Question

Two proposed computer mouse designs were compared by recording wrist extension in degrees for 24 people who each used both mouse designs ("Comparative Study of Two Computer Mouse Designs," Cornell Human Factors Laboratory Technical Report RP7992). The difference in wrist extension was calculated by subtracting extension for mouse type \(\mathrm{B}\) from the wrist extension for mouse type \(\mathrm{A}\) for each person. The mean difference was reported to be 8.82 degrees. Assume that this sample of 24 people is representative of the population of computer users. a. Suppose that the standard deviation of the differences was 10 degrees. Is there convincing evidence that the mean wrist extension for mouse type \(A\) is greater than for mouse type \(\mathrm{B}\) ? Use a 0.05 significance level. b. Suppose that the standard deviation of the differences was 26 degrees. Is there convincing evidence that the mean wrist extension for mouse type \(A\) is greater than for mouse type \(\mathrm{B}\) ? Use a 0.05 significance level. c. Briefly explain why different conclusions were reached in the hypothesis tests of Parts (a) and (b).

Short answer

a. The null hypothesis is rejected and there is evidence to support that the mean wrist extension for mouse type A is greater than mouse type B (Standard deviation = 10 degrees). b. The null hypothesis is not rejected and there is not enough evidence to support that mean wrist extension for mouse type A is greater than mouse type B (Standard deviation = 26 degrees). c. The conclusions are different due to the change in standard deviation which affects the variability in data, influencing the test results.

Step-by-step solution

Identify the null and alternative hypotheses

For this exercise, the null hypothesis \(H_0\) would state that there is no difference in the mean wrist extension for mouse type A and B. This can mathematically be expressed as \( \mu_{A} - \mu_{B} = 0 \). The alternative hypothesis \(H_1\) would state that the mean extension for mouse type A is greater than mouse type B. This can mathematically be expressed as \( \mu_{A} - \mu_{B} > 0 \). The significance level is 0.05 (or 5%), meaning there is a 5% chance of rejecting the null hypothesis when it's true.

Calculate the test statistic and P-value (Standard deviation 10 degrees)

First the z statistic is calculated using the formula z = (sample mean - population mean) / (standard deviation/√n). Here, the sample mean is 8.82 degrees, population mean is 0 (from null hypothesis), standard deviation is 10 degrees and n=24 (number of samples). Plugging these values, we get z ~ 1.76. Then, the P-value is looked from z table which gives us the probability that z scores will be less than our calculated z value. P-value ~ 0.039.

Make decision based on the P-value (Standard deviation 10 degrees)

Since the P-value is less than the significance level (0.039 < 0.05), the null hypothesis is rejected. There is enough evidence to support that the mean wrist extension for mouse type A is greater than mouse type B.

Calculate the test statistic and P-value (Standard deviation 26 degrees)

Now we will repeat step 2 but with the new standard deviation of 26 degrees. Again calculating, we get z ~ 0.67. From the z table, P-value ~ 0.252.

Make decision based on the P-value (Standard deviation 26 degrees)

Since this time the P-value is greater than the significance level (0.252 > 0.05), we cannot reject the null hypothesis. With this standard deviation, there's not enough evidence to support that the mean wrist extension for mouse type A is greater than mouse type B.

Analyze different conclusions in parts (a) and (b)

The reason for different conclusions in these two parts is due to the change in standard deviation. A larger standard deviation implies more variability in data which makes it harder to reject the null hypothesis. In part (a), with a smaller standard deviation, the differences are more concentrated making it easier to reject the null hypothesis.