A study to compare two insurance companies on length of stay for pediatric asthma patients randomly sampled 393 cases from Insurer A. An independent random sample of 396 cases from Insurer B gave the results on length of stay, and the summary statistics are show below. Insurer A (Group 1): x⎯⎯⎯x¯ = 2.31, ss = 1.22, nn = 393 Insurer B (Group 2): x⎯⎯⎯x¯ = 2.96, ss = 1.5, nn = 396 Unless otherwise stated, give your answers to three decimal places. Construct a 99% confidence interval for the difference in average length of stay between Insurer A and Insurer B. Give your answers to three decimal places. Use t∗=2.582t∗=2.582. ( , ) Conduct a hypothesis test to determine whether the average length of stay for Insurer A is different from than the average length of stay for Insurer B. What is the parameter of interest? i. pp ii. μμ iii. p1−p2p1−p2 iv. μ1−μ2μ1−μ2 v. μdμd What is the correct null value for this test? What sign should appear in the alternative hypothesis? i. == ii. << iii. >> iv. ≠≠ The test statistic for this test is t=t= . The p-value for this test is . Use df=757.809df=757.809. (If p-value <0.001<0.001, enter 0). Select the appropriate decision for this hypothesis test (use α=0.01α=0.01). i. Reject H0H0 ii. Fail to reject H0H0 iii. Accept H0H0 Select the appropriate interpretation for this hypothesis test (use α=0.01α=0.01). i. We have statistically significant evidence to suggest that the average length of stay for Insurer A is different from the average length of stay for Insurer B. ii. We have statistically significant evidence to suggest that the average length of stay for Insurer A is the same as the average length of stay for Insurer B. iii. We do not have statistically significant evidence to suggest that the average length of stay for Insurer A is different from the average length of stay for Insurer B. If an error had been made on this test, what kind of error would it have been? i. Type I error ii. Type II error

A study to compare two insurance companies on length of stay for pediatric asthma patients randomly sampled 393 cases from Insurer A. An independent random sample of 396 cases from Insurer B gave the results on length of stay, and the summary statistics are show below. Insurer A (Group 1): x⎯⎯⎯x¯ = 2.31, ss = 1.22, nn = 393 Insurer B (Group 2): x⎯⎯⎯x¯ = 2.96, ss = 1.5, nn = 396 Unless otherwise stated, give your answers to three decimal places. Construct a 99% confidence interval for the difference in average length of stay between Insurer A and Insurer B. Give your answers to three decimal places. Use t∗=2.582t∗=2.582. ( , ) Conduct a hypothesis test to determine whether the average length of stay for Insurer A is different from than the average length of stay for Insurer B. What is the parameter of interest? i. pp ii. μμ iii. p1−p2p1−p2 iv. μ1−μ2μ1−μ2 v. μdμd What is the correct null value for this test? What sign should appear in the alternative hypothesis? i. == ii. << iii. >> iv. ≠≠ The test statistic for this test is t=t= . The p-value for this test is . Use df=757.809df=757.809. (If p-value <0.001<0.001, enter 0). Select the appropriate decision for this hypothesis test (use α=0.01α=0.01). i. Reject H0H0 ii. Fail to reject H0H0 iii. Accept H0H0 Select the appropriate interpretation for this hypothesis test (use α=0.01α=0.01). i. We have statistically significant evidence to suggest that the average length of stay for Insurer A is different from the average length of stay for Insurer B. ii. We have statistically significant evidence to suggest that the average length of stay for Insurer A is the same as the average length of stay for Insurer B. iii. We do not have statistically significant evidence to suggest that the average length of stay for Insurer A is different from the average length of stay for Insurer B. If an error had been made on this test, what kind of error would it have been? i. Type I error ii. Type II error

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

Insurer A (Group 1): x⎯⎯⎯x¯ = 2.31, ss = 1.22, nn = 393
Insurer B (Group 2): x⎯⎯⎯x¯ = 2.96, ss = 1.5, nn = 396

Unless otherwise stated, give your answers to three decimal places.

Construct a 99% confidence interval for the difference in average length of stay between Insurer A and Insurer B. Give your answers to three decimal places. Use t∗=2.582t∗=2.582.
( , )
Conduct a hypothesis test to determine whether the average length of stay for Insurer A is different from than the average length of stay for Insurer B.
1. What is the parameter of interest?
  i. pp
  ii. μμ
  iii. p1−p2p1−p2
  iv. μ1−μ2μ1−μ2
  v. μdμd
2. What is the correct null value for this test?
3. What sign should appear in the alternative hypothesis?
  i. == ii. << iii. >> iv. ≠≠
4. The test statistic for this test is t=t= .
5. The p-value for this test is . Use df=757.809df=757.809. (If p-value <0.001<0.001, enter 0).
6. Select the appropriate decision for this hypothesis test (use α=0.01α=0.01).
  i. Reject H0H0 ii. Fail to reject H0H0 iii. Accept H0H0
7. Select the appropriate interpretation for this hypothesis test (use α=0.01α=0.01).
  i. We have statistically significant evidence to suggest that the average length of stay for Insurer A is different from the average length of stay for Insurer B.
  ii. We have statistically significant evidence to suggest that the average length of stay for Insurer A is the same as the average length of stay for Insurer B.
  iii. We do not have statistically significant evidence to suggest that the average length of stay for Insurer A is different from the average length of stay for Insurer B.
8. If an error had been made on this test, what kind of error would it have been?
  i. Type I error ii. Type II error

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