Question

Nadia wants to estimate the mean driving range for her company's new electric vehicle. She'll sample vehicles and measure each of their driving ranges to construct a confidence interval for the mean driving range. She wants the margin of error to be no more than 10 kilometers at a 90% level of confidence. A pilot study suggests
that the driving ranges for this type of vehicle have a standard deviation of 15 kilometers.

Which of these is the smallest approximate sample size required to obtain the desired margin of error?

A) 5 vehicles
B) 7 vehicles
C) 10 vehicles
D) 15 vehicles
E) 30 vehicles

Answer 1

Using the z-distribution, it is found that the smallest sample size required is given by:

B) 7 vehicles.

What is the margin of error for a z-distribution confidence interval?

It is given by:

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

In which:

• z is the critical value.

• $\sigma$ is the population standard deviation.

• n is the sample size.

In this problem, we have a 90% confidence level, hence$\alpha = 0.9$, z is the value of Z that has a p-value of $\frac{1+0.9}{2} = 0.95$, so the critical value is z = 1.645.

The margin of error and the population standard deviation are given by, respectively:

$M = 10, \sigma = 15$.

Then, we solve for n to find the needed sample size.

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

$10 = 1.645\frac{15}{\sqrt{n}}$

$10\sqrt{n} = 1.645 \times 15$

$\sqrt{n} = 1.645 \times 1.5$

$(\sqrt{n})^2 = (1.645 \times 1.5)^2$

$n = 6.1$

Rounding up, 7 vehicles have to be sampled, hence option B is correct.

More can be learned about the z-distribution at brainly.com/question/25890103