# Given below are the analysis of variance results from a minitab

1. Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis.
A= 0.05 for a left-tailed test.

Find the value of the test statistic z using z = P-P/?
pq/n The claim is that the proportion of accidental

2. deaths of the elderly attributable to residential falls is more than 0.10, and the sample statistics include n = 800 deaths of the elderly with 15% of them attributable to residential falls.

3. Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis).
The test statistic in a right-tailed test is z = 1.43.

4. Find the number of successes x suggested by the given statement.
Among 720 people selected randomly from among the residents of one city, 13.89% were found to be living below the official poverty line.

Assume that you plan to use a significance level of alpha = 0.05 to test the claim that p1 = p2. Use the given sample sizes and numbers of successes to find the z test statistic for the hypothesis test.
A

5. random sampling of sixty pitchers from the National League and fifty-two pitchers from the American League showed that 19 National and 11 American League pitchers had E.R.A’s below 3.5.

6. Assume that you plan to use a significance level of alpha = 0.05 to test the claim that p1 = p2. Use the given sample sizes and numbers of successes to find the P-value for the hypothesis test.
n1 = 50; n2 = 50
x1 = 8; x2 = 7

7. Construct the indicated confidence interval for the difference between population proportions . Assume that the samples are independent and that they have been randomly selected.
In a random sample of 300 women, 49% favored stricter gun control legislation. In a random sample of 200 men, 28% favored stricter gun control legislation. Construct a 98% confidence interval for the difference between the population proportions p1 – p2.

8. Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Also assume that the population standard deviations are equal (sigma1 = sigma2), so that the standard error of the difference between means is obtained by pooling the sample variances. A researcher was interested in comparing the resting pulse rates of people who exercise regularly and people who do not exercise regularly. Independent simple random samples were obtained of 16 people who do not exercise regularly and 12 people who do exercise regularly. The resting pulse rate (in beats per minute) of each person was recorded. The summary statistics are as follows

Do not Exercise | Does Exercise

X1=73.4 beats/per min X2=69.7 beats/per min

S1=10.3 beats/per min S2=8.6 beats/ per min

N1=16 N2=12

Construct a 90% confidence interval for the difference between the mean pulse rate of people who do not exercise regularly and the mean pulse rate of people who exercise regularly.

9. The two data sets are dependent. Find d= to the nearest tenth x=11.2, 11.3,11.6,12.9, 10.6.

Y=11.1, 12.6, 12.9, 10.7, 13.3

10. Given the linear correlation coefficient r and the sample size n, determine the critical values of r and use your finding to state whether or not the given r represents a significant linear correlation. Use a significance level of 0.05.
r = 0.843, n = 5

11. Find the value of the linear correlation coefficient r.
The paired data below consist of the test scores of 6 randomly selected students and the number of hours they studied for the test. Hours= 5, 10, 4, 6, 10, 9…..Score=64,86,69,86,59,87

12. Suppose you will perform a test to determine whether there is sufficient evidence to support a claim of a linear correlation between two variables. Find the critical values of r given the number of pairs of data n and the significance level alpha.
n = 17, alpha = 0.05

13. Use the given data to find the best predicted value of the response variable.

Nine pairs of data yield r=0.867 and the regression equation y=19.4 + 0.93x. Also, y=64.7. What is the best predicted value of y for x=50?

14. 25,28,21,21,25,26,30,34,36 Dexterity 49, 53,59,42,47,53,55,63,67,75

Use the given data to find the equation of the regression line. Round the final values to three significant digits, if necessary.
Two different tests are designed to measure employee productivity and dexterity. Several employees are randomly selected and tested with these results. Productivity 23,

15. Use the given information to find the coefficient of determination.
A regression equation is obtained for a collection of paired data. It is found that the total variation is 130.3, the explained variation is 79.3, and the unexplained variation is 51. Find the coefficient of determination.

16. Find the explained variation for the paired data.

17. Find the unexplained variation for the paired data.

18. Find the total variation for the paired data.

19. Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean.
Identify the value of the test statistic.

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# Given below are the analysis of variance results from a minitab

1. Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis.
A= 0.05 for a left-tailed test.

Find the value of the test statistic z using z = P-P/?
pq/n The claim is that the proportion of accidental

2. deaths of the elderly attributable to residential falls is more than 0.10, and the sample statistics include n = 800 deaths of the elderly with 15% of them attributable to residential falls.

3. Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis).
The test statistic in a right-tailed test is z = 1.43.

4. Find the number of successes x suggested by the given statement.
Among 720 people selected randomly from among the residents of one city, 13.89% were found to be living below the official poverty line.

Assume that you plan to use a significance level of alpha = 0.05 to test the claim that p1 = p2. Use the given sample sizes and numbers of successes to find the z test statistic for the hypothesis test.
A

5. random sampling of sixty pitchers from the National League and fifty-two pitchers from the American League showed that 19 National and 11 American League pitchers had E.R.A’s below 3.5.

