Question

1. The corporation believes that the population standard deviation of employee salaries is $18,100. How large of a sample would be required in order to estimate the mean salary at the 98% confidence level with an error of at most $150?

2. Construct a 97.5% confidence interval for population mean ?? of employee salaries.

3. An internal corporate report claims that the population mean salary is actually greater than $36,00. Can you verify this at ?? = .03. Construct an appropriate full hypothesis test.

ID	Salary
1	$57,000
2	$40,200
3	$21,450
4	$21,900
5	$45,000
6	$32,100
7	$36,000
8	$21,900
9	$37,900
10	$24,000
11	$30,300
12	$28,350
13	$27,750
14	$35,100
15	$27,300
16	$40,800
17	$46,000
18	$83,750
19	$42,300
20	$26,250
21	$38,850
22	$21,750
23	$24,000
24	$16,950
25	$21,150
26	$31,050
27	$60,375
28	$32,550
29	$85,000
30	$31,200

Answer 1

1.

we have given that

Population Standard Deviation =SD=18100

confidence level =% =98 %

Hence  

margin of error =E=150

Margin of error is given by

so n~79048

2.

from the given data we have

sample size =n=30

sample mean =m=36274

sample SD=S=16688

now 97.5% confidence interval is given by

so interval is (29071.37,43476.63)

3.

we have to test that

we have

from the given data we have

sample size =n=30

sample mean =m=36274

sample SD=S=16688

now test statistics is given by

now P value is given by

P-Value =p(t>0.0)=0.4645

since P value is more than level of significance hence we failed to reject H0 so we haven enough evidence to conclude that actual salary is more than 36000