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Answer # 6 please

Answer # 6 please

MATLAB: An Introduction with Applications
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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Answer # 6 please

pm 13.20 1.3.20 2 stat: x-p/o/ 2,20-13,20/2,50/40 2.52982 Part 2. Mean 4. The hourly wages in a particular industry are normally distributed with mean $13.20 and standard deviation $2.50. A company in this industry employs 40 workers, paying them an average of $12.20 per hour. Can this company be accused of paying substandard wages? Use an a = .01 level test. 0005706 Popl, mean hourly wages in particular industry (u) = $13.20 Popl, std, deviation of hourly wages in a particular industry (0) = $2.50 Randomly selected no, of workers (n) = 40 alue: ce alt, hypo, has sign, so it is left tailed test alue = P(Z & Zstat) = P(Z < -2.52982) ng excel command, NORMSDIST (2 score) AI & NORMSDIST(-2.52902) Sample mean hourly wages (x) = $12.20 Level of signif. (a) = 0,01 1: u > 15 That is, there is evidence to conclude that the sales eople are adding more than 5 sales contacts per week 6. 50, p value = 0,005706 5. A vice president in charge of sales for a large corporation claims that salespeople are averaging no more than 15 sales contacts per week. (He would like to increase this figure.) As a check on his claim, n = 36 salespeople are selected at random, and the number of contacts made by each is recorded for a single randomly selected week. The mean and variance of the 36 measurements were 17 and 9, respectively. Does the evidence contradict the vice president's claim? Use a test with level a = .05. at is, there is no evidence N = 36, X = 17, S = 9 conclude that the sales zople are averaging more man 15 sales contacts per P: u 15 week T= x-u/s/n = 17-15/ 3/36 = 2/0,50 df= n-1 Decision rule; If p value level of signif (a) then reject Ho otherwise fail to reject Ho Here, p value < a Hence, reject Ho = 35-1 = 34 Conclusion: There is suffi. evid, to support that the company can be accused of paying substd. wages = 4,00 Thus, the value of the test statistic (t) is 4.00. Since the P value (0,002) is Less than (<) the significance level (0.05) therefore, by the rejection rule, we reject the null hypothesis. That there is evidence to conclude that the sales people are averaging more than 15 sales contacts per week. Yes there's evidence to contradict the vice presidents claim. Listed below are brain volumes (in centimeters cubed) of unrelated subjects used in a study. Use a 0.05 significance level to test the claim that the population of brain volumes has a mean equal to 1100.8 centimeters cubed. 963 1027 1271 1078 1070 There is suffi. evid, to support that the company can be accused of paying substd. wages The level of significance is 0,05. From the XL output, the P value is 0.0002. Thus, the P value is 0.002. 1174 1066 1348 1099 1204
Transcribed Image Text:pm 13.20 1.3.20 2 stat: x-p/o/ 2,20-13,20/2,50/40 2.52982 Part 2. Mean 4. The hourly wages in a particular industry are normally distributed with mean $13.20 and standard deviation $2.50. A company in this industry employs 40 workers, paying them an average of $12.20 per hour. Can this company be accused of paying substandard wages? Use an a = .01 level test. 0005706 Popl, mean hourly wages in particular industry (u) = $13.20 Popl, std, deviation of hourly wages in a particular industry (0) = $2.50 Randomly selected no, of workers (n) = 40 alue: ce alt, hypo, has sign, so it is left tailed test alue = P(Z & Zstat) = P(Z < -2.52982) ng excel command, NORMSDIST (2 score) AI & NORMSDIST(-2.52902) Sample mean hourly wages (x) = $12.20 Level of signif. (a) = 0,01 1: u > 15 That is, there is evidence to conclude that the sales eople are adding more than 5 sales contacts per week 6. 50, p value = 0,005706 5. A vice president in charge of sales for a large corporation claims that salespeople are averaging no more than 15 sales contacts per week. (He would like to increase this figure.) As a check on his claim, n = 36 salespeople are selected at random, and the number of contacts made by each is recorded for a single randomly selected week. The mean and variance of the 36 measurements were 17 and 9, respectively. Does the evidence contradict the vice president's claim? Use a test with level a = .05. at is, there is no evidence N = 36, X = 17, S = 9 conclude that the sales zople are averaging more man 15 sales contacts per P: u 15 week T= x-u/s/n = 17-15/ 3/36 = 2/0,50 df= n-1 Decision rule; If p value level of signif (a) then reject Ho otherwise fail to reject Ho Here, p value < a Hence, reject Ho = 35-1 = 34 Conclusion: There is suffi. evid, to support that the company can be accused of paying substd. wages = 4,00 Thus, the value of the test statistic (t) is 4.00. Since the P value (0,002) is Less than (<) the significance level (0.05) therefore, by the rejection rule, we reject the null hypothesis. That there is evidence to conclude that the sales people are averaging more than 15 sales contacts per week. Yes there's evidence to contradict the vice presidents claim. Listed below are brain volumes (in centimeters cubed) of unrelated subjects used in a study. Use a 0.05 significance level to test the claim that the population of brain volumes has a mean equal to 1100.8 centimeters cubed. 963 1027 1271 1078 1070 There is suffi. evid, to support that the company can be accused of paying substd. wages The level of significance is 0,05. From the XL output, the P value is 0.0002. Thus, the P value is 0.002. 1174 1066 1348 1099 1204
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