1. Suppose you are in charge of inventory maintenance at a bicycle shop. One of
ID: 3331438 • Letter: 1
Question
1. Suppose you are in charge of inventory maintenance at a bicycle shop. One of your jobs is to ensure that the tire pressure in each of the display bicycles is between 65-85 PSI (pounds per square inch). If the pressure is too low, then there is a risk of wheel damage when a customer rides out on one. On the other end, if the pressure is too high, there is a (small) risk of the tire exploding! Of course, you don't know the true pressure of any particular tire. Instead, you have the output from your pressure gauge. This will be similar to the true pressure, but not necessarily the same. Laboratory testing of the particular gauge you use has shown that there is a 2 PSI error margin, and so to be careful, you decide that you will adjust the pressure on any tire that has a measured pressure above 81 PSI or below 69 PSI, giving you 2x the error margin on either side. Previous testing shows that this procedure will give the following results: Measure inside 69-81PSI Measure outside 69-81 PSI PSI within 65-85 PSI PSI outside 65-85 PSI 5000 105 216 (a) State (in words) the appropriate null and alternate hypothesis (b) Calculate the value of a. What situation does the Type I error rate represent here? (c) Calculate the value of B. What situation does the Type II error rate represent here? (d) Calculate the Power of this test. What situation does test Power represent here? (e) What is the rejection region for the test you are conducting? (f) Practically speaking, which error seems more problematic: a Type I error or a Type II error? (g) If we wanted to decrease the Type I error rate, should we increase or decrease the rejection region? h) If we wanted to decrease the Type II error rate, should we increase or decrease the rejection region?Explanation / Answer
Answer to part a)
Null hypothesis: The tire pressure in in the range 65 to 85 PSI
Alternate hypothesis: The Tire pressure is NOt in the range 65 to 85
.
Answer to part b)
Alpha is type I error; this is when the null is true but we reject it
this implies that the actual pressure is within 65 to 85 , but the gauge reads it wrong
the number of chaces for this are 105 /(5000+105) = 0.0206
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Answer to part c)
Type II error is when the null is not true , that is the actual pressure is not between 65 to 85 , but the gauge tells us that it lies in this raneg
Beta = 3 / (216+3) = 0.01370
.
Answer to part d)
Power of the test is 1-type II error
Power = 1 - 0.01370 = 0.9863
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