Assignment Sample on ACC01169 Quantitative Methods

Question 2)

  1. ii) Linear Correlation Coefficient
Obs Income (X1) Sales (Y) (Xi-Mean(X)) (Yi-Mean(Y)) (Xi-Mean(X))^2 (Yi-Mean(Y))^2 (Xi-Mean(X))*(Yi-Mean(Y))
1       2,450         162        (512)           11          262,144            121           (5,632)
2       3,254         120          292          (31)            85,264            961           (9,052)
3       3,802         223          840           72          705,600         5,184           60,480
4       2,838         131        (124)          (20)            15,376            400            2,480
5       2,347           67        (615)          (84)          378,225         7,056           51,660
6       3,782         169          820           18          672,400            324           14,760
7       3,008           81            46          (70)             2,116         4,900           (3,220)
8       2,450         192        (512)           41          262,144         1,681         (20,992)
9       2,137         116        (825)          (35)          680,625         1,225           28,875
10       2,560           55        (402)          (96)          161,604         9,216           38,592
11       4,020         252       1,058         101       1,119,364       10,201         106,858
12       4,427         232       1,465           81       2,146,225         6,561         118,665
13       2,660         144        (302)            (7)            91,204             49            2,114
14       2,088         103        (874)          (48)          763,876         2,304           41,952
15       2,605         212        (357)           61          127,449         3,721         (21,777)
Average       2,962         151    Total       7,473,616       53,904         405,763

 

Correlation Co-efficient =

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Correlation Co-efficient = 405,763 / Sqrt(7,473,616 * 53,904)

= 0.6393

iii) Calculate Regression Equation

SUMMARY OUTPUT                
                 
Regression Statistics                
Multiple R          0.64              
R Square          0.41              
Adjusted R Square          0.37              
Standard Error       47.71              
Observations       16.00              
                 
ANOVA                
  df SS MS F Significance F      
Regression 1   22,029.89   22,029.89          9.68          0.01      
Residual 14   31,871.71     2,276.55          
Total 15   53,901.60            
                 
  Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%
Intercept      (10.21)          53.05           (0.19)          0.85    (123.99)     103.58   (123.99)     103.58
Income (X1)          0.05             0.02             3.11          0.01          0.02          0.09          0.02          0.09

 

The Regression Equation

Sales (Y) = 0.054 (X1) – 10.21

  1. iv) The Regression Equation for the given data is Sales (Y) = -10.21 + 0.054 (X1). This states that for every increase of one unit of discretionary income, the sales is likely to go up by $ 54.

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Moreover, it is also found that the R square value stands at 0.41 and this means that the variable discretionary income is able to define the sale 41% of the total times.

 

Question 3)

 

Question 4)

A)

  1. i) The probability distribution of the sample is considered to be Normal distribution. The Central Limit Theorem states that as the sample size increases, then the distribution will tend to be normal. Any data for which the same size is greater than or equal to 30, then it is understood that it would follow Normal distribution.

Over here, it is believe that the probability distribution is Normal with mean of £ 40,000 and standard deviation of £ 5,000. The number of samples is 80.

  1. ii) Probability of the sample mean being over £ 41,000 is

P(X>Z) = P(X-Mean(X)/SD) = P((£ 41,000 – £ 40,000)/ £ 5,000)

= P(0.2 > Z) = 1 – P(Z<0.2) = 1- 0.5793 = 0.4207

The probability of the same mean being over £ 41,000 is 42.07 %

iii) Probability of the sample mean being below £ 39,000 is

P(X<Z) = P(X-Mean(X)/SD) = P((£ 39,000 – £ 40,000)/ £ 5,000)

= P(-0.2 > Z) = 0.4207

The probability of the same mean being less than £ 39,000 is 42.07 %

  1. iv) If the sample size is less than 20, then it means that there is no sufficient sample to consider the distribution of sample to be normally distributed. In such a scenario, we have to go with Poisson distribution or binomial distribution and thus the normal distribution cannot be applied.
  2. B) Two Sample Mean Test
  3. a) Over here, the sample size in both the samples is more than 30. Hence, we go with the Two Sample Mean Test using Z (Normal Distribution).
Samples Mean SD Sample Size
X1 31 7.6 33
X2 32.2 5.8 40

 

H0 : The means of two samples is considered to be same.

H1: The means of two sample is different.

Z = {(Mean 1 – Mean2) / sqrt (SD1^2/n1 + SD2^2/n2)}

Z = {32.2 – 31} / sqrt(1.751+0.841)

Z = 1.2/sqrt(2.592)

Z = 0.745

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