Question

4. In order to test whether brand-name printer cartridges produce more printed pages, on average, than generic cartridges, a research firm has 6 randomly selected printer users use both types of cartridges and record how many pages were printed with each. The number of pages printed for each user by each type of cartridge are shown.
	Use the 0.01 significance level to test whether the brand-name cartridges print more pages on average than the generic cartridges.
	Identify and interpret the p-value for the test. User Name Brand Generic difference 1 306 300 2 256 260 3 402 357 4 299 286 5 306 290 6 257 260

Answer 1

#No	Name	Generic	Difference
1	306	300	6
2	256	260	-4
3	402	357	45
4	299	286	13
5	306	290	16
6	257	260	-3

The calculations for the mean and the standard deviations are given after the test.

The mean of the difference = 12.1667, and the standard deviation of the difference = 18.0157

= 0.01 and the degrees of freedom = number of pairs - 1 = 6 - 1 = 5

Let be the difference in scores of the 2 populations.

(i) The Hypothesis:

H0: = 0 : The Number of pages printed by the brand name cartridge and the generic cartridge are equal.

Ha: > 0: The Number of pages printed by the brand name cartridges is greater than the generic cartridge.

(ii)The Test Statistic: Since sample size is small, and population std. deviation is unknown, we use the students t test.

(iii) The Critical value: (Right tailed) for = 0.01, df = 5 is +3.365

The p value: (Right tailed) at t = 1.65, degrees of freedom = 5 is 0.0799

(iv) The Decision Rule:

If t test is > 3.365, Reject H0

If p value is < (0.01), Reject H0.

(v) The Decision:   Since tobserved (1.65) is < 3.365, We Fail to Reject H0.

Also since P value (0.0799) is > (0.01) , We Fail to Reject H0.

(vi) The Conclusion: There is insufficient evidence at the 99% significance level to conclude that the Number of pages printed by the brand name cartridges is greater than the generic cartridge.

__________________________________________________________

Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where

SS = SUM(X - Mean)².

#	Difference	Mean	(X-Mean)2
1	6	12.1667	38.0281889
2	-4	12.1667	261.362189
3	45	12.1667	1078.02559
4	13	12.1667	0.69438889
5	16	12.1667	14.6941889
6	-3	12.1667	230.028789

n	6
Sum	73
Mean	12.1667
SS	1622.833333
Variance	324.5667
Std Dev	18.0157