Answer to Question #226064 in Statistics and Probability for Carrie

Question

Random samples from two normal populations produced the following statistics:

S

2

1 D 350 n1 D 31 S

2

2 D 700 n2 D 31 D 0:05 H0 : 

2

1 D 

2

2

vs H1 : 

2

1 > 2

2

The statistician wants to test the quality of the two population variances. The 95% confidence

interval of ratio of two population variances is

1. .0:2415I 1:035/

2. .1:2415I 2:035/

3. .

Answer 1

$$s_1^2=350, n_1=31, s_2^2=700, n_2=31$$

The critical values for $\alpha=0.05$ and $df_1=n_1-1=31-1=30,$

$df_2=n_2-1=31-1=30$ degrees of freedom are:

$$F_L=F_{1-\alpha/2, n_2-1, n_1-1}=F_{1-0.05/2, 31-1, 31-1}=0.4822$$

$$F_U=F_{\alpha/2, n_2-1, n_1-1}=F_{0.05/2, 31-1, 31-1}=2.0739$$

The corresponding 95% confidence interval is computed as follows:

$$CI=(\dfrac{s_1^2}{s_2^2}F_L, \dfrac{s_1^2}{s_2^2}F_U)$$

$$=(\dfrac{350}{700}\times0.4822, \dfrac{350}{700}\times2.0739)$$

$$=(0.24110, 1.03695)$$

Therefore, based on the data provided, the 95% confidence interval for the ratio of the population variances is $0.2411<\dfrac{\sigma_1^2}{\sigma_2^2}<1.0370.$ Therefore, we are 95% confident that the true ratio of population variances $\dfrac{\sigma_1^2}{\sigma_2^2}$ is contained by the interval $(0.2411, 1.0370).$

The 95% confidence interval of ratio of two population variances is

$(0.2411, 1.0370).$