Answer to Question #147926 in Statistics and Probability for Star Eduah

Question

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

Answer 1

$σ^2_1$ - variance of heart rates of smokes

$σ^2_2$ - variance of heart rates of non-smokes

Null hypothesis H₀: $σ^2_1 = σ^2_2$

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

Test statistic will be:

$\frac{m(n – 1)S^2_1}{n(m-1)S^2_2}$ ~ $F_{(m-1, n-1)}df$

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

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

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