I just need number 14. Any help is super apprecited, let me know how you solved
ID: 3325611 • Letter: I
Question
I just need number 14. Any help is super apprecited, let me know how you solved it and if you used any shortcuts with the calculator, I have a Ti-83
One dispensed between the two machines. conclude that there is a significant difference in the variability of liquid 14. Independent random samples from two normal populations produced the variances listed here: Sample SizeSample Variance Gender Male Female 21 25 69.6 37.7 Do the data provide sufficient evidence to indicate that ' differs from :? Test using = .05. Test Statistic Reject Region: Reject Ho if F> Conclusion: One conclude that the variances are different. Find the approximate p-value for the test and interpret its value. Enter (nl, n2) Develop a 95% confidence interval for . CI = Enter (nl, n2) An experimenter is concerned that variability of responses using two different experimental procedures may not be the same. He randomly selects two samples of 16 and 14 responses from two normal populations and gets the statistics: s1-55, and si 118, respectively. Do the sample variances provide enough evidence at the 10% significance level to infer that the two population variances differ? 15.Explanation / Answer
14.
a.
Given that,
sample 1
s1^2=69.6, n1 =21
sample 2
s2^2 =37.7, n2 =25
null, Ho: ^2 = ^2
alternate, H1: ^2 != ^2
level of significance, = 0.05
from standard normal table, two tailed f /2 =2.327
since our test is two-tailed
reject Ho, if F o < -2.327 OR if F o > 2.327
we use test statistic fo = s1^1/ s2^2 =69.6/37.7 = 1.846
| fo | =1.846
critical value
the value of |f | at los 0.05 with d.f f(n1-1,n2-1)=f(20,24) is 2.327
we got |fo| =1.846 & | f | =2.327
make decision
hence value of |fo | < | f | and here we do not reject Ho
ANSWERS
---------------
null, Ho: ^2 = ^2
alternate, H1: ^2 != ^2
test statistic: 1.846
critical value: -2.327 , 2.327
decision: do not reject Ho
we do not have enough evidence to support the claim that variances are different
b.
the ratio of sample variances s1^2/s2^2 =69.6/37.7 =1.8461
assume take sample size is 25 above given data,
CONFIDENCE INTERVAL FOR VARIANCE
ci = (n-1) s^2 / ^2 right < ^2 < (n-1) s^2 / ^2 left
where,
s^2 = variance
^2 right = (1 - confidence level)/2
^2 left = 1 - ^2 right
n = sample size
since aplha =0.05
^2 right = (1 - confidence level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
^2 left = 1 - ^2 right = 1 - 0.025 = 0.975
the two critical values ^2 left, ^2 right at 24 df are 39.3641 , 12.401
variacne( s^2 )=1.8461
sample size(n)=25
confidence interval = [ 24 * 1.8461/39.3641 < ^2 < 24 * 1.8461/12.401 ]
= [ 44.3064/39.3641 < ^2 < 44.3064/12.4012 ]
[ 1.1256 , 3.5728 ]
15.
Given that,
sample 1
s1^2=118, n1 =14
sample 2
s2^2 =55, n2 =16
null, Ho: ^2 = ^2
alternate, H1: ^2 != ^2
level of significance, = 0.1
from standard normal table, two tailed f /2 =2.448
since our test is two-tailed
reject Ho, if F o < -2.448 OR if F o > 2.448
we use test statistic fo = s1^1/ s2^2 =118/55 = 2.145
| fo | =2.145
critical value
the value of |f | at los 0.1 with d.f f(n1-1,n2-1)=f(13,15) is 2.448
we got |fo| =2.145 & | f | =2.448
make decision
hence value of |fo | < | f | and here we do not reject Ho
ANSWERS
---------------
null, Ho: ^2 = ^2
alternate, H1: ^2 != ^2
test statistic: 2.145
critical value: -2.448 , 2.448
decision: do not reject Ho