Question

Battery packs in radio-controlled racing cars need to be able to last pretty long. The distribution of the lifetimes of battery packs made by Lectric Co. is slightly left skewed. Assume that the standard deviation of the lifetime distribution is σ = 2.5 hours. A simple random sample of 75 battery packs results in a mean of = 29.6 hours. What is a 90% confidence interval for μ, the true average lifetime of the battery packs made by Lectric Co.? (29.13, 30.07) (29.03, 30.17) (28.86, 30.34) The confidence interval cannot be calculated because the population distribution is not Normal.

Answer 1

Answer:

(29.13, 30.07)

Step-by-step explanation:

The first step is finding 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 $1-\alpha$.

So it is z with a pvalue of $1-0.05 = 0.95$, that is between $Z = 1.64$ and $Z = 1.65$, so we use $z = 1.645$.

Now, find M as such:

$M = z*\frac{\sigma}{\sqrt{n}} = 1.645*\frac{2.5}{\sqrt{75}} = 0.47$

In which $\sigma$ is the standard deviation and n is the length of the sample

The lower end of the interval is the sample mean subtracted by M. So it is 29.6 - 0.47 = 29.13 hours.

The upper end of the interval is the sample mean added to M. So it is 29.6 + 0.47 = 30.07 hours

The 90% confidence interval is (29.13 hours, 30.07 hours).

The correct answer is:

(29.13, 30.07)