Problem 16 - Basic Statistics for Business and Economics with Student

Question

A computer manufacturer is about to unveil a new, faster personal computer. The new machine clearly is faster, but initial tests indicate there is more variation in the processing time. The processing time depends on the particular program being run, the amount of input data, and the amount of output. A sample of 16 computer runs, covering a range of production jobs, showed that the standard deviation of the processing time was 22 (hundredths of a second) for the new machine and 12 (hundredths of a second) for the current machine. At the .05 significance level can we conclude that there is more variation in the processing time of the new machine?

Step-by-Step Solution

Answer

Yes, there is more variation in the new machine's processing time.

Step 1: Define Hypotheses

We need to compare the variances of the two machines. Let \( \sigma_1^2 \) be the variance of the new machine and \( \sigma_2^2 \) be the variance of the current machine. We state our null hypothesis \( H_0: \sigma_1^2 \leq \sigma_2^2 \) and the alternative hypothesis \( H_a: \sigma_1^2 > \sigma_2^2 \) indicating the new machine has more variation.

Step 2: Set Significance Level

The significance level \( \alpha \) is given as 0.05.

Step 3: Test Statistic Calculation

We use the F-test to compare variances. The test statistic is \( F = \frac{s_1^2}{s_2^2} \), where \( s_1 = 22 \) and \( s_2 = 12 \). Thus, \( F = \frac{22^2}{12^2} = \frac{484}{144} \approx 3.361 \).

Step 4: Determine Degrees of Freedom

The degrees of freedom for the numerator \( v_1 = n_1 - 1 = 16 - 1 = 15 \) and for the denominator \( v_2 = n_2 - 1 = 16 - 1 = 15 \).

Step 5: Critical Value and Decision Rule

Using an F-distribution table or software for \( \alpha = 0.05 \), \( v_1 = 15 \), and \( v_2 = 15 \), the critical F-value \( F_{critical} \approx 2.54 \). If \( F > F_{critical} \), reject \( H_0 \).

Step 6: Conclusion

Since \( F = 3.361 \) is greater than the critical value 2.54, we reject the null hypothesis. There is significant evidence to conclude that the new machine has more variation in processing time.

Key Concepts

F-test

The F-test is a statistical test used to compare two sample variances and determine if they come from populations with the same variance. It is particularly useful when you want to establish whether one set of processes exhibits more variability than another.

The F-test follows the F-distribution, and it is calculated by dividing the variance of one sample by the variance of another sample. This yields the F-statistic, expressed as:

• Test Statistic: \( F = \frac{s_1^2}{s_2^2} \)

When applying the F-test, always ensure the data sets, or sample variances, are normally distributed and the samples are independent. This test is powerful but requires careful selection of the correct critical value from the F-distribution table.

For the exercise, we computed the F-statistic to be approximately 3.361. The critical value that needs to be compared is based on the chosen significance level and degrees of freedom.

Variance Comparison

Variance comparison involves analyzing the scatter or spread of data points in a distribution. When you compare variances, you're essentially examining how much data points differ from the mean of the data set.

In hypothesis testing, comparing the variances between two groups can reveal whether the variability differs significantly. This is crucial in deciding if a particular treatment or change had a differing effect on group variability.

In the given exercise, the variances for two groups of computers were being compared. The standard deviation for the new machine was 22, and for the current machine, it was 12. To find the variance, you simply square the standard deviation:

• Variance of new machine: \( 22^2 = 484 \)
• Variance of current machine: \( 12^2 = 144 \)

This set-up introduces us to a clear variance comparison between the two machines to determine if the new machine exhibits significantly more variation in processing time.

Significance Level

The significance level, often represented by \( \alpha \), is a threshold chosen in hypothesis testing that determines the probability of rejecting the null hypothesis when it is actually true. It's often set to 0.05, which means there is a 5% risk of concluding that a difference exists when there is no actual difference.

Choosing an appropriate significance level is important because it affects the likelihood of making Type I errors, which is what happens when you wrongly reject a true null hypothesis. In general, a lower significance level indicates a stricter criterion for rejecting the null hypothesis.

In our exercise, the significance level was set at 0.05. This helped define the critical F-value, which was approximately 2.54, and guided us in making the final conclusion of the hypothesis test. If our calculated F-statistic exceeds this critical value, the result is considered statistically significant, leading us to reject the null hypothesis and affirm there is a notable variance increase in the new machine.