Please help me out, Thank you! Actuaries use various parameters when evaluating

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Please help me out, Thank you!

Actuaries use various parameters when evaluating the cost of a life insurance policy. The variance of the life spans of a population is one of the parameters used for the evaluation. Each year, the actuaries at a particular insurance company randomly sample 30 people who died during the year (with the samples chosen independently from year to year) to see whether the variance of life spans has changed. The life span data from this year and from last year are summarized below Current Year Last Year = 76.3 s:-44.89 | si-17.64 (The first row gives the sample means and the second row gives the sample variances.) Assume that life spans are approximately normally distributed for each of the populations of people who died this year and people who died last year. Can we conclude, at the 0.01 significance level, that the variance of the life span for the current year, . , differs from the variance of the life span for last year, ? Perform a two-tailed test Then fill in the table below Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary. consult a list of formulas.) The null hypothesis The alternative hypothesis: The type of tet staistic (Choose one) The value of the test statistic: Round to at least three decimal places.) The p-value (Round to at least three decimal places.) Can we conclude that the variance of the life span for the current year differs from the variance of the life span for last year? es 0 Clear Undo Help

Explanation / Answer

Given that,
sample 1
s1^2=44.89, n1 =30
sample 2
s2^2 =17.64, n2 =30
null, Ho: ^2 = ^2
alternate, H1: ^2 != ^2
level of significance, = 0.01
from standard normal table, two tailed f /2 =2.674
since our test is two-tailed
reject Ho, if F o < -2.674 OR if F o > 2.674
we use test statistic fo = s1^1/ s2^2 =44.89/17.64 = 2.545
| fo | =2.545
critical value
the value of |f | at los 0.01 with d.f f(n1-1,n2-1)=f(29,29) is 2.674
we got |fo| =2.545 & | f | =2.674
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: 2.545
critical value: -2.674 , 2.674
decision: do not reject Ho

we do not have enough evidence to support the variance of the life spans for the current year differ from the last year

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