Question

The lifetime of a certain brand of battery is known to have a standard deviation of 9 hours. Suppose that a random sample of 150 such batteries has a mean lifetime of 40.5 hours. Based on this sample, find a 90% confidence interval for the true mean lifetime of all batteries of this brand. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)

Answer 1

Answer:

The 90% confidence interval for the true mean lifetime of all batteries of this brand is between 39.3 hours and 41.7 hours. The lower limit is 39.3 hours while the upper limit is 41.7 hours.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$.

That is z with a pvalue of $1 - 0.05 = 0.95$, so Z = 1.645.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.645\frac{9}{\sqrt{150}} = 1.2$

The lower end of the interval is the sample mean subtracted by M. So it is 40.5 - 1.2 = 39.3 hours

The upper end of the interval is the sample mean added to M. So it is 40.5 + 1.2 = 41.7 hours

The 90% confidence interval for the true mean lifetime of all batteries of this brand is between 39.3 hours and 41.7 hours. The lower limit is 39.3 hours while the upper limit is 41.7 hours.