In: Statistics and Probability
A shoe manufacturer compared material A and material B for the soles of shoes. Twelve volunteers each got two shoes. The left was made with material A and the right with material B. On both shoes, the material was 1 inch thick. Volunteers used the shoes normally for two months and returned them to the manufacturer. Technicians then measured the thickness of the soles and recorded the amount of wear (in microns):
Volunteer | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Sole A | 379 | 378 | 328 | 372 | 325 | 304 | 356 | 309 | 354 | 318 | 355 | 392 |
Sole B | 372 | 376 | 328 | 368 | 283 | 252 | 369 | 321 | 379 | 303 | 328 | 411 |
They wish to test H0: μA −μB = 0 against HA: μA −μB ̸= 0, where μA and μB are the unknown populations mean amounts of sole wear, using α = 0.05.
(a) Are the two samples paired or independent? Explain your answer.
(b) Make a normal QQ plot of the differences within each pair. Is it reasonable to assume a normal population of differences?
(c) Choose a test appropriate for the hypotheses above and justify your choice based on your answers to parts (a) and (b). Perform the test by computing a p-value, make a test decision, and state your conclusion in the context of the problem.
a) TWO SAMPLES ARE PAIRED . BECAUSE THIS IS A REPEATED MEASURE EXPERIMENT.
b)
From above q-q plot we can conclude that data is approximately normal.
c) 1)H0: μA −μB = 0
HA: μA −μB ̸= 0
alpha=0.05
(2) Rejection Region
Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=11.
Hence, it is found that the critical value for this two-tailed test is tc =2.201, for α=0.05 and df=11.
The rejection region for this two-tailed test is R={t:∣t∣>2.201}.
(3) Test Statistics
The t-statistic is computed as shown in the following formula:
(4) Decision about the null hypothesis
Since it is observed that ∣t∣=0.968≤tc=2.201, it is then concluded that the null hypothesis is not rejected.
Using the P-value approach: The p-value is p=0.3537, and since p=0.3537≥0.05, it is concluded that the null hypothesis is not rejected.
(5) Conclusion: It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1 is different than μ2 , at the 0.05 significance level.