ASS-2(1-4)

7/29/2019 ASS-2(1-4)

1/8

Answer to the Ques. No-1

A) 99% Confidence interval for the mean annual wages of the employees.

One-Sample Statistics

51 30886.55 17091.2236 2393.2500Annual wage in dollars

N Mean Std. Deviation

Std. Error

Mean

One-Sample Test

12.906 50 .000 30886.549 24477.92 37295.18Annual wage in dol larst df Sig. (2-tailed)

Mean

Difference Lower Upper

99% Confidence

Interval of the

Difference

Test Value = 0

Comment: Here we can say that the employees mean annual wage liesbetween $ 24477.92 to $ 37295.18 and we are 95% confident about it.

b) 95% Confidence interval for the mean work experience of the employees.


51 22.0196 13.5403 1.8960Years of work experience

N Mean Std. DeviationStd. Error

Mean

One-Sample Test

11.614 50 .000 22.0196 18.2113 25.8279Years of work experiencet df Sig. (2-tailed)

Mean


95% Confidence

Interval of the

Difference

Test Value = 0

Comment: Here we can say that the employees mean work experience liesbetween $ 18.2113 to 25.8279 and we are 99% confident about it.

7/29/2019 ASS-2(1-4)

2/8


a)

Step-1

H0 : 40000H1 : >40000

Step-2 =.01

Step-3

t-statistics will be used.

Step-4P< = reject HoP> = accept Ho

Step-5


51 30886.55 17091.2236 2393.2500Annual wage in dollarsN Mean Std. Deviation

Std. Error

Mean

One-Sample Test

-3.808 50 .000 -9113.4510 -15522.1 -2704.82Annual wage in dollarst df Sig. (2-tailed)

Mean


99% Confidence

Interval of the

Difference

Test Value = 40000

Here P value is 0.000/2 = 0.000

Decision: Since P value (0.00) < value (0.01); So, Ho is rejected & H 1 isaccepted. That is > 40000. At the 0.01 level of significance we canconclude that mean wage of employer in the city is more than 40000.

7/29/2019 ASS-2(1-4)

3/8

b)

Step-1

H0 : 12H1 : = accept Ho

Step-5


51 12.5294 2.7666 .3874Years of education


Std. Error

Mean

One-Sample Test

1.367 50 .178 .5294 -.5080 1.5668Years of educationt df Sig. (2-tailed)

Mean


99% Confidence

Interval of the

Difference

Test Value = 12

Here P value is 0.178/2 = 0.089

Decision: Since P value (0.089) is > Value (0.05); So, Ho is accepted & H1is rejected. That is 12. At the 0.05 level of significance we can concludethat mean experience of employer is not less than 12 years.

7/29/2019 ASS-2(1-4)

4/8

Answer to the Ques. No-3a)

Step-1H0 : m= fH1 : m f

Step-2 =.05

Step-3t-statistics will be used.Step-4

P< = reject HoP> = accept Ho

Step-5

Group Statistics

28 35260.14 17977.6378 3397.454223 25562.17 14603.2367 3044.9853

Sex of respondent

MaleFemale

Annual wage in dollars


Std. Error

Mean

I n d e p e n d e n t S a m p l e s T e s t

2 .0 7 2 . 1 5 6 2 . 0 8 3 4 9 .0 4 3 9 6 9 7 .9 6 8 94 6 5 6 . 7 8 3 23 3 9 . 8 1 2 81 9 0 5 6 . 1 3

2 . 1 2 6 4 8 . 9 9 8 .0 3 9 9 6 9 7 .9 6 8 94 5 6 2 . 3 0 5 45 2 9 . 6 6 1 31 8 8 6 6 . 2 8

E q u a l v a r ia n c e s

a s s u m e d


n o t a s s u m e d

A n n u a l w a g e in d o ll a r sF S ig .

L e v e n e ' s T e s t fo r

E q u a l it y o f V a r ia n c e s

t d f S ig . ( 2 - t a ile d )

M e a n

D i f f e r e n c e

S t d . E r r o r

D i f f e r e n c e L o w e r U p p e r

9 5 % C o n f id e n c

In t e r v a l o f t h e

D i f f e r e n c e

t - te s t f o r E q u a l it y o f M e a n s

7/29/2019 ASS-2(1-4)

5/8

Decision:

Since P value (0.156) of F statistic is greater than value (0.05); so, weassumed equal variance. In equal variance we found that p value is 0.156and it is greater than value. So Ho is accepted & H1 is rejected. That meansm = f.

