An article in Cancer Research [“Analyses of Litter-Matched Time-to-Response Data, with Modifications for Recovery of Interlitter
Information” (1977, Vol. 37, pp. 3863–3868)] tested the tumorigenesis of a drug. Rats were randomly selected from litters and given the drug. The times of tumor appearance were recorded as follows: 101, 104, 104, 77, 89, 88, 104, 96, 82, 70, 89, 91, 39, 103, 93, 85, 104, 104, 81, 67, 104, 104, 104, 87, 104, 89, 78, 104, 86, 76, 103, 102, 80, 45, 94, 104, 104, 76, 80, 72, 73 Calculate a 95% confidence interval on the standard deviation of time until a tumor appearance. Check the assumption of normality of the population and comment on the assumptions for the confidence interval
1 answer:
Answer:
2.51 < σ < 3.92
Step-by-step explanation:
We will use Chi-Square distribution to create the interval. We have the following information:
n = 41
Level of Confidence : 95%,
This corresponds to Chi-Square values of:
24.433 for the left tail and 59.342 for the right tail
We need to calculate the sample Variance. This is done on attached photo 1 (the shortcut formula for finding variance was used)
We get s = 9.3866.
See the construction of the confidence interval on the second attached photo
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<u>Answer:</u>
1 , (Answer)
<u>Step-by-step explanation:</u>
Since, 8 chips are selected at random and 5 are there defectives in the lot, So, at least (8 - 5) = 3 chosen chips will be non - defective.
So, P ( At least two of selected part is non defective ) = 1 .
P(first and second samples are defective)
=
P ( first, second and third samples are defective)
=
So,
=
=
= (Answer)