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 of eight teenage smokers during the month before and the month after the fi lm presentation, with the following results: (Note: When deciding on the form of the alternative hypothesis, H 1 , remember that a positive difference score ( D 5 X 1 2 X 2 ) reflects a decline in cigarette consumption.)

Smoker Number	Before Film	After Film
1	28	26
2	29	27
3	31	32
4	44	44
5	35	35
6	20	16
7	50	47
8	25	23

1) Using t, Test the null hypothesis at the 0.5 level of significance

a)what is the critical t?

b) What is the value of t?

2) specify the p-value for this test result

3) If appropriate (because the null hypothesis was rejected), construct a 95 percent confidence interval for the true population mean for all difference scores.

4) If appropriate, use Cohen’s d to obtain a standardized estimate of the effect size.

Answer 1

Given that,
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.365
since our test is two-tailed
reject Ho, if to < -2.365 OR if to > 2.365
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 1.5
We have d = 1.5
pooled variance = calculate value of Sd= √S^2 = sqrt [ 38-(12^2/8 ] / 7 = 1.69
to = d/ (S/√n) = 2.51
critical Value
the value of |t α| with n-1 = 7 d.f is 2.365
we got |t o| = 2.51 & |t α| =2.365
make Decision
hence Value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.51 ) = 0.0404
hence value of p0.05 > 0.0404,here we reject Ho
ANSWERS
---------------
1.
a.
null, H0: Ud = 0
alternate, H1: Ud != 0
b.
test statistic: 2.51
critical value: reject Ho, if to < -2.365 OR if to > 2.365
decision: Reject Ho
2.
p-value: 0.0404
we have enough evidence to support the claim that whether a new antismoking film actually reduces the daily consumption of cigarettes by teenage smokers.
3.
Confidence Interval
CI = d ± t a/2 * (Sd/ Sqrt(n))
Where,
d = ∑ di/n
Sd = Sqrt( ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
d = ( ∑ di/n ) =12/8=1.5
Pooled Sd( Sd )= Sqrt [ 38- (12^2/8 ] / 7 = 1.69
Confidence Interval = [ 1.5 ± t a/2 ( 0.976/ Sqrt ( 8) ) ]
= [ 1.5 - 2.365 * (0.598) , 1.5 + 2.365 * (0.598) ]
= [ 0.087 , 2.913 ]
4.
cohen's d size = mean difference / pooled standard deviation
cohen's d size = 1.5/1.69 = 0.887