Question

You measure the lifetime of a random sample of 64 tires of a certain brand. The sample mean is x = 56.7 months and the sample variance is 39.1 months squared. Suppose that the lifetimes for tires of this brand follow a normal distribution, with unknown mean µ and a population variance is 25 months squared. Construct a 99% confidence interval for µ, the average lifetime for the tires.

Answer 1

n = sample size = 64

Here we have given the population standard deviation ()so we use Z test to find the confidence interval.

When the population standard deviation ( ) is not known then we use T test.

Formula for confidence interval

E = margin of error

Z: Z critical value

C = level of confidence = 0.99 ( we convert 99% in to decimal

We use the above area =0.9950 to find the z critical value

We search for the area inside the body of the table

We get the two area which are very close to 0.9950 as 0.9949 and 0.9951

We look for row headed and column headed number and add then

We get the z score for area 0.9949 as 2.57    and the z score for 0.99451 as 2.58

We find the average of the both z score

So we get Z critical value as 2.575

We plug the values in formula.

99% confidence interval for µ, the average lifetime for the tires

Hint :-

You can use Ti -83 or Ti -84

press " STAT"

select " TESTS"

Select "Z Interval"

Select "Stats " from "Z Interval"

c-level = 0.99

select "calculate " and press enter

You can get the same confidence interval.

I hope this will help you :)