Question

A company makes auto batteries. They claim that 86% of their LL70 batteries are good for 70 months or longer. Assume that this claim is true. Let p be the proportion in a random sample of 80 such batteries For a populations that are good for 70 months or more.
What is the probability that this sample proportion is within 0.03 of the population proportion?

Answer 1

Answer:

The probability that the sample proportion of 80 LL70 batteries is within 0.03 of the population proportion is 0.44

Step-by-step explanation:

Sample proportion being within margin, or margin of error (ME) around the mean can be found using the formula

ME=$\frac{z*\sqrt{p*(1-p)}}{\sqrt{N} }$ where

• z is the corresponding statistic of the probability that the sample proportion is within the 0.03 of the population proportion

• p is the claimed proportion (86% or 0.86)

• N is the sample size (80)

Then 0.03=$\frac{z*\sqrt{0.86*0.14}}{\sqrt{80} }$ from this we get:

z≈0.773 and the p(z)≈0.439

Therefore, the probability that the sample proportion is within 0.03 of the population proportion is 0.44