A recent study compared the life expectancy of individuals i

A recent study compared the life expectancy of individuals in the UK with individuals in the US. They found that the average life expectancy in the UK was 76 and the US had an average life expectancy of 72.5. But now they want to know if the social medicine helps the variance. For a sample of 15 US individuals the sample standard deviation of life expectancy is 5. The Sample standard deviation of a sample of 13 UK citizens was 3.5. At the significance level of 0.05 can we conclude that there is a difference in the variance of the life expectancy of the two countries?

What is the null hypothesis?

what is the alternative hypothesis

what is the appropriate test statistic

what is the critical value of the test staistic?

what is your decsison based on the results

interpret your results

Solution

alpha,a = 0.05
n1 = 15 , s1 = 5
n2 = 13 , s2 = 3.5

null hypothesis
H0: Sigma1^2 = Sigma2^2

alternative hypothesis
H1: Sigma^2 is not equal to Sigma2^2

test statistic
F = s1^2 / s2^2
F = 5^2 / 3.5^2
F = 2.041

critical values are
F(a/2, n1-1, n2-1) = F(0.025, 14, 12) = 3.206
F(1 - a/2, n1-1, n2-1) = 1 / F(a/2, n2-1, n1-1) = 1/F(0.025, 12, 14) = 0.328

decsison based on the results
non-critical region is interval: (0.328 , 3.206)
since F lies inside the interval, hence H0 can not be rejected

interpretation
So, we can not conclude that there is a difference in the variance of the life expectancy of the two countries.