The manufacturer of an MP3 player wanted to know whether a 10 percent reduction
ID: 3312308 • Letter: T
Question
The manufacturer of an MP3 player wanted to know whether a 10 percent 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 sampled outlets. Regular price 137 122 88 116 143 123 96 Reduced price 123 133 152 136 116 109 115 116 At the .005 significance level, can the manufacturer conclude that the price reduction resulted in an increase in sales? Hint: For the calculations, assume the "Reduced price" as the first sample. The pooled variance is . (Round your answer to 2 decimal places.) The test statistic is . (Round your answer to 2 decimal places.) H0.
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: Reduction - Regular< 0
Alternative hypothesis: Reduction - Regular > 0
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 9.12
DF = 13
t = [ (x1 - x2) - d ] / SE
t = - 0.783
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
The observed difference in sample means produced a t statistic of - 0.783. We use the t Distribution Calculator to find P(t > - 0.783) = 0.7762
Therefore, the P-value in this analysis is 0.7762.
Interpret results. Since the P-value (0.7762) is greater than the significance level (0.05), we cannot reject the null hypothesis.
From the above test we do not have sufficient evidence in the favor of the claim that the price reduction resulted in an increase in sales.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.