a) dentify the explanatory/independent variable(X)and response/dependent variabl
ID: 2859611 • Letter: A
Question
a) dentify the explanatory/independent variable(X)and response/dependent variable (Y)
(b)Calculate the estimated regression line.
(c)Interpret the slope of the estimated regression line in terms of the problem.
(d)Interpret the intercept of the estimated regression line in terms of the problem (if appropriate).
(e)Predict the peak ow of a man who is 174 cm tall
(f)Find the 99% condence interval for the slope.
(g)Interpret your condence interval from (f) in terms of the problem.
(h)If we are comparing two men, and their dierence in heights is 10cm, what can we expect their average dierence in peak ow to be?
(i)If there was an unusually tall man in the dataset,should we keep his data or remove it? Explain.
(j)Calculate the value of Se , and interpret it in terms of the problem.
7. The peak flow rate of a person is the fastest rate at which ing a deep breath. Pealk a person can expel air after taking a deep breath. Peak flow 1s measured in u , and gives an in dication of a persons respiratory health. 17 men were randomly sampled, and information on their peak flow nits of liters/min and height (in cm) follows and height in Cm) folloWS: mean 180.4118 std.dev height peak flow 660 8.5591117.9952 The thought is that the height of a man should effect their peak flow. In addition, we are given that the sample correlation is 0.32725, and SS(resid) 198.909Explanation / Answer
x = height and y = peak flow
Slope = correlation (std dev y)/std dev x
= 4.5114
The reg line passes through (x bar, y bar)
Hence y- 660 = 4.5114(x-180.4118)
y = 4.5114 x-153.9183
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Slope is the rate of change of y wrt x
and intercept is the y value for x =0
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h =174
peak flow y = 4.5114(174)-153.9182
= 631.0654
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