Question

The manufacturer of an MP3 player wanted to know whether a 10% reduction in price is enough to increase the sales of its product. To investigate, the owner randomly selected eight outlets and sold the MP3 player at the reduced price. At seven randomly selected outlets, the MP3 player was sold at the regular price. Reported below is the number of units sold last month at the regular and reduced prices at the randomly selected outlets. Regular price 138 121 123 116 116 122 132 Reduced price 143 133 151 136 144 123 130 131 Click here for the Excel Data File At the 0.010 significance level, can the manufacturer conclude that the price reduction resulted in an increase in sales? Hint: For the calculations, assume reduced price as the first sample. Compute the pooled estimate of the variance. (Round your answer to 3 decimal places.) Compute the test statistic. (Round your answer to 2 decimal places.) State your decision about the null hypothesis. Reject H0 Fail to reject H0

Answer 1

Let be the population average for sales at the regular price.

Let be the population average for sales at the reduced price.

From the data:

For Reduced Price: = 124, s₁ = 8.185, n₁ = 8

For: = 136.375, s₂ = 9.07, n₂ = 8

Since s₁/s₂ = 8.185/9.07 = 0.9 (it lies between 0.5 and 2) we used the pooled variance.

Since we use the pooled variance, the degrees of freedom = n1 + n2 - 2 = 7 + 8 - 2 = 13

The Hypothesis:

H0: = : The average for sales at the regular price is equal to the average for sales at the reduced price.

Ha: < : The average for sales at the regular price is lesser than the average for sales at the reduced price.

This is a Left tailed test.

The Test Statistic:

The p Value: The p value (Left Tail) for t = -2.76,df = 13,is; p value = 0.0081

The Critical Value:   The critical value (Left tail) at = 0.01, df = 13,tcritical = -2.65

The Decision Rule:

By The Critical Value Method: If tobserved is < -tcritical, Then Reject H0.

By The p - Value Method: If the P value is < , Then Reject H0

The Decision:

By The Critical Value Method: Since t observed (-2.76) is < tcritical (-2.65), We Reject H0.

By The p - Value Method: Since P value (0.0081) is < (0.01), We Reject H0.

The Conclusion: Reject H0. There is sufficient evidence at the 99% significance level to conclude that the average for sales at the regular price is lesser than the average for sales at the reduced price.

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Calculations for the mean and the 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.

Outlet	Regular	Reduced
1	138	143
2	121	133
3	123	151
4	116	136
5	116	144
6	122	123
7	132	130
8		131

Regular	X	Mean	(x - mean)2		Reduced	X	Mean	(x - mean)2
1	138	124	196		1	143	136.375	43.890625
2	121	124	9		2	133	136.375	11.390625
3	123	124	1		3	151	136.375	213.890625
4	116	124	64		4	136	136.375	0.140625
5	116	124	64		5	144	136.375	58.140625
6	122	124	4		6	123	136.375	178.890625
7	132	124	64		7	130	136.375	40.640625
					8	131	136.375	28.890625
Total	868		402			1091		575.875

	Regular	Reduced
n	7	8
Sum	868	1091
Average	124.00	136.375
SS(Sum of squares)	402	575.875
Variance = SS/n-1	67	82.26785714
Std Dev=Sqrt(Variance)	8.185	9.070