Answer:
If we examine a sample of tires used for 60,000 miles and determine the proportion that are worn out, how likely is that proportion to be 3.6% or less?
Step-by-step explanation:
First of all, when you want to contrast two hypothesis you need a confidence level p% without which the comparison makes no sense.
In this case, you want to refuse the null hypothesis (the average is 5%) with an alternative one (the average is less than 5%).
You propose a new average (3.6%) based on the evidence.
You cannot prove that the average is actually 3.6%, <em>but if you somehow determine that the probability the average is greater than 3.6% is lower than 100% minus your confidence level p% ,then you can refute the null hypothesis and accept the alternative one as new average.
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So, from the possible choices, the one that fits the best is
If we examine a sample of tires used for 60,000 miles and determine the proportion that are worn out, how likely is that proportion to be 3.6% or less?
https://www.tiger-algebra.com/drill/1.4dot5(x~2-4xy_4y~2)/(2x-4):(x-2y)/(x-2)$(x~2-4)/(4)/
The answer is choice A. d=60t
