Question

FEV (forced expiratory volume) is an index of pulmonary function that measures the volume of air expelled after one second of constant effort. 654 children ages 3-19 who were seen in the Childhood Respiratory Disease Study in East Boston, Massachusetts. Variables:

Age: age in years

FEV: FEV in liters

Age group 3-8: n=215 mean = 1.859 sd=.421

Age group 9-12: n=322 mean= 2.81 sd=.640

Age group 13-19: n=117 mean= 3.59 sd= .796

1. Is FEV the same across the three age groups? Perform a hypothesis test to answer the question. Use ? = 0.05.

2. Rank the three age groups using the multiple comparison approach

3. What assumptions are made with regard to the analysis in part 1? Check whether these assumptions are violated.

Answer 1

1)

H0 = m1 = m2=m3

H1: at least one of the means are different

comparing m1, m2

t = (1.859-2.81)/ sqrt(421^2/215 + .640^2/322) =-20.7702

as t at 95% ~1.96

ao |t| > 1.96 so the means are not equal

so FEV is not same

2)

as t above is negative and significant we see m2>m1

no t for m2and m2

t = (3.59 -2.81)/ sqrt(.796^2/117+.640^2/322) = 9.538068

as t >1.96

wecan say m3 >m2

ao m3>m2>m1

3)

assumption is 1) all have equal variance

we perform the Bartlett test

for us k =3

n = 654

sp= (214 *(.421^2) + 321* (.640^2) + 116 * (.796^2)) / 651

0.3731346

T = (651* ln (0.3731346) - 214 *ln (.421^2) - 321* ln (.640^2) - 116 *ln (.796^2) )/ 1+ 1/6 ( 1/ 214 + 1/321 + 1/116 - 1/ 651))

=67.95468 /1.002478792 =67.78665

qchisq at .05 and 2 Df is 5.991465

as T > t we reject the null hypothesis that the varainces are equal so yes the assumptions are violated