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 on the answers sheet in cells F92 to G98. 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.

DATA:

User	Name Brand	Generic
1	306	300
2	256	260
3	402	357
4	299	286
5	306	290
6	257	260

ANSWER FORMAT:

Define H0 :

Define H1 :

Test statistic

Critical value of test statistic

Decision rule

Calculated value of test statistic

Reject or fail to reject H0?

Conclusion about the cartridges

Find the p-value

Interpret p-value

Answer 1

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.

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

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

Calculated Value of the test statistic: 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

Reject Or Fail to Reject: Since t observed is < +tcritical (3.365), We Fail to Reject H0.

Conclusion about the cartridge: 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.

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

Interpretation of the p - value: The p value is the probability of getting a test statistic as extreme as that which was obtained, assuming that the null hypothesis is true. If the p value is very low, then the probability of such an even happenning is very low and the null hypothesis would get rejected.

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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