Here are heart rates for a sample of 30 students before and after a class break. At α = .05, was
there a significant difference in the mean heart rate? (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
Student Before After Student Before After
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
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To determine if there is a significant difference in the mean heart rate before and after a class break, we can perform a paired sample t-test. Let's go through the steps:
(a) State the hypotheses:
- Null Hypothesis (H0): The mean heart rate before and after the class break is the same.
- Alternate Hypothesis (Ha): There is a significant difference in the mean heart rate before and after the class break.
(b) State the decision rule and sketch it:
- Since the significance level (α) is given as 0.05, we will use a two-tailed t-test.
- The decision rule is: If the calculated t-statistic falls within the critical region (beyond the critical values), we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
(c) Find the test statistic:
- To find the test statistic, we need to calculate the paired differences and then find the mean and standard deviation of those differences.
- Let's calculate the paired differences:
Student | Before | After | Difference
--------------------------------------
1 | 60 | 62 | 2
2 | 70 | 76 | 6
3 | 77 | 78 | 1
4 | 80 | 83 | 3
5 | 82 | 82 | 0
6 | 82 | 83 | 1
7 | 41 | 66 | 25
8 | 65 | 63 | -2
9 | 58 | 60 | 2
10 | 50 | 54 | 4
11 | 82 | 93 | 11
12 | 56 | 55 | -1
13 | 71 | 67 | -4
14 | 67 | 68 | 1
15 | 66 | 75 | 9
16 | 70 | 64 | -6
17 | 69 | 66 | -3
18 | 64 | 69 | 5
19 | 70 | 73 | 3
20 | 59 | 58 | -1
21 | 62 | 65 | 3
22 | 66 | 68 | 2
23 | 81 | 77 | -4
24 | 56 | 57 | 1
25 | 64 | 62 | -2
26 | 78 | 79 | 1
27 | 75 | 74 | -1
28 | 66 | 67 | 1
29 | 59 | 63 | 4
30 | 98 | 82 | -16
- Now we calculate the mean and standard deviation of the differences:
Mean = Σdifferences / n = (-2 + 6 + 1 + 3 + 0 + ... -16) / 30 = -0.8333 (rounded to 4 decimal places)
Standard Deviation = √[ Σ(differences - Mean)^2 / (n - 1) ] = √[ (4.75 + 30.9167 + ... + 273.4167) / 29 ] = 6.0777 (rounded to 4 decimal places)
- Since the sample size (n) is relatively small (30) and the population standard deviation is unknown, we will use a t-distribution for the test.
(d) Make a decision:
- To make a decision, we compare the calculated t-statistic with the critical values from the t-distribution.
- The calculated t-statistic is obtained as: t = (Mean - μ0) / (Standard Deviation / √n) = (-0.8333 - 0) / (6.0777 / √30) = -0.5033 (rounded to 4 decimal places)
- We need to find the critical values from the t-distribution table or use a statistical software. For α = 0.05 and with 29 degrees of freedom (n - 1 = 30 - 1 = 29), the critical values are ±2.0452 (rounded to 4 decimal places).
- Since the calculated t-statistic (-0.5033) falls within the non-critical region (-2.0452 to 2.0452), we fail to reject the null hypothesis.
(e) Estimate the p-value and interpret it:
- The p-value represents the probability of obtaining a test statistic as extreme as the one observed under the null hypothesis.
- To estimate the p-value, we can use statistical software or a t-distribution table. In this case, the p-value is approximately 0.6160 (rounded to 4 decimal places).
- Since the p-value (0.6160) is greater than the significance level (α = 0.05), we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that there is a significant difference in the mean heart rate before and after the class break.
In summary, based on the results of the paired sample t-test, we fail to reject the null hypothesis and conclude that there is no significant difference in the mean heart rate before and after the class break.