Question

Battery life for a hand-held computer is normally distrituted and has a population standard deviation of 8 hours. Suppose you need to estimate a confidence interval estimate at the 80% level of confidence for the mean life of these batteries. Determine the sample size required to have a margin of error of 0.332 hours. Round up to the nearest whole number

Answer 1

Answer:

Sample size= 956

Step-by-step explanation:

Hello!

You need to construct an 80% CI for the average life of batteries. (hours)

The study variable is:

X: Lifespan of a battery.

X~N(μ; σ²)

The population standard devition isσ= 8

The interval to estimate the population mean follows the structure:

"Point estimator" ± "margin of error"

X[bar] ± $Z_{1-\alpha /2}$* (σ/√n)

X[bar] is the sample mean

$Z_{1-\alpha /2}$* (σ/√n) is the margin of error of the interval (d)

Using the margin of error you can calculate the sample size:

d= $Z_{1-\alpha /2}$* (σ/√n)

n= (σ* $\frac{Z_{1-\alpha /2}}{d}$)²

n= $(8*(\frac{1.283}{0.332} ))^2$

n= 955.77 ≅ 956

I hope it helps!