Question 41 A company that owns department stores advertised for one month on te
ID: 3255938 • Letter: Q
Question
Question 41
A company that owns department stores advertised for one month on television and radio and in newspapers and spend the same amount of money on each of these three media. A random sample of 900 customers who visited this company's department stores during that month as a result of these promotional advertisements was asked which advertising medium was most effective in getting each of them to visit the stores. The responses of these customers are recorded in the following table.
Advertising Medium
Television
Radio
Newspaper
Medium that was most effective
330
295
275
Using 5% level of significance, we would like to test the null hypothesis that these three media sources have the same promotional impact on customers. The test and critical statistics are:
Select one:
A. 5.991 and 5.1667 respectively
B. 5.1667 and 7.815 respectively
C. 5.1667 and 5.991 respectively
D. 7.815 and 5.1667 respectively
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Question 42
Select one:
A. LCL=85 UCL=95
B. LCL=74 UCL=86
C. LCL=80 UCL=110
D. LCL=20 UCL=170
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Question 43
Select one:
A. [0.851 - 0.949]
B. [0.876 - 0.924]
C. [0.898 - 0.902]
D. [0.10 - 0.90]
Advertising Medium
Television
Radio
Newspaper
Medium that was most effective
330
295
275
Explanation / Answer
41)
applying chi square goodness of fit:
here test stat =5.1667
and for 2 degree of freedom and 5% level of significance ; critical value =5.991
therefore C. 5.1667 and 5.991 respectively
42)
here std error of mean =std deviation/(n)1/2 =5
hence control limits =mean -/+ 3*std error =80 ; 110
C. LCL=80 UCL=110
43)
here p=0.9 ; n=144
std error =(p(1-p)/n)1/2 =0.025
for 95% CI; z=1.96
hence 95% confidence interval =estimated proportion -/+ z*std error =0.851 ; 0.949
option A. [0.851 - 0.949]
observed Expected Chi square medium Probability O E=total*p =(O-E)^2/E television 1/3 330.000 300.00 3.00 Radio 1/3 295.000 300.00 0.08 Newspaper 1/3 275.000 300.00 2.08 1 900 900 5.1667Related Questions
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