In: Statistics and Probability
Two major computer companies manufacture and sell computer accessories. The prices of 5 randomly selected accessories are listed below:
Computer Accessory Price Data | |||
---|---|---|---|
Computer Accessory | Macrohard Price ($) | Pear Computer Price ($) | |
keyboard | 67.33 | 65 | |
mouse | 54.78 | 40.24 | |
speaker | 135.53 | 134.82 | |
modem | 100.31 | 95.71 | |
monitor | 377.62 | 352.36 |
The quality of the computer accessories for both companies are practically the same because both companies purchase the parts from the same wholesaler. Since the market for these accessories is very competitive, the main reason why one product would be more expensive than another would be due to brand name and the advertizing related to it.
A computer periodical claims that Macrohard products are generally more expensive than Pear Computer products. You plan to do a hypothesis test on this claim where:
H0: Macrohard accessories are equal in price to Pear
Computer products
Ha: Macrohard accessories are more expensive
You may assume that the differences in prices are normally distributed.
a)Calculate the test statistic (t) that is best suited in conducting this hypothesis test. Give your answer to 2 decimal places.
t =
b)According to the above data and at a level of significance of 0.01, the claim that Macrohard accessories are more expensive is justifiednot justified given the results of the hypothesis test.
Solution:
Here, we have to use paired t test for the difference between population means. The null and alternative hypotheses for this test are given as below:
H0: Macrohard accessories are equal in price to Pear
Computer products.
Ha: Macrohard accessories are more expensive.
H0: µd = 0 vs. Ha: µd > 0
This is a one tailed test. (Right/Upper tailed)
Part a)
Test statistic for paired t test is given as below:
t = (Dbar - µd)/[Sd/sqrt(n)]
From given data, we have
Dbar = 9.4880
Sd = 10.3255
n = 5
df = n – 1 = 4
α = 0.01
Upper Critical t value = 3.7469
(by using t-table)
t = (Dbar - µd)/[Sd/sqrt(n)]
t = (9.4880 - 0)/[ 10.3255/sqrt(5)]
t = 9.4880/4.6177
t = 2.0547
t = 2.05
Part b)
We are given
α = 0.01
P-value = 0.0546
(By using t-table)
P-value > α = 0.01
So, we do not reject the null hypothesis
There is insufficient evidence to conclude that Macrohard accessories are more expensive.
According to the above data and at a level of significance of 0.01, the claim that Macrohard accessories are more expensive is not justified given the results of the hypothesis test.