Essay Title: 

Quantatitive Studies

March 24, 2016 | Author: | Posted in health and medicine

PART A ) and Standard Deviation (S ) of a sample are given by where Xi is the individual member of the sample

and n is number of elements in the sample Using these formula for given sample on sales we get

1 . Mean value of sales 1225 .1364

Standard deviation of sales sample 345 .5701

2 . Radio Ads sample mean value and standard deviation 43 .1818 19 .7933

3 . News Ads sample mean value and standard deviation 30 .0000 11 .7514 (b ) Calculating median for samples on Sales , Radio Ads and News Ads 1148 [banner_entry_middle]

45 30

Comparing mean and median values for different samples it can be concluded that

Sales data sample is skewed right (positive ) and Radio Ads data sample is skewed left (negative , while the News Ads data sample is symmetrical or zero skewed

For Sales sample Sk 0 .6696

For Radio Ads sample Sk -0 .2756

Therefore , Sales data sample is more skewed

In the sample on sales

No . of cities with sales figure of 1500 or more x 6

Sample size n 22 0 .2727 0 .0949 8 .7775 This means , that with 95 confidence we can state the mean of the Radio Ads lies in the range given by [34 .4043 , 51 .9593] 7 .0903 This means , that with 99 confidence we can state the mean of the News Ads lies in the range given by [22 .9097 , 37 .0903] (iii ) In the sample on sales

No . of cities with sales figure of 1500 or more x 6

Sample size n 22 0 .2727 0 .0949

0 .1632 Required confidence level (90 ) 1- 0 .90

For 90 confidence level z 1 .645 Squaring both sides and rearranging the terms gives The value of sample size depends on the value of , but we do not have any previous point estimate value . Therefore , the worst case scenario should be assumed . The worst case scenario is the one in which value of standard deviation ?p is the maximum and this corresponds to

0 .5 Using this value in the above formula , the required sample size is Therefore , the minimum sample size should be 68 (f ) To test this claim For News Ads sample 30 S 11 .7514 n 22

As population is normal ? is unknown and n 30 , therefore , t distribution

Degrees of freedom df n-1 21

Confidence level 1- 0 .95 , therefore 0 .05 t .05 , 21df 2 .08 As , ttest t .05 , 21df (figure above ) therefore , the Null Hypothesis (H0 ) is accepted and the claim of the advertising companies about average news ads or the Alternative Hypothesis (H1 ) is rejected (g ) To test this claim For Sales sample

p 0 .2727 S 0 .0949 n 22 ? is unknown and n t .05 , 21df (figure above ) therefore , the Null Hypothesis (H0 ) is accepted and the claim of the `Top Gun Adverts ‘ magazine or the Alternative Hypothesis (H1 ) is rejected (h ) To test this claim For News Ads sample 43 .1818 S 19 .7933 n 22 ? is unknown… [banner_entry_footer]

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