6. Assume that you plan to use a significance level of alpha = 0.05 to test the claim that p1 = p2. Use the given sample sizes and numbers of successes to find the P-value for the hypothesis test.
n1 = 50; n2 = 50
x1 = 8; x2 = 7

7. Construct the indicated confidence interval for the difference between population proportions . Assume that the samples are independent and that they have been randomly selected.
In a random sample of 300 women, 49% favored stricter gun control legislation. In a random sample of 200 men, 28% favored stricter gun control legislation. Construct a 98% confidence interval for the difference between the population proportions p1 – p2.

8. Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Also assume that the population standard deviations are equal (sigma1 = sigma2), so that the standard error of the difference between means is obtained by pooling the sample variances. A researcher was interested in comparing the resting pulse rates of people who exercise regularly and people who do not exercise regularly. Independent simple random samples were obtained of 16 people who do not exercise regularly and 12 people who do exercise regularly. The resting pulse rate (in beats per minute) of each person was recorded. The summary statistics are as follows

Do not Exercise | Does Exercise

X1=73.4 beats/per min X2=69.7 beats/per min

S1=10.3 beats/per min S2=8.6 beats/ per min

N1=16 N2=12

Construct a 90% confidence interval for the difference between the mean pulse rate of people who do not exercise regularly and the mean pulse rate of people who exercise regularly.

9. The two data sets are dependent. Find d= to the nearest tenth x=11.2, 11.3,11.6,12.9, 10.6.

Y=11.1, 12.6, 12.9, 10.7, 13.3

10. Given the linear correlation coefficient r and the sample size n, determine the critical values of r and use your finding to state whether or not the given r represents a significant linear correlation. Use a significance level of 0.05.
r = 0.843, n = 5

11. Find the value of the linear correlation coefficient r.
The paired data below consist of the test scores of 6 randomly selected students and the number of hours they studied for the test. Hours= 5, 10, 4, 6, 10, 9…..Score=64,86,69,86,59,87

12. Suppose you will perform a test to determine whether there is sufficient evidence to support a claim of a linear correlation between two variables. Find the critical values of r given the number of pairs of data n and the significance level alpha.
n = 17, alpha = 0.05

13. Use the given data to find the best predicted value of the response variable.

Nine pairs of data yield r=0.867 and the regression equation y=19.4 + 0.93x. Also, y=64.7. What is the best predicted value of y for x=50?

14. 25,28,21,21,25,26,30,34,36 Dexterity 49, 53,59,42,47,53,55,63,67,75

Use the given data to find the equation of the regression line. Round the final values to three significant digits, if necessary.
Two different tests are designed to measure employee productivity and dexterity. Several employees are randomly selected and tested with these results. Productivity 23,

15. Use the given information to find the coefficient of determination.
A regression equation is obtained for a collection of paired data. It is found that the total variation is 130.3, the explained variation is 79.3, and the unexplained variation is 51. Find the coefficient of determination.

16. Find the explained variation for the paired data.

17. Find the unexplained variation for the paired data.

18. Find the total variation for the paired data.

19. Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean.
Identify the value of the test statistic.

Pages (550 words)
Approximate price: -

Why Work with Us

Top Quality and Well-Researched Papers

We always make sure that writers follow all your instructions precisely. You can choose your academic level: high school, college/university or professional, and we will assign a writer who has a respective degree.

We have a team of professional writers with experience in academic and business writing. Many are native speakers and able to perform any task for which you need help.

Free Unlimited Revisions

If you think we missed something, send your order for a free revision. You have 10 days to submit the order for review after you have received the final document. You can do this yourself after logging into your personal account or by contacting our support.

Prompt Delivery and 100% Money-Back-Guarantee

All papers are always delivered on time. In case we need more time to master your paper, we may contact you regarding the deadline extension. In case you cannot provide us with more time, a 100% refund is guaranteed.

Original & Confidential

We use several writing tools checks to ensure that all documents you receive are free from plagiarism. Our editors carefully review all quotations in the text. We also promise maximum confidentiality in all of our services.

Our support agents are available 24 hours a day 7 days a week and committed to providing you with the best customer experience. Get in touch whenever you need any assistance.

Try it now!

## Calculate the price of your order

Total price:
\$0.00

How it works?

Fill in the order form and provide all details of your assignment.

Proceed with the payment

Choose the payment system that suits you most.

Our Services

No need to work on your paper at night. Sleep tight, we will cover your back. We offer all kinds of writing services.

## Essay Writing Service

No matter what kind of academic paper you need and how urgent you need it, you are welcome to choose your academic level and the type of your paper at an affordable price. We take care of all your paper needs and give a 24/7 customer care support system.