The accompanying data on food intake (in Kcal) for 15 men on the day following t
ID: 3324510 • Letter: T
Question
The accompanying data on food intake (in Kcal) for 15 men on the day following two nights of only 4 hours of sleep each night and for 15 men on the day following two nights of 8 hours of sleep each night is consistent with summary quantities in the paper "Short-Term Sleep Loss Decreases Physical Activity under Free-Living Conditions But Does Not Increase Food Intake under Time Deprived Laboratory Conditions in Healthy Men" (American Journal of Clinical Nutrition [2009]: 1476–1482). The men participating in this experiment were randomly assigned to one of the two sleep conditions.
Verify the assumption of approximate normality for each of the populations. Carry out a two-sample t test with = 0.05 to determine if there is a significant difference in mean food intake for the two different sleep conditions. (Use a statistical computer package to calculate the P-value. Use 4-hour 8-hour. Round your test statistic to two decimal places, your df down to the nearest whole number, and your P-value to three decimal places.)
t=
df=
P-value=
4-hour sleep group: 3585 4470 3068 5338 2221 4791 4435 3099 3187 3901 3868 3869 4878 3632 4518 8-hour sleep group 496539181987 49935220 3653 3510 3338 4100 5792 45473319 3336 4304 4057Explanation / Answer
Given that,
mean(x)=3924
standard deviation , s.d1=829.6681
number(n1)=15
y(mean)=4069.267
standard deviation, s.d2 =952.8958
number(n2)=15
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.145
since our test is two-tailed
reject Ho, if to < -2.145 OR if to > 2.145
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =3924-4069.267/sqrt((688349.15616/15)+(908010.40566/15))
to =-0.445
| to | =0.445
critical value
the value of |t | with min (n1-1, n2-1) i.e 14 d.f is 2.145
we got |to| = 0.44529 & | t | = 2.145
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.4453 ) = 0.663
hence value of p0.05 < 0.663,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -0.445
critical value: -2.145 , 2.145
decision: do not reject Ho
p-value: 0.663
df = 2.145
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