For any hypothesis tests, you must use the p-value approach and 5-step procedure

Question

For any hypothesis tests, you must use the p-value approach and 5-step procedure discussed in class. In order to receive credit, you must separately list each step in the hypothesis test Step 1: State the null and alternative hypotheses Step 2: State the decision rule for rejecting the null hypothesis Step 3: Report the value of the test statistic and p-value directly from the Excel output Step 4: Evaluate the null hypothesis Step 5: State the practical conclusion to be drawn from the hypothesis test, in the context of the problem, in plain English. All requested confidence intervals must be in the correct final format (as discussed in class) and be fully interpreted in the context of the problem.

Explanation / Answer

Q1.
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 =0.04
^2 right = (1 - confidence level)/2 = (1 - 0.96)/2 = 0.04/2 = 0.02
^2 left = 1 - ^2 right = 1 - 0.02 = 0.98
the two critical values ^2 left, ^2 right at 22 df are 37.6595 , 10.6
variacne( s^2 )=25
sample size(n)=23
confidence interval = [ 22 * 25/37.6595 < ^2 < 22 * 25/10.6 ]
= [ 550/37.6595 < ^2 < 550/10.6 ]
[ 14.6045 , 51.8868 ]

Q2.
Given that,
population variance (sigma^2) =0.73
sample size (n) = 12
sample variance (s^2)=0.99242
null, Ho: sigma^2 =0.73
alternate, H1 : sigma^2 <0.73
level of significance, alpha = 0.05
from standard normal table,left tailed chisqr^2 alpha/2 =19.675
since our test is left-tailed
reject Ho, if chisqr^2 o < -19.675
we use test statistic chisquare chisqr^2 =(n-1)*s^2/o^2
chisqr^2 cal=(12 - 1 ) * 0.99242 / 0.73 = 11*0.99242/0.73 = 14.95
| chisqr^2 cal | =14.95
critical value
the value of |chisqr^2 alpha| at los 0.05 with d.f (n-1)=11 is 19.675
we got | chisqr^2| =14.95 & | chisqr^2 alpha | =19.675
make decision
hence value of | chisqr^2 cal | < | chisqr^2 alpha | and here we do not reject Ho
chisqr^2 p_value =0.1848
ANSWERS
---------------
null, Ho: sigma^2 =0.73
alternate, H1 : sigma^2 <0.73
test statistic: 14.95
critical value: -19.675
p-value:0.1848
decision: do not reject Ho
no evidence to support the variance in the number of vehicles owned or leased by
subscribers of the other magazine is less than the variance of .73

Q3.
Given that,
population variance (sigma^2) =0.00041
sample size (n) = 27
sample variance (s^2)=0.00063
null, Ho: sigma^2 =0.00041
alternate, H1 : sigma^2 < 0.00041
level of significance, alpha = 0.05
from standard normal table,left tailed chisqr^2 alpha/2 =38.885
since our test is left-tailed
reject Ho, if chisqr^2 o < -38.885
we use test statistic chisquare chisqr^2 =(n-1)*s^2/o^2
chisqr^2 cal=(27 - 1 ) * 0.00063 / 0.00041 = 26*0.00063/0.00041 = 39.95
| chisqr^2 cal | =39.95
critical value
the value of |chisqr^2 alpha| at los 0.05 with d.f (n-1)=26 is 38.885
we got | chisqr^2| =39.95 & | chisqr^2 alpha | =38.885
make decision
hence value of | chisqr^2 cal | > | chisqr^2 alpha| and here we reject Ho
chisqr^2 p_value =0.0395
ANSWERS
---------------
null, Ho: sigma^2 = 0.00041
alternate, H1 : sigma^2 <0.00041
test statistic: 39.95
critical value: -38.885
p-value:0.0395
decision: reject Ho

have evidence ti support
maximum variance in the length of the part is specified to 0.0041

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