Researchers at a university have conducted a sleep study and measured the number
ID: 3217571 • Letter: R
Question
Researchers at a university have conducted a sleep study and measured the number of hours an individual has slept over the period of a week. They need to conduct a hypothesis test to determine if there is a difference in sleeping patterns between males and females.
a) What hypothesis test should be conducted for this scenario (assuming the data are normal)? ____________________________________________________________________________________
b) Is this test one-sided or two-sided? _______________________________________________________
c)What would the hypotheses for this study be? Identify the hypotheses in symbols and a statement.
statement
HO: ______________ ______________________________________________________________
H1: ______________ ______________________________________________________________
d) If the p-value for this study was 0.34998 (p=0.34998), how would you conclude? Would you reject or FTR Ho? Explain.
______________________________________________________________________________________________________________________________________________________________________________
e) What possible error could have been made from the conclusion in part d? ________________________
f) Given the dataset below, conduct the hypothesis test using = 0.01, assuming unequal variances. Again, also assume the data are normal.
Males
Females
72
32
70
40
28
20
48
72
42
26
61
50
22
67
75
35
Males
Females
72
32
70
40
28
20
48
72
42
26
61
50
22
67
75
35
Explanation / Answer
Given that,
mean(x)=52.25
standard deviation , s.d1=20.4433
number(n1)=8
y(mean)=42.75
standard deviation, s.d2 =18.813
number(n2)=8
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.01
from standard normal table, two tailed t /2 =3.499
since our test is two-tailed
reject Ho, if to < -3.499 OR if to > 3.499
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =52.25-42.75/sqrt((417.92851/8)+(353.92897/8))
to =0.967
| to | =0.967
critical value
the value of |t | with min (n1-1, n2-1) i.e 7 d.f is 3.499
we got |to| = 0.96716 & | t | = 3.499
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.9672 ) = 0.34998
hence value of p0.01 < 0.34998,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.967
critical value: -3.499 , 3.499
decision: do not reject Ho
p-value: 0.366
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