Answer # 6 please
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1fe2c370-0169-4708-afcf-2bc9d4c62801%2F2eae7195-0516-4904-8166-9df3ae739342%2F7dsjsu_processed.jpeg&w=3840&q=75)
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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