Question #147926

A medical researcher wishes to see whether the variance of the heart beats (in beats per minute) of smokers is different from the variance of heart rates of people who do not smoke. Two samples are selected and the data shows that 26 smokers have a variance of 36 in their heart beats and 18 non smokers have a variance of 10. using a=0.05, is there enough evidence to support the claim that the variances are equal? Assume the variable is normally distributed


1
Expert's answer
2020-12-01T19:48:59-0500

σ12σ^2_1 - variance of heart rates of smokes

σ22σ^2_2 - variance of heart rates of non-smokes

Null hypothesis H0: σ12=σ22σ^2_1 = σ^2_2

Alternative hypotheses H1: σ12σ22σ^2_1 ≠ σ^2_2

Test statistic will be:

m(n1)S12n(m1)S22\frac{m(n – 1)S^2_1}{n(m-1)S^2_2} ~ F(m1,n1)dfF_{(m-1, n-1)}df

Where m is size of first sample and n is size of second sample.

m=26n=18s12=36S22=10F=m(n1)S12n(m1)S22=159124500=3.536F(m1,n1)df=F(25,17)(0.05)=2.05Fcal>ftabm = 26 \\ n = 18 \\ s^2_1 = 36 \\ S^2_2 = 10 \\ F = \frac{m(n – 1)S^2_1}{n(m-1)S^2_2} = \frac{15912}{4500} = 3.536 \\ F_{(m-1, n-1)}df = F_{(25,17)}(0.05) = 2.05 \\ F_{cal} > f_{tab}

Here we reject our null hypothesis. There is significant difference in the variance as heart rates of smokes and non-smokes.


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