3.4 Table 3.9 shows the tidal volume of 37 adults sufering from atcial seotal de
ID: 2933628 • Letter: 3
Question
3.4 Table 3.9 shows the tidal volume of 37 adults sufering from atcial seotal defest. in 26 of essl et al. (E5). Do these data provide sufficient evidence to indicate a lower these, pulmonary hypertension was absent, and in 11 it was present. The da ta were re- rted by R tidal volume in subjects without pul P value? lmonary hypertension? Let x = 0.05. What is the Tidal volume, in milliliters, in two groups of subjects TABLE 3.9 Pulmonary hypertension absent Pulmonary hypertension absent Case Case (Y) 876 652 556 618 500 500 526 511 538 440 547 556 493 348 530 780 569 546 4 5 766 9 10 9 10 819 710 500 12 13 14 15 16 17 18 19 20 21 437 481 572 589 605 436 515 552 722 23 24 25 26 778 677 680 428Explanation / Answer
Null hypothesis H0: The tidal volume in subjects without pulmonary hypertension is same as the tidal volume in subjects with pulmonary hypertension.
Alternative hypothesis H1: The tidal volume in subjects without pulmonary hypertension is less than the tidal volume in subjects with pulmonary hypertension.
Tidal volume in subjects without pulmonary hypertension are
652, 556, 618, 500, 500, 526, 511, 538, 440, 547, 605, 500, 437, 481, 572, 589, 605, 436, 724, 515, 552, 722, 778, 677, 680, 428
Tidal volume in subjects with pulmonary hypertension are
876, 556, 493, 348, 530, 780, 569, 546, 766, 819, 710
Mean Tidal volume in subjects without pulmonary hypertension is 564.96
Mean Tidal volume in subjects with pulmonary hypertension is 635.73
Standard deviation of Tidal volume in subjects without pulmonary hypertension is 96.05
Standard deviation Tidal volume in subjects with pulmonary hypertension is 163.65
For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(96.052/26) + (163.652/11)] = 52.82
Degree of freedom, DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (96.052/26 + 163.652/11)2 / { [ (96.052 / 26)2 / (26 - 1) ] + [ (163.652 / 11)2 / (11 - 1) ] }
= 13 (Rounding to nearest integer)
Test statistic, t = (Difference in means) / SE = (635.73 - 564.96) / 52.82 = 1.34
P-value for t = 1.34 and DF = 13 is 0.1016
The P-value of the test is 0.1016
As, P-value is greater than the significance level (0.05), we fail to reject the null hypothesis and conclude that there is not suffcient evidence at 0.05 significance level that the tidal volume in subjects without pulmonary hypertension is less than the tidal volume in subjects with pulmonary hypertension.
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