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Monday 12 September 2016

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DRM02

Hypothesis Testing
   
Assignment

Assignment Code: 2016DRM02A1                                        Last Date of Submission: 26th May 2016
                                                                                       Maximum Marks 100


SECTION – A

(Each question is of 25 marks)

Q1. a)     For the purpose of conducting a test of difference of two means, explain the levels of     measurement for the different variables that are involved.
       b) A retailer that sells home entertainment systems accumulated 10451 sales invoices     during     the previous year. The total of the sales amount on these invoices as claimed by     the company is Rs. 63,84,675. In order to estimate the true total sales for last year, an     independent auditor randomly selects 350 of the invoices and determines the actual     sales amounts by contacting the purchasers. The mean and standard deviation of the     350 samples sales amounts are Rs.532 and standard deviation 168. Find the point and     interval estimate given that a 95% confidence is required. Do you think that the amount     of invoices claimed by the accounts department is correct?

Q2.      FedEx and UPS are the world’s two leading cargo carriers by volume and revenue.     According to the Airports Council International, the Memphis International Airport     (FedEx) and Louisville International airport (UPS) are two of the ten largest cargo     airports in the world. A sample of cargo handled in the two airport were recorded and     based on the same can we conclude that there is a difference in the cargo handled by     the two airports.

Memphis    9.1    15.1    8.8    10    7.5    10.5    8.3    9.1    6    5.8    12.1    9.3
Louisville     4.7    5.0    4.2    3.3    5.5    2.2    4.1    2.6    3.4    7.0    6.3    5.8
    Use 5% level of significance to verify the above question.

SECTION – B
Mr. LePew, the advertising manager of a medium sized manufacturer of rug and room deodorizers, has developed three preliminary advertising campaigns for the company’s line of deodorizers. The three campaigns are tested in an independent sample of 24 cities across the US and the sales in each city are monitoried.( note: cities are randomly assigned to each campaign and cities are comparable in terms of various socioeconomic and demographic variables)
    Sales in the cities
Adv 1    10    6    8    12    6    8    9    7
Adv 2    9    7    6    10    6    4    5    5
Adv 3    12    10    8    13    11    10    9    7

The data for the above were analyzed for basic descriptive statistics and the results are given below:
                

Figure 1


    Table 1        Tests of Normality

     type of ad campaign    Kolmogorov-Smirnov(a)    Shapiro-Wilk
          Statistic    df    Sig.    Statistic    df    Sig.
sales    adv1    .173    8    .200(*)    .933    8    .542
     adv2    .220    8    .200(*)    .917    8    .408
     adv3    .125    8    .200(*)    .983    8    .975
*  This is a lower bound of the true significance.
a  Lilliefors Significance Correction

    Descriptives  Table 2

     type of ad campaign         Statistic    Std. Error
sales    adv1    Mean    8.2500    .72580
          Median    8.0000   
          Variance    4.214   
          Skewness    .743    .752
          Kurtosis    .142    1.481
     adv2    Mean    6.5000    .73193
          Median    6.0000   
          Variance    4.286   
          Skewness    .773    .752
          Kurtosis    -.448    1.481
     adv3    Mean    10.0000    .70711
          Median    10.0000   
          Skewness    .000    .752
          Kurtosis    -.700    1.481
   
                                      Table 3
Levene Statistic    df1    df2    Sig.
.023    2    21    .978

    Multiple Comparisons

Dependent Variable: sales
Tukey HSD                                           Table 4
(I) type of ad campaign    (J) type of ad campaign    Mean Difference (I-J)    Std. Error    Sig.    95% Confidence Interval
                         Lower Bound    Upper Bound
adv1    adv2    1.75000    1.02062    .223    -.8225    4.3225
     adv3    -1.75000    1.02062    .223    -4.3225    .8225
adv2    adv1    -1.75000    1.02062    .223    -4.3225    .8225
     adv3    -3.50000(*)    1.02062    .007    -6.0725    -.9275
adv3    adv1    1.75000    1.02062    .223    -.8225    4.3225
     adv2    3.50000(*)    1.02062    .007    .9275    6.0725
*  The mean difference is significant at the .05 level.

Case Questions:
a)     State the null and alternative hypothesis for the above problem.
b)      Carry out the appropriate hypothesis at 5% level of significance. ( table value  = 3.47)
c)      State all the assumptions for carrying out the above test. Verify based on the output     given in the question whether each of the assumptions are satisfied and justify your     answer.
d)      Based on the data given in table 4 as well as that obtained by you in part (b) above ,     interpret the complete result and state your recommendation.  



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