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?

Thanks

You may want to check this with a statistics text, but here is one way you might approach this problem.

This looks like a hypothesis test involving two variances (standard deviation is the square root of the variance). The null hypothesis would be the ratio of the two variances less than or equal to 1. The alternative or alternate hypothesis would be the ratio of the two variances greater than 1.

Test statistic = 22^2/12^2 = ?

Finish the calculation. Using an F-distribution table at .05 level of significance, determine the critical or cutoff value to reject the null. (You will need to use degrees of freedom to determine the value.) If the test statistic exceeds the critical value from the table, the null is rejected in favor of the alternative hypothesis. If the test statistic does not exceed the critical value from the table, then the null is not rejected.

Thank you.

To determine whether there is more variation in the processing time of the new machine compared to the current machine, we can conduct a hypothesis test.

Let's define the null and alternative hypotheses:

Null hypothesis (H₀): There is no difference in the variation of the processing time between the new and current machines.
Alternative hypothesis (H₁): There is more variation in the processing time of the new machine compared to the current machine.

To test this, we can use a one-tailed F-test for comparing the variances of two independent samples.

Here's how we can perform the hypothesis test:

Step 1: Set up the hypotheses:

H₀: σ₁² = σ₂² (The variances are equal)
H₁: σ₁² > σ₂² (The variance of the new machine is greater than the current machine)

Step 2: Select the significance level (α). In this case, the significance level is given as 0.05.

Step 3: Compute the test statistic:

We can use the formula for the F-test for comparing variances:

F = s₁² / s₂²

Where:
- s₁² = sample variance of the new machine
- s₂² = sample variance of the current machine

In this case, we are given that the standard deviation of the processing time for the new machine is 22 hundredths of a second, which we can convert to a variance by squaring it:

s₁² = (22/100)² = 0.0484

Similarly, the standard deviation of the processing time for the current machine is 12 hundredths of a second:

s₂² = (12/100)² = 0.0144

So the test statistic is:

F = 0.0484 / 0.0144 = 3.361

Step 4: Determine the critical value:

The critical value is found using the F-distribution with degrees of freedom equal to the number of samples - 1. In this case, we have 16 computer runs, so the degrees of freedom are 16 - 1 = 15.

Using a significance level of 0.05 and 15 degrees of freedom, we find the critical value to be F₀.₀₅(15, ∞) = 2.845.

Step 5: Compare the test statistic with the critical value:

If the test statistic is greater than the critical value, we reject the null hypothesis in favor of the alternative hypothesis. Otherwise, we fail to reject the null hypothesis.

In our case, F = 3.361 > 2.845, so we reject the null hypothesis.

Step 6: Make a conclusion:

Based on the results of the hypothesis test, we can conclude that there is more variation in the processing time of the new machine compared to the current machine at the 0.05 significance level.

Therefore, we can conclude that the new machine exhibits greater variation in processing time compared to the current machine.