In a study of the lifetimes of electronic components, a random sample of 400 components are tested until they fail to function. The sample mean lifetime was 370 hours and the standard deviation was 650 hours. True or false:

a. An approximate 95% confidence interval for the mean lifetime of this type of component is from 306.3 to 433.7 hours.

b. About 95% of the sample components had lifetimes between 306.3 and 433.7 hours.

c. If someone takes a random sample of 400 components, divides the sample standard deviation of their lifetimes by 20, and then adds and subtracts that quantity from the sample mean, there is about a 68% chance that the interval so constructed will cover the mean lifetime of this type of component.

d. The z table can't be used to construct confidence intervals here, because the lifetimes of the components don't follow the normal curve.

e. About 68% of the components had lifetimes in the interval 370 ± 650 hours.

Answer:

Step 1 of 5:

(a)

In this question, we are asked to state the True or false for given statements.

Statement: an approximate confidence interval for the mean lifetime of this type of component is from .

Given a random sample of 400 components are tested. The sample mean lifetime was and the standard deviation was .

Let be a large sample of size from a population with mean and standard deviation

then a level confidence interval for is

Since we want 95% confidence interval, then a level is

and

Then for is , hence

Hence an approximate confidence interval for the mean lifetime of this type of component is from .

The statement is true.

Step 2 of 5:

(b)

Statement: about 95 of the sample components had lifetimes between .

The statement is false. Because a specific confidence interval is given. We are 95 confident that the population mean is either in the interval or it isn’t. We can not say that 95% of the sample components had lifetimes between

Step 3 of 5:

(c)

Statement: if someone takes a random sample of 400 components, divides the sample standard deviation of their lifetimes by 20, and then adds and subtracts the quantity from the sample mean.

Then there is 68% chance that the interval so constructed will cover the mean lifetime of this type of component.

lets divide the sample standard deviation of their lifetimes by 20, and then adds and subtracts the quantity from the sample mean.

We get,

………(3)

The confidence interval for is

………(4)

Compare equation (3) and (4), we have,

Using the , the area to the left of is approximately .

Therefore , so the level is,

or approximately .

Hence there is 68% chance that the interval so constructed will cover the mean lifetime of this type of component.

The statement is true.