Question

A school psychologist wishes to determine whether a new antismoking film actually reduces the daily consumption of cigarettes by teenage smokers. The mean daily cigarette consumption is calculated for each eight teenage smokers during the month before and the month after the film presentation, with the following results:

MEAN DAILY CIGARETTE CONSUMPTION

SMOKER NUMBER	BEFORE FILM (X1)	AFTER FILM (X2)
1	28	26
2	29	27
3	31	31
4	44	44
5	35	35
6	20	16
7	50	47
8	25	23

(Note: when deciding on the form of the alternative hypothesis, H1, remember that a positive difference score (D=X1-X2) reflects a decline in cigarette consumption.)

Using t, test the null hypothesis at the .05 level of significance.

A)What is the research problem in this scenario?

B)Which of the following is the appropriate pair of statistical hypotheses for this study?

C)Compute the degrees of freedom for this scenario.

D)What is the decision rule in this scenario?

E)Calculate the value of the t test.

F)What is the decision about the null hypothesis in this scenario?

H)What is the interpretation in this scenario?

I)If appropriate (because the null hypothesis was rejected), construct a 95 percent confidence interval for the true population mean for all difference scores and use Cohen’s d to obtain a standardized of the effect size. Lower bound, upper bound, or 0 if null hypothesis is retained

J)Enter the estimate of the standardized effect size (Cohen’s d).

K)What might be done to improve the design of this experiment?

Answer 1

as per company policies, I am answering 4 parts only

a)
ho: u1=u2 ; there is no significant difference in the mean daily cigarette consumption between before and after film. u1 = u2
h1: u1>u2 ; the mean daily cigarette consumption between before film is less than after film. u1<u2

b)
a paired t-test is appropriate here as 2 groups of before and after film are related to each other by the same smoker.

c)

part1	part2	di
28	26	2
29	27	2
31	31	0
44	44	0
35	35	0
20	16	4
50	47	3
25	23	2

df = n-1= 8-1 = 7

d)

t(a,n-1) = t(0.05,8-1) = abs(T.INV(0.05,8-1)) =  1.895
decision rule is to reject HO if t > t(a,n-1)

e)

mean_diff=   1.625   [Excel function used -> AVERAGE]
sd_diff=   1.506   [Excel function used -> STDEV]
n=   8   [Excel function used -> COUNT]
      
t= (mean_diff)/(sd_diff/sqrt(n))      
t=   1.625/(1.50594061730772/sqrt(8))  
t=   3.0520  

f)

Sincce t > t(a,n-1), i reject ho at 5% level of significance

g)

there is sufficient evidence to conclude that the mean daily cigarette consumption between before film is less than after film. u1<u2.

h)

t(a/2,n-1)   =t(0.05/2,8-1)   =T.INV.2T(0.05,8-1) =   2.365
          
lower=   1.625-2.36462*1.50594/sqrt(8))=   0.366  
upper=   1.625+2.36462*1.50594/sqrt(8))=   2.884

i am 95% confidenet that estimated mean difference in the daily cigarette consumption between before film is less lie in the interval (0.366, 2.884).