Question

For the two large newspapers in your city, you are interested in knowing whether there is a significant difference in the average number of pages in each dedicated solely to advertising. You randomly select 10 editions of newspaper A and 6 editions of newspaper B (excluding weekend editions). The data follow. Use = 0.01 to test whether there is a significant difference in averages. Assume the number of pages of advertising per edition is normally distributed and the population variances are approximately equal.

A:

18 18 21 15 11 18 18 22 26 18

B:

7 14 11 11 10 6

1. Observed T= ? (Round the intermediate values to 3 decimal places. Round your answer to 2 decimal places.)

2. The decision is to fail to reject the null hypothesis or to reject the null hypothesis?

Answer 1

	Values ( A )	Σ ( Xi- X̅ )2	Values ( B )	Σ ( Yi- Y̅ )2
	18	0.25	7	8.0276
	21	6.25	11	1.3612
	11	56.25	10	0.0278
	18	0.25	14	17.3614
	26	56.25	11	1.3612
	18	0.25	6	14.6942
	15	12.25	59	42.8334
	18	0.25
	22	12.25
	18	0.25
Total	185	144.5

Mean X̅ = Σ Xi / n
X̅ = 185 / 10 = 18.5
Sample Standard deviation S_A = √ ( (Xi - X̅ )2 / n - 1 )
S_A = √ ( 144.5 / 10 -1 ) = 4.0069

Mean Y̅ = ΣYi / n
Y̅ = 59 / 6 = 9.8333
Sample Standard deviation S_B = √ ( (Yi - Y̅ )2 / n - 1 )
S_B = √ ( 42.8334 / 6 -1) = 2.9269

To Test :-

H0 :- µ_A = µ_B
H1 :- µ_A ≠ µ_B

Test Statistic :-
t = (X̅1 - X̅2) / SP √ ( ( 1 / n1) + (1 / n2))



t = ( 18.5 - 9.8333) / 3.658 √ ( ( 1 / 10) + (1 / 6 ))
t = 4.588

Test Criteria :-
Reject null hypothesis if | t | > t(α/2, n1 + n2 - 2)
t(α/2, n1 + n1 - 2) = t(0.01 /2, 10 + 6 - 2) = 2.977
| t | > t(α/2, n1 + n2 - 2) = 4.5881 > 2.977
Result :- Reject Null Hypothesis

Decision based on P value
P - value = P ( t > 4.5881 ) = 0.0004
Reject null hypothesis if P value < α = 0.01 level of significance
P - value = 0.0004 < 0.01 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

Decision

reject the null hypothesis.