(15) 5a) A brightness of a television picture tube can be evaluated by measuring
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Question
(15) 5a) A brightness of a television picture tube can be evaluated by measuring the amount of current required to achieve a particular brightness level. A sample of 10 tubes results in s- 15.7. (i) Do the data support the claim that the standard deviation of current is less than 20 microamps? Set up a hypothesis statement and perform test using -0.05 State: Ho: and Hi: Perform test using appropriate test statistic: Write conclusion in the problem context: (i) Should p-value greater than in this case? Yes No (Circle one) (iii) Find the P-value for this test. 5b) In a comparison study of two independent normal populations, the sample (10) variances were s-1.1 and s -3.4, and the corresponding sample sizes were nl = 6 and n2 = 10. Calculate a 95% confidence interval on the ratio of the two population variances. Check if the two population variances are equal using the confidence interval calculated above.Explanation / Answer
5.
a.
Given that,
population standard deviation ()=20
sample standard deviation (s) =15.7
sample size (n) = 10
we calculate,
population variance (^2) =400
sample variance (s^2)=246.49
null, Ho: =20
alternate, H1 : <20
level of significance, = 0.05
from standard normal table,left tailed ^2 /2 =16.919
since our test is left-tailed
reject Ho, if ^2 o < -16.919
we use test statistic chisquare ^2 =(n-1)*s^2/o^2
^2 cal=(10 - 1 ) * 246.49 / 400 = 9*246.49/400 = 5.546
| ^2 cal | =5.546
critical value
the value of |^2 | at los 0.05 with d.f (n-1)=9 is 16.919
we got | ^2| =5.546 & | ^2 | =16.919
make decision
hence value of | ^2 cal | < | ^2 | and here we do not reject Ho
^2 p_value =0.7844
ANSWERS
---------------
i.
null, Ho: =20
alternate, H1 : <20
test statistic: 5.546
critical value: -16.919
ii.
Yes,
p-value is greater than the level of significance
iii.
p-value:0.7844
decision: do not reject Ho
we do not have enough evidence to support the claim that standard deviation of current is less than 20
b.
i.
ratio of variances s1^2/s2^2 = 1.2/3.4 =0.3235
confidence interval 95%
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 9 df are 19.0228 , 2.7
variacne( s^2 )=0.3235
sample size(n)=10
confidence interval = [ 9 * 0.3235/19.0228 < ^2 < 9 * 0.3235/2.7 ]
= [ 2.9115/19.0228 < ^2 < 2.9115/2.7004 ]
[ 0.1531 , 1.0782 ]
ii.
confidence interval for two population variances are not equal because there are two independent sample variances
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