth graders were randomly selected from a school district. Then, they were divided into 20 matched pairs, each pair having equivalent IQ's. One member of each pair was randomly selected to receive special problem solving training. Then, all of the students were given an IQ test based on logical problem solving abilities. Test results are summarized below.
Pair
Training
No training
1
95
90
2
89
85
3
76
73
4
92
90
5
91
90
6
53
53
7
67
68
8
88
90
9
75
78
10
85
89
11
90
95
Pair
Training
No training
12
85
83
13
87
83
14
85
83
15
85
82
16
68
65
17
81
79
18
84
83
19
71
60
20
46
47
Please show work
Pair
Training
No training
1
95
90
2
89
85
3
76
73
4
92
90
5
91
90
6
53
53
7
67
68
8
88
90
9
75
78
10
85
89
11
90
95
Pair
Training
No training
12
85
83
13
87
83
14
85
83
15
85
82
16
68
65
17
81
79
18
84
83
19
71
60
20
46
47
Explanation / Answer
The solution to this problem takes four steps:
(1) state the hypotheses,
(2) formulate an analysis plan,
(3) analyze sample data,
(4) interpret results.
We work through those steps below:
Step 1:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: d = 0
Alternative hypothesis: d 0
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.
Step 2:
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-testof the null hypothesis
Step 3:
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-score test statistic (t).
s = sqrt * ((di - d)2 / (n - 1) ] = sqrt[ 270/(22-1) ] = sqrt(12.857) = 3.586
SE = s / sqrt(n) = 3.586 / [ sqrt(22) ] = 3.586/4.69 = 0.765
DF = n - 1 = 22 -1 = 21
t = [ (x1 - x2) - D ] / SE = (d - D)/ SE = (1 - 0)/0.765 = 1.307
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-score having 21 degrees of freedom is more extreme than 1.307; that is, less than -1.307 or greater than 1.307.
We use the t Distribution Calculatorto find P(t < - 1.307) = 0.103, and P(t > 1.307) = 0.103. Thus, the Pvalue = 0.103 + 0.103 = 0.206.
Step 4:
Interpret results. Since the P-value (0.206) is greater than the significance level (0.05), we cannot reject the null hypothesis.
this approach on an exam, you may alsowant to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the samples consisted of paired data, and the mean differences were normally distributed. In addition, we used the approximation formula to compute the standard error, since the sample size was small relative to the population size
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