1. Suppose you are in charge of inventory maintenance at a bicycle shop. One of
ID: 3332503 • 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. 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 This will be similar to the 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 (a) State (in words) the appropriate mull and alternate hypothesis. (b) Calculate the value of a. What situation does the Type I error rate represent here? (c) Calculate the value of . 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? (8) 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
a) null - PSI with in 65-85 PSI
alternate - PSI outside 65-85
b)
In statistical hypothesis testing, a type I error is the incorrect rejection of a true null hypothesis (a "false positive"), while a type II error is incorrectly retaining a false null hypothesis (a "false negative").
type i error = 105 / ( 5000 + 105)
= 0.02056
c)
type ii error = 3/(3 +216) = 0.01369
d)
power = 1 -type ii error 1 -0.01369
= 0.986301
Please post rest parts again as we have to solve first 4) subparts
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.