In: Statistics and Probability
Aviation and high-altitude physiology is a specialty in the study of medicine. Let x = partial pressure of oxygen in the alveoli (air cells in the lungs) when breathing naturally available air. Let y = partial pressure when breathing pure oxygen. The (x, y) data pairs correspond to elevations from 10,000 feet to 30,000 feet in 5000 foot intervals for a random sample of volunteers. Although the medical data were collected using airplanes, they apply equally well to Mt. Everest climbers (summit 29,028 feet). x 7.2 4.8 4.2 3.3 2.1 (units: mm Hg/10) y 43.0 31.5 26.2 16.2 13.9 (units: mm Hg/10)
(a) Verify that Σx = 21.6, Σy = 130.8, Σx2 = 107.82, Σy2 = 3983.34, Σxy = 653.49, and r ≈ 0.980. Σx Σy Σx2 Σy2 Σxy r
(b) Use a 1% level of significance to test the claim that ρ > 0. (Use 2 decimal places.) t critical t
(c) Verify that Se ≈ 2.7422, a ≈ -0.173, and b ≈ 6.096. Se a b
(d) Find the predicted pressure when breathing pure oxygen if the pressure from breathing available air is x = 3.7. (Use 2 decimal places.)
(e) Find a 95% confidence interval for y when x = 3.7. (Use 1 decimal place.) lower limit upper limit
(f) Use a 1% level of significance to test the claim that β > 0. (Use 2 decimal places.) t critical t
(g) Find a 95% confidence interval for β and interpret its meaning. (Use 2 decimal places.) lower limit upper limit

a)
| ΣX = | 21.600 | 
| ΣY= | 130.800 | 
| ΣX2 = | 107.820 | 
| ΣY2 = | 3983.340 | 
| ΣXY = | 653.490 | 
| r = | 0.980 | 
b)
| test statistic t = | r*(√(n-2)/(1-r2))= | 8.47 | ||
| t crit = | 4.54 | |||
c)
| Se =√(SSE/(n-2))= | 2.74220 | |||
| a= | -0.173 | |||
| b= | 6.096 | |||
d)
| predicted value = | 22.38 | |||
e)
| std error of confidence interval = | s*√(1+1/n+(x0-x̅)2/Sxx)= | 3.0369 | |||||
| for 95 % confidence and 3degree of freedom critical t= | 3.182 | ||||||
| lower limit = | 12.7 | ||||||
| uppr limit = | 32.0 | ||||||
f)
| test statistic t = | r*(√(n-2)/(1-r2))= | 8.47 | ||
| t crit = | 4.54 | |||
g)
| std error of slope sb1 = | s/√SSx= | 0.7199 | ||||
| for 95 % confidence and -2degree of freedom critical t= | 3.1820 | |||||
| 95% confidence interval =b1 -/+ t*standard error= | (3.805,8.386) | |||||
| lower limit = | 3.81 | |||||
| uppr limit = | 8.39 | |||||