Application Exercise: A national organization ran an advertising campaign design

Question

Application Exercise:
A national organization ran an advertising campaign designed to impact smoking. To evaluate their campaign, they had subjects record the average number of cigarettes smoked per day in the week before and the week after exposure to the ad. What can be concluded with ? = 0.05. The data are below.

a) What is the appropriate test statistic?
---Select--- na z-test One-Sample t-test Independent-Samples t-test Related-Samples t-test

b)
Condition 1:
---Select--- advertising campaign after smoking before cigarettes
Condition 2:
---Select--- advertising campaign after smoking before cigarettes

c) Compute the appropriate test statistic(s) to make a decision about H0.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value =  ; test statistic =  
Decision:  ---Select--- Reject H0 Fail to reject H0

d) If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.
[  ,  ]

e) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and/or select "na" below.
d =  ;   ---Select--- na trivial effect small effect medium effect large effect
r2 =  ;   ---Select--- na trivial effect small effect medium effect large effect

f) Make an interpretation based on the results.

-There was a significant decrease in cigarettes smoked after the ad

-.There was a significant increase in cigarettes smoked after the ad.    

-There was no significant difference between the number of cigarettes smoked before and after the ad.

Explanation / Answer

a) Here population standard deviation is unknown.
Thus we use t-test for equality of two means.

b)Here let
mu1=average number of cigaretts smoked before ad.
mu2=average number of cigaretts smoked after ad.

c) We have to test,
H : mu1=mu2
H : mu1>mu2 or mu1<mu2

The R-code for that is:

x=c(35,17,18,29,18,19,27,26,16,41)
y=c(46,17,21,34,31,20,34,26,27,41)
t.test(x,y)
qt(0.025,18) # tcritical

The output is :

> t.test(x,y)

Welch Two Sample t-test

data: x and y
t = -1.2757, df = 17.855, p-value = 0.2184
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-13.504207 3.304207
sample estimates:
mean of x mean of y
24.6 29.7

> qt(0.025,18) # tcritical
[1] -2.100922

Here value of test statistic is -1.2757
Critical value is -2.100922

here |tcal|<|tcritical|
Thus we accept null hypothesis at 5% l.o.s.

d) 95% C.I. is given by,
[ -13.504207 ,3.304207]

e) Thus conclusion is;
average number of cigaretts smoked before ad is same as average number of cigaretts smoked after ad.

f) Thus There was no significant difference between the number of cigarettes smoked before and after the ad.

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