Question

1.A tyre manufacturer produces tyres that have a mean life of 20,000 km when the production is working properly. Based on past experience, the standard deviation of the life of tyres is 2,800 km. The operations manager stops the production process if there is evidence that the mean life is different from 20,000 km. If you select a random sample of 100 tyres and you are willing to have a 5% level of significance (i.e. risk of committing a Type I error), the power of the test if the population mean is actually 19,920 km would be around __________ percent. a. 29 b. 71 c. 6 d. 94

2.

An economics professor believes that debt payments are influenced by income and the unemployment rate. She collects data on debt payments (in $), income (in $1,000s), and unemployment rate (in %) from 26 metropolitan areas and obtains the following regression results:

	Coefficients	Standard error
Intercept	199	156.36
Income	10.51	1.48
Unemployment rate	0.62	6.87

Based on this information, the regression coefficient on income can be interpreted as follows.

	a.	If income increases by $0.62, then mean debt payments are estimated to increase by $10.51, holding the unemployment rate constant.
	b.	If income increases by $1,000, then mean debt payments are estimated to increase by $10.51, holding the unemployment rate constant.
	c.	If income increases by $1,000, then mean debt payments are estimated to increase by $209.51, holding the unemployment rate constant.
	d.	If income increases by $10.51, then mean debt payments are estimated to increase by $0.62, holding the unemployment rate constant.

Answer 1

(2) If income increases by $1,000, then mean debt payments are estimated to increase by $10.51, holding the unemployment rate constant, because it means that if the value of X1 increases by 1, the value of Y will increase by 10.51 units, if we keep value of X2 constant.

(1) We use an R command to solve this problem. Here, n = sample size = 100, significance level = 0.05, two sided alternative, one sample t test, d = effect size = (20000 - 19920)/2800 = 0.0286.
Now, we run these R commands.
install.packages("pwr")
library(pwr)
pwr.t.test(n=100,d=(20000-19920)/2800,sig.level=0.05,type="one.sample",alternative="two.sided")
We get the value of power to be 0.0592 or 5.92% or 6%.