Question

Two coworkers commute from the same building. They are interested in whether or not there is any variation in the time it takes them to drive to work. They each record their times for 20 commutes. The first worker's times have a variance of 12.2. The second worker's times have a variance of 16.9. The first worker thinks that he is more consistent with his commute times and that his commute time is shorter. Test the claim at the 10% level.

Answer 1

Answer:

Step-by-step explanation:

Hello!

You have two samples.

Sample 1 (time that takes one commute of worker 1)

n₁= 20

S₁²= 12.2

Sample 2 (time that takes one commute of worker 2)

n₂= 20

S₂²= 16.9

The hypothesis is that the first coworker is more consistent with his commute times, this means that the variability of his commute times is less than the variability of the commute times of worker 2. With this in mind, the hypothesis for a variance ratio test is:

H₀: σ₁² ≥ σ₂²

H₁:: σ₁² < σ₂²

α: 0.10

The statistic for this test is:

F= (S₁²/S₂²) * (σ₁²/σ₂²) ~ F$_{n1-1;n2-1}$

The test is one-tailed (left) with just one critical value:

$F_{n_1-1;n_2-1;\alpha } = \frac{1}{F_{n_2-1;n_1-1;1-\alpha }} = \frac{1}{F_{19;19;0.90}} = \frac{1}{1.82} =0.549$

(The F-table I use only has the values for high probabilities so I've used a conversion method to obtain the value for the small probability)

Your decision rule is:

If F ≤ 0.549, you reject the null hypothesis.

if F > 0.549, you support the null hypothesis.

The calculated value:

F= (S₁²/S₂²) * (σ₁²/σ₂²) = (12.2/16.9) * 1= 0.722

Since the calculated F-value is greater than the critical value, the decision is to not reject the null hypothesis. So you can say that the variability of the commute times of worker 1 is less than the variability of the commute times of worker 2.

I hope you have a SUPER day!