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. HeartRate
10.59 Emergency room arrivals in a large hospital showed the statistics below for 2 months. At α = .05,
has the variance changed? Show all steps clearly, including an illustration of the decision rule.
10.61 A certain company will purchase the house of any employee who is transferred out of state and
will handle all details of reselling the house. The purchase price is based on two assessments, one
assessor being chosen by the employee and one by the company. Based on the sample of eight assessments
shown, do the two assessors agree? Use the .01 level of significance, state hypotheses
clearly, and show all steps. HomeValue
Statistic October November
Mean arrivals 177.0323 171.7333
Standard deviation 13.48205 15.4271
Days 31 30
Note: Thanks to colleague Gene Fliedner for having his evening students take their own pulses before and after the 10-minute class break.
Heart Rate Before and After Class Break
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
(a) The hypotheses for testing the significant difference in mean heart rate before and after the class break are:
Null hypothesis (H0): There is no significant difference in the mean heart rate before and after the class break.
Alternative hypothesis (Ha): There is a significant difference in the mean heart rate before and after the class break.
(b) The decision rule can be determined by conducting a t-test for dependent samples and comparing the calculated t-value with the critical t-value at α = 0.05 level of significance. The decision rule can be represented by a t-distribution curve with critical regions on both sides.
(c) To find the test statistic, you need to calculate the paired differences between the before and after heart rates for each student, then find the mean and standard deviation of these differences. Finally, use the formula for the t-test for dependent samples:
t = (mean difference - hypothesized difference) / (standard deviation / √n)
In this case, the hypothesized difference is 0 (since we are testing if there is a significant difference) and n is the number of pairs (30 in this case).
(d) After calculating the test statistic, you compare it with the critical t-value from the t-distribution table. If the test statistic falls within the rejection region (beyond the critical t-value), you reject the null hypothesis. Otherwise, fail to reject the null hypothesis.
(e) The p-value can be estimated by finding the area under the t-distribution curve beyond the absolute value of the test statistic. The p-value represents the probability of observing a test statistic as extreme as the calculated value, assuming the null hypothesis is true. If the p-value is smaller than the chosen significance level (α = 0.05), you reject the null hypothesis.
For the remaining questions regarding emergency room arrivals and house assessments, the same steps will apply: stating the hypotheses, determining the decision rule, finding the test statistic, making a decision, and estimating the p-value.