Vishwesh QMB

7/30/2019 Vishwesh QMB

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QMB Assignment

IIM RANCHI

PGEXP-2012-14 Batch

Name : - Vishwesh Koundilya

Roll No : - PGEXP/004/2012

Solution- 7-55

X= the life span of the employee in southern textileIt is given the total life span varies between 550 weeks.

6 = 550 = 550/5= 91.66

Also the confidence level is 98%. = .02 /2 = 0.01 then z= 2.33

The required equation isn> (Z /2 X /B) > (2.33 X 91.67 / 30) 7.120 50.69

So, the sample size should be minimum 51.

Solution 7-69

X = Average no. of hours television watched in a week by a household = 1.1 hour, B = 0.3 hours

CI = 98%, so = 0.02 , So /2 = 0.01

So, n (Z /2 X /B) ( 2.33 X 1.1 / 0.3) 8.543 72.98

So, minimum sample size should be 73 for Harris Polls

Solution 8-25

a) Dr. Ross would be interesting to know that the heart beat of smoker is more than 78 beats permin during the day time compare to the non smoker which has the average heart beat 78 per min.

The hypotheses are

H0 : = 78 beats per min

H1 : > 78 beats per min


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b) Dr. Ross would be interesting to know that the average number of cigarettes consumed by thesmoker is more than the 15 cigarette if he or she switches to another brand which is having low

nicotine content.

The hypotheses are

H0 : = 15 high-nicotine cigarettes per day

H1 : > 15 cigarettes per day

Solution 8-29

a) X = Average weekly hot chocolate sales after the recent year long coffee advertisementcampaign.

Null Hypothesis is H0 : = 984.7 pounds

Alternate hypothesis is H1 : < 984.7 pounds

Reject H0 , if x < SV

b) = 72.6 , n = 30, x = 912.1 , = 0.02

S V = - Z X / n = 984.7 2.06 X 72.6 / 30 = 957.39

As x, i.e. 912.1 is less than 957.39, H0 is rejected.

That means average weekly sales has come down at significance level of 2%.

Solution SC12-6

Y X

10 11

12 15

8 3

15 18

9 10

11 12

8 6

10 7

13 18

11 13


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SUMMARY OUTPUT

Regression Statistics

Multiple R 0.924433

R Square 0.854576Adjusted R

Square 0.836398StandardError 0.89535Observations 10

ANOVA

df SS MS FSignifican

ce F

Regression 137.6867

937.6867

947.0114

7 0.00013

Residual 86.41320

80.80165

1

Total 9 44.1

Coefficients

Standard Error t Stat P-value

Lower95%

Upper95%

Lower95.0%

Upper95.0%

Intercept 6.0660420.73276

18.27834

13.41E-

05 4.3762937.75579

14.37629

37.75579

1X Variable1 0.410085 0.05981

6.856491 0.00013 0.272163

0.548006

0.272163

0.548006

Now, B = 0.4101

To test beta at significance level of 0.05.

H0 : B = 1

Alternate hypothesis is H1 : B < 1

Reject H0 , if B > t

t = (b- BH0)/ Sb = (0.4101 1)/ .0598 = -9.864

Degree of freedom for residue is n-k-1 = 10-1-1 = 8

Standard t value is -1.860, which is GREATER than -9.864.

So, we reject H0, i.e. beta of stock is significantly less than 1.

Stock is insensitive to changes in the market.


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Solution: - 13-48.

price wt sq ft

225 94 37

240 90 36

225 112 35220 92 40

167 93 48

212 98 40

217 114 40

200 108 35

SUMMARY OUTPUT

Regression Statistics

Multiple R 0.807524

R Square 0.652095

Adjusted RSquare 0.512933StandardError 15.30732Observations 8

ANOVA

df SS MS F

Significan

ce F

Regression 2 2195.931097.96

5 4.68587 0.071392

Residual 5 1171.57 234.314

Total 7 3367.5

Coefficients

Standard Error t Stat P-value

Lower95%

Upper95%

Lower95.0%

Upper95.0%

Intercept 444.718397.4120

64.56533

10.00602

8 194.3126 695.124194.312

6 695.124X Variable1 -0.61239

0.633392

-0.96684

0.378028 -2.24058

1.015795

-2.24058

1.015795

X Variable2 -4.37692

1.429979

-3.06083

0.028076 -8.0528

-0.70104 -8.0528

-0.70104

The least squares equation would be

Price = 441.718 0.6124(Weight) 4.377(Sq ft).


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b) Weight = 100 ounces , space= 46 sq ft , what will be price .

Putting values in above equation , we get

Price = 441.718 0.6124 (100) 4.377 (46)

= 301.616 .

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