Asked by Cindy
Here are heart rates for a sample of 30 students before and after a class break. At á = .05, was
there a significant difference in the mean heart rate?
Heart Rate Before and After Class Break
(a) State the hypotheses. (b) State the decision
rule and sketch it. (c) Find the test statistic. (d) Make a decision. (e) Estimate the p-value and
interpret it.
S B A S B A
1 60 62 16 70 64
2 70 76 17 69 66
3 77 78 18 64 69
4 80 83 19 70 73
5 82 82 20 59 58
6 82 83 21 62 65
7 41 66 22 66 68
8 65 63 23 81 77
9 58 60 24 56 57
10 50 54 25 64 62
11 82 93 26 78 79
12 56 55 27 75 74
13 71 67 28 66 67
14 67 68 29 59 63
15 66 75 30 98 82
there a significant difference in the mean heart rate?
Heart Rate Before and After Class Break
(a) State the hypotheses. (b) State the decision
rule and sketch it. (c) Find the test statistic. (d) Make a decision. (e) Estimate the p-value and
interpret it.
S B A S B A
1 60 62 16 70 64
2 70 76 17 69 66
3 77 78 18 64 69
4 80 83 19 70 73
5 82 82 20 59 58
6 82 83 21 62 65
7 41 66 22 66 68
8 65 63 23 81 77
9 58 60 24 56 57
10 50 54 25 64 62
11 82 93 26 78 79
12 56 55 27 75 74
13 71 67 28 66 67
14 67 68 29 59 63
15 66 75 30 98 82
Answers
Answered by
PsyDAG
We do not do your work for you. Once you have answered your questions, we will be happy to give you feedback on your work. Although it might require more time and effort, you will learn more if you do your own work. Isn't that why you go to school?
However, I will start you out.
Ho: mean1 = mean2
Ha: mean1 ≠ mean2
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion equivalent to the Z score.
However, I will start you out.
Ho: mean1 = mean2
Ha: mean1 ≠ mean2
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion equivalent to the Z score.
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