Aviation and high altitude physiology is a specialty in the study or mediene. Le

Question

Aviation and high altitude physiology is a specialty in the study or mediene. Let . Partial pressure o, o ven in the alveen ar cels the ings) een breathing abery a stable ar an- partial pressure when breathing pure oxygen. The (x, y) data pairs correspond to elevations from 10,000 feet to 30,000 feet in s000 foot intervals for a random sample of volunteers. Although the medical data were collected using airplanes, they apply equally well to Mt. Everest dlimbers (summit 29,028 feet) (a) Verify that Er-21.3, 132.8, Zr, 105.03. Er, 4143.86, 656.82, and r . 0.970. Sx 21.3 y 1328 106.03 y 4143.86 y 656.82 .970 (b) use a 5% level of significance to test the daim that 0, (Use 2 decimal places.) critical Conclusion Reject the null hy hesis, there is sufficient evidence that > 0. Reject the num hypothesis, there is insumelent evidence that Fail to reject the rul hypothesis, there is insumeert evidence that > 0, Fall to regect the null hypothesis, there is sutncient evidence that 0. (e) Verify that S 3.4691-0.592, andb6374 S 3.4691 0 582 b6.374 (d) Find the predicted pressure when breathing pure oxypen if the pressure from breathing avaliabie air i29.Uuse 2

Explanation / Answer

(b) t = r sqrt [(n-2)/(1-r2)]] = 0.970/ sqrt [(5-2)/ (1-0.9702)] = 6.911

Here dF = 5 - 2 = 3 ; alpha = 0.05 ; one tailed test

tcritical  = 2.3534

so here t > tcritical  

We rejct the null hypothesis and can concludwe that rho > 0

(d) Here if x = 2.9

y^ = -0.592 + 6.374 x = -0.592 + 6.374 * 2.9 = 17.89

(e) Here 95% confidence interval = y^ +- tcritical se * sqert [1/n + (x - xo)2 /SSx]

To find the mean of sample x and SSx the tableis given below

x = x/n = 21.3/5 = 4.26

SSxx = x2 -(x)2/n = 105.03 * - 21.32 /5 = 14.292

tcritical = t3,0.05 = 3.18245

= 17.892 +- 3.18245 * 3.4691 * sqrt [1/5 + (4.26 - 2.9)2/14.292]

= 17.892 +- 11.0402 * 0.5739

= 17.892 + 6.3365

= (11.5555, 24.2285) or (11.6, 24.2)

(f) Here = 0

> 0

^ = 6.374

se(^) = 3.4691/ sqrt(ssxx) = 3.4691 / sqrt(14.292) = 0.9176

t = 6.374/0.9176 = 6.95

here t (critical) = 2.3534 < t

So we reject the null hypothesis . There is sufficient evidence that > 0

(g) 95% confidenc einterval of

95% CI = ^ +- tcritical se(^)

= 6.3736 +- 3.1824 * 0.9176

= (3.453, 9.294)

(g) Here option A is correct.

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