Need help with both problems 3&4 . Thanks 3) A study of estrogen levels in women
ID: 2935827 • Letter: N
Question
Need help with both problems 3&4 . Thanks 3) A study of estrogen levels in women find the following results (in pg/mL): 309.1 226.4 141.6 309.9 198.2 268.5 During ovulation: Beginning of menstruation: 58.9 109. 83.7 82.1 95.9 63.7 101.2 107. 60.2 Is there a difference in estrogen levels? (Note that this time you need to do everything (including all the calculations) yourself). 4) Repeat (3), but this time assume equal variances (a) Use the same level of a you used in (3) (b) Do you think this test is appropriate? Why or why not (hint: are the sample variances similar)?
Explanation / Answer
Q3.
Given that,
mean(x)=242.2833
standard deviation , s.d1=66.4177
number(n1)=6
y(mean)=84.6556
standard deviation, s.d2 =20.0295
number(n2)=9
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.571
since our test is two-tailed
reject Ho, if to < -2.571 OR if to > 2.571
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =242.2833-84.6556/sqrt((4411.31087/6)+(401.18087/9))
to =5.645
| to | =5.645
critical value
the value of |t | with min (n1-1, n2-1) i.e 5 d.f is 2.571
we got |to| = 5.64472 & | t | = 2.571
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 5.6447 ) = 0.002
hence value of p0.05 > 0.002,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 5.645
critical value: -2.571 , 2.571
decision: reject Ho
p-value: 0.002
we claim that diffrence in the estrogen levels
Q4.
Given that,
mean(x)=242.2833
standard deviation , s.d1=66.4177
number(n1)=6
y(mean)=84.6556
standard deviation, s.d2 =20.0295
number(n2)=9
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.16
since our test is two-tailed
reject Ho, if to < -2.16 OR if to > 2.16
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (5*4411.310873 + 8*401.18087) / (15- 2 )
s^2 = 1943.538564
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=242.2833-84.6556/sqrt((1943.538564( 1 /6+ 1/9 ))
to=157.6277/23.235142
to=6.784021
| to | =6.784021
critical value
the value of |t | with (n1+n2-2) i.e 13 d.f is 2.16
we got |to| = 6.784021 & | t | = 2.16
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != 6.784 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 6.784021
critical value: -2.16 , 2.16
decision: reject Ho
p-value: 0
we claim that diffrence in the estrogen levels