may developed this program supplied materials and teacher training for a large-scale test involving nearly 9,600 employees in several different sites. Statistical analysis of the results showed that the percentage of employees who did not pass the course was reduced from 15.2% to 14.6%. The hypothesis that the program produced no improvement was rejected with a P-value of 0.028. Complete parts a) and b) below. a) Explain what the P-value means in this context. Choose the correct answer below. O A. There is only a 2.8% chance of seeing a sample proportion of 15.2% (or more) of employees failing the course by natural sampling variation if 14.6% is the true population value. OB. There is only a 2.8% chance of seeing a sample proportion of 14.6% (or less) of employees failing the course by natural sampling variation if 15.2% is the true population value. OC. There is only a 97.2% chance of seeing a sample proportion of 14.6% (or less) of employees failing the course by natural sampling variation if 15.2% is the true population value. O D. There is a 97.2% chance of seeing a sample proportion of 15.2% (or more) of employees failing the course by natural sampling variation if 14.6% is the true population value. b) Even though this program has been shown to be significantly better statistically, why might you not recommend that your company adopt it? O A. You might not recommend that your company adopt the new program because the sample was not chosen at random, thus compromising the results of the O B. You might not recommend that your company adopt the new program because the sample size of the test was not large enough to come to a decision. OC. Under the old methods, 1,402 employees would be expected to fail. With the new program, 1,459 failed. The company would not want to adopt a new program that increases the number of people who fail. OD. Under the old methods, 1,459 employees would be expected to fail. With the new program, 1,402 failed. This is only a decrease of 57 employees. It would depend on the costs of switching to the new program
may developed this program supplied materials and teacher training for a large-scale test involving nearly 9,600 employees in several different sites. Statistical analysis of the results showed that the percentage of employees who did not pass the course was reduced from 15.2% to 14.6%. The hypothesis that the program produced no improvement was rejected with a P-value of 0.028. Complete parts a) and b) below. a) Explain what the P-value means in this context. Choose the correct answer below. O A. There is only a 2.8% chance of seeing a sample proportion of 15.2% (or more) of employees failing the course by natural sampling variation if 14.6% is the true population value. OB. There is only a 2.8% chance of seeing a sample proportion of 14.6% (or less) of employees failing the course by natural sampling variation if 15.2% is the true population value. OC. There is only a 97.2% chance of seeing a sample proportion of 14.6% (or less) of employees failing the course by natural sampling variation if 15.2% is the true population value. O D. There is a 97.2% chance of seeing a sample proportion of 15.2% (or more) of employees failing the course by natural sampling variation if 14.6% is the true population value. b) Even though this program has been shown to be significantly better statistically, why might you not recommend that your company adopt it? O A. You might not recommend that your company adopt the new program because the sample was not chosen at random, thus compromising the results of the O B. You might not recommend that your company adopt the new program because the sample size of the test was not large enough to come to a decision. OC. Under the old methods, 1,402 employees would be expected to fail. With the new program, 1,459 failed. The company would not want to adopt a new program that increases the number of people who fail. OD. Under the old methods, 1,459 employees would be expected to fail. With the new program, 1,402 failed. This is only a decrease of 57 employees. It would depend on the costs of switching to the new program
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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A new program may reduce the proportion of employees who fail a computer skills course. The company that developed this program supplied materials and teacher training for a large-scale test involving nearly 9,600 employees in several different sites. Statistical analysis of the results showed that the percentage of employees who did not pass the course was reduced from 15.2% to 14.6%. The hypothesis that the program produced no improvement was rejected with a P-value of 0.028. Complete parts a) and b) below.
---
a) Explain what the P-value means in this context. Choose the correct answer below.
- **A.** There is only a 2.8% chance of seeing a sample proportion of 15.2% (or more) of employees failing the course by natural sampling variation if 14.6% is the true population value.
- **B.** There is only a 2.8% chance of seeing a sample proportion of 14.6% (or less) of employees failing the course by natural sampling variation if 15.2% is the true population value.
- **C.** There is only a 97.2% chance of seeing a sample proportion of 14.6% (or less) of employees failing the course by natural sampling variation if 15.2% is the true population value.
- **D.** There is a 97.2% chance of seeing a sample proportion of 15.2% (or more) of employees failing the course by natural sampling variation if 14.6% is the true population value.
b) Even though this program has been shown to be significantly better statistically, why might you not recommend that your company adopt it?
- **A.** You might not recommend that your company adopt the new program because the sample was not chosen at random, thus compromising the results of the test.
- **B.** You might not recommend that your company adopt the new program because the sample size of the test was not large enough to come to a decision.
- **C.** Under the old methods, 1,402 employees would be expected to fail. With the new program, 1,459 failed. The company would not want to adopt a new program that increases the number of people who fail.
- **D.** Under the old methods, 1,459 employees would be expected to fail. With the
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