Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Imagine an automobile company looking for additives that might increase gas mile

ID: 3221199 • Letter: I

Question

Imagine an automobile company looking for additives that might increase gas mileage. As a pilot study, they send 30 cars fueled with a new additive on a road trip from Boston to Los Angeles. Without the additive, those cars are known to average 25.0mpg with a standard deviation of 2.4 mpg. Suppose it turns out that the thirty cars averaged 26 3 mpg with the additive. b) Find the power of the test when mu is actually (i) 25.750 (by hand + Minitab) (ii) 26.8 (Minitab only) (iii) 28 (Minitab only) What effect does increasing the distance between the true value of mu and hypothesized value mu = 25 c) Find the power of the test when mu is actually 25.750 and n = 100. What effect does increasing the sample size have on the power of the test? (Minitab only) d) Find the power of the lest when mu is actually 25.750 and n = 30. What effect does increasing the sample size have on the power of the test? Use alpha = 0.05(Minitab only) and alpha = 0. l(MATLAB only) e) What would be the effect on power when mu is actually 25.750 (n = 30, alpha = 0.01) if sigma could be reduced from 2.4 mpg to 1.2 mpg? (Minitab only)

Explanation / Answer

Part-b(i)

Power=0.380343 as shown below:

Power and Sample Size

1-Sample t Test

Testing mean = null (versus null)
Calculating power for mean = null + difference
= 0.05 Assumed standard deviation = 2.4


Sample
Difference Size Power
0.75 30 0.380343

Part-b(ii)

Power=0.977704 as shown below:

Power and Sample Size

1-Sample t Test

Testing mean = null (versus null)
Calculating power for mean = null + difference
= 0.05 Assumed standard deviation = 2.4


Sample
Difference Size Power
1.8 30 0.977704

Part-b(iii)

Power=1.000 as shown below:

Power and Sample Size

1-Sample t Test

Testing mean = null (versus null)
Calculating power for mean = null + difference
= 0.05 Assumed standard deviation = 2.4


Sample
Difference Size Power
3 30 1.00000

Part-c

Power=0.871725 as shown below. We conclude that increasing the sample size resulted in increase in power.

Power and Sample Size

1-Sample t Test

Testing mean = null (versus null)
Calculating power for mean = null + difference
= 0.05 Assumed standard deviation = 2.4


Sample
Difference Size Power
0.75 100 0.871725

Part-d

Power=0.168399 as shown below for alpha=0.01. Increasing alphs (part-b(i) ) leads to increased power

Power and Sample Size

1-Sample t Test

Testing mean = null (versus null)
Calculating power for mean = null + difference
= 0.01 Assumed standard deviation = 2.4


Sample
Difference Size Power
0.75 30 0.168399

Part-e

Power=0.742201 as shown below from where we conclude that decreasing the standard deviation increased the power

Power and Sample Size

1-Sample t Test

Testing mean = null (versus null)
Calculating power for mean = null + difference
= 0.01 Assumed standard deviation = 1.2


Sample
Difference Size Power
0.75 30 0.742201