Pediatricians wish to look at the relationship between the number of hours per d

Question

Pediatricians wish to look at the relationship between the number of hours per day a child watches televison and the number of pounds the child is overweight. Summary measures for a random sample of fifteen 10-year-old children are given below.

LEGEND Answer must be typed to either 2 or 3 decimals depending on whether the F or t table is appropriate. Please use the tables provided in STAT 3331 (not the tables in the textbook) Answer must be typed to four decimals Answer must be chosen from drop down menu Answer must be derived using cell references There is significant evidence to conclude There is not significant evidence to conclude TISETC TINSETO

Explanation / Answer

Question 6

x0 = 6.2

y^(6.2) = -15.0909 + 3.3636x = -15.0909 + 3.3636 * 6.2 = 5.7634 pounds

Here the 80% confidence interval = y^ +- tcritical se0

Here y^(6.2) = 5.7634 pounds

Here alpha = 0.20 , dF = 15 - 2 = 13

tcritical  = 1.3502

Here se0 = se * sqrt [ 1/n + (x - x0)2 /SSXX]

se = standard error of the estimate = sqrt [SSE/ (n-2)] = sqrt [352.5454/13] = 5.2076

SSXX = 11

se0 = se * sqrt [ 1/n + (x - x0)2 /SSXX] = 5.2076 * sqrt [1/15 + (5.2 - 6.2)2 /11] = 2.0672

80% confidence interval = y^ +- tcritical se0

= 5.7634 +- 1.3502 * 2.0672

= (2.9723 pounds , 8.5545 pounds)

Question 7A

at x0  = 4.8

y^(4.8) = -15.0909 + 3.3636x = -15.0909 + 3.3636 * 4.8 = 1.0544 pounds

QUestion 7(B)

95% prediction interval = y^ +- tcritical sep

Here alpha = 0.05 , dF = 15 - 2 = 13

tcritical  = 2.1604

Here sep = se * sqrt [ 1 + 1/n + (x - x0)2 /SSXX]

se = 5.2076

SSXX = 11

se0 = se * sqrt [ 1 + 1/n + (x - x0)2 /SSXX] = 5.2076 * sqrt [1 + 1/15 + (5.2 - 4.8)2 /11] = 5.4149

95% prediction interval = y^ +- tcritical sep

= 1.0544 + 2.1604 * 5.4149

= (-10.644 pounds, 12.753 pounds)

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