Refer to Case Study 10.1, found on page 371 of your text. The full data set (ann

Question

Refer to Case Study 10.1, found on page 371 of your text. The full data set (annual returns) is shown below. Use the full data set on n = 20 for each group. It is recommended strongly that you use Minitab to calculate the test and confidence interval results. Class policy requires that you print out a hard copy of your Minitab results (hand-written results will not received credit), therefore, print out the results and complete the rest of the answers on your printout. Answer the following questions: Part A) Report the point estimate for the difference between the 2 means and interpret this value. Be specific in your interpretation - do not explain how it is computed. Part B) Perform the appropriate test described in the case. Make sure you include all required steps for the Ho testing process. Part C) Identify the confidence interval provided by Minitab. Briefly explain how the confidence interval supports the results of the hypothesis test.

Health: 10.29, 32.17, 34.27, 10.57, 20.27, 31.43, 15.4, 0.88, 17.85, 28.98, 6.49, 28.73, 8.17, 14.83, 12.89, 23.51, 29.14, 4.42, 34.67, 13.21

Info Tech: 4.77, 1.14, -12.72, 48.1, 6.69, 11.65, 53.28, -10.41, 50.88, 48.63, 24.67, 23.88, -14.8, 6.94, 29.52, 43.42, 23.56, -6.62, 45.55, 22.61

Explanation / Answer

please note that we can provide solutions using the open source statistical package R , the complete snippet is as follows

Health <- c( 10.29, 32.17, 34.27, 10.57, 20.27, 31.43, 15.4, 0.88, 17.85, 28.98, 6.49, 28.73, 8.17, 14.83, 12.89, 23.51, 29.14, 4.42, 34.67, 13.21)
InfoTech <- c( 4.77, 1.14, -12.72, 48.1, 6.69, 11.65, 53.28, -10.41, 50.88, 48.63, 24.67, 23.88, -14.8, 6.94, 29.52, 43.42, 23.56, -6.62, 45.55, 22.61)

## t test of the data

t.test(Health,InfoTech)

## confidence interval 95%

mean(Health) + qnorm(1-0.05/2)*sd(Health)/sqrt(length(Health))
mean(Health) - qnorm(1-0.05/2)*sd(Health)/sqrt(length(Health))

## confidence interval 95%

mean(InfoTech) + qnorm(1-0.05/2)*sd(InfoTech)/sqrt(length(InfoTech))
mean(InfoTech) - qnorm(1-0.05/2)*sd(InfoTech)/sqrt(length(InfoTech))

The results are

> t.test(Health,InfoTech)

   Welch Two Sample t-test

data: Health and InfoTech
t = -0.19975, df = 26.985, p-value = 0.8432 , as the p value is not less than 0.05 , hence we fail to reject thenull hypothesis and conclude that there is no difference in the mean values of health and Infotech
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-12.72068 10.46368
sample estimates:
mean of x mean of y
18.9085 20.0370

The confidence intervals are

> mean(Health) + qnorm(1-0.05/2)*sd(Health)/sqrt(length(Health))
[1] 23.61339
> mean(Health) - qnorm(1-0.05/2)*sd(Health)/sqrt(length(Health))
[1] 14.20361

range is 14.2 - 23.6

> mean(InfoTech) + qnorm(1-0.05/2)*sd(InfoTech)/sqrt(length(InfoTech))
[1] 30.06059
> mean(InfoTech) - qnorm(1-0.05/2)*sd(InfoTech)/sqrt(length(InfoTech))
[1] 10.01341

10.01 to 30.06

this means that we are 95% confident that the true value for health and infotech would lie in the respective ranges.

as the interval for Infotech covers the interval for Health , hence we can conclude that the true mean for both is not statistically different

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