At 0.05 level of confidence we can conclude that there is a no difference inmean wage among male & female employer.

b)

Step-1

H0 : m= u

H1 : m u Here, 1 = Married2 = Unmarried

Step-2

=.01

Step-3


Step-4P< = reject HoP> = accept Ho

Step-5

Group Statistics

17 28590.82 15752.1050 3820.4466

34 32034.41 17838.5361 3059.2837

Marital status

Unmarried

Married



Std. Error

Mean

7/29/2019 ASS-2(1-4)

6/8


. 9 9 3 . 3 2 4 - . 6 7 5 4 9 .5 0 3 - 3 4 4 3 . 5 8 8 25 1 0 4 . 7 4 3 6- 1 3 7 0 2 . 06 8 1 4 . 7 7 8

- . 7 0 4 3 5 . 9 3 4 .4 8 6 - 3 4 4 3 . 5 8 8 24 8 9 4 . 3 8 7 5- 1 3 3 7 0 . 56 4 8 3 . 3 2 0


a s s u m e d


n o t a s s u m e d

A n n u a l w a g e in d o ll a r s

F S ig .


E q u a l i t y o f V a r i a n c e s

t d f S ig . ( 2 - t a i l e d )M e a n

D i f f e r e n c eS t d . E r r o r D i f f e r e n c eL o w e r U p p e r

9 5 % C o n f id e n c


D i f f e r e n c e

t - te s t f o r E q u a l i t y o f M e a n s

Decision:

Since P value (0.324) of F statistic is greater than value (0.01); so, weassumed equal variance. In equal variance we found that p value is 0.324and it is greater than value. So, Ho is accepted. That means 1 = 2.

At 0.01 level of confidence we can conclude that there exists no differencein mean wage among married & unmarried employer.


a)

Step-1

H0 : m2= f

2 here, f = femaleH1 : m

2 f2 m = male

Step-2

=.01

Step-3

t- statistics will be used.

Step-4


7/29/2019 ASS-2(1-4)

7/8

Step-5

Group Statistics

28 35260.14 17977.6378 3397.4542

23 25562.17 14603.2367 3044.9853

Sex of respondent

Male

Female


N Mean Std. DeviationStd. Error

Mean


2 .0 7 2 .1 5 6 2 . 0 8 3 4 9 . 0 4 3 9 6 9 7 . 9 6 8 94 6 5 6 . 7 8 3 2- 2 7 8 1 . 9 92 2 1 7 7 . 9 2

2 . 1 2 6 4 8 . 9 9 8 . 0 3 9 9 6 9 7 .9 6 8 94 5 6 2 . 3 0 5 4- 2 5 2 8 . 8 22 1 9 2 4 . 7 5

E q u a l va r ia n c e s

a s s u m e d

E q u a l va r ia n c e s

n o t a s s u m e d

A n n u a l w a g e in d o ll a r sF S ig .


E q u a l it y o f V a r ia n c e s

t d f S ig . ( 2 - t a ile d )

M e a n

D i f f e r e n c e

S t d . E r r o r

D i f f e r e n c e L o w e r U p p e r

9 9 % C o n f id e n c e


D i f f e r e n c e

t - te s t f o r E q u a l it y o f M e a n s

Decision:Since P value (0.156) of F statistic is greater than value (0.01); So, Ho isaccepted. That means 2f= 2m.At the 0.01 significance level, the annual wage does not varies for thevariable sex of the respondent.

7/29/2019 ASS-2(1-4)

8/8

b)

Step-1

Ho: p = se = c = m = sa = o Here, p = ProfessionalH1: Not all the occupation means are equal se = Service

c = Clearicalm = Managementsa = Sales

o = Others

Step-2

=.05

Step-3


Step-4


Step-5

ANOVA


3.20E+09 5 640097392.0 2.526 .042

1.14E+10 45 253444651.1

1.46E+10 50

Between Groups

Within Groups

Total

Sum of

Squares df Mean Square F Sig.

Decision:

Since, P value (0.042) < value (0.05); So, Ho is rejected.

ASS-2(1-4)

Documents

Transcript of ASS-2(1-4)