5.13 Car insurance savings. A market researcher wants to evaluate car insurance savings at a competing company. Based on past studies he is assuming that the standard deviation of savings is $100. He wants to collect data such that he can get a margin of error of no more than $10 at a 95% confidence level. How large of a sample should he collect?

Solution

What do we need to find?
\(n\) - sample size

What have we got?
\(\sigma = 100\)
\(\alpha = 0.05\)
\(Margin Of Error \le 10\)

Chapter 4 definition: In a confidence interval, z* x SE is called the margin of error.

Then let’s solve for margin of error less or equal 10:

\(z^*\) x \(SE \le 10\)

\(z^*\) = qnorm(0.025, mean=0, sd=1, lower.tail=F) = 1.959964

\(SE = \frac{\sigma}{\sqrt n}\)

1.96 * \(\frac{100}{\sqrt n} \le 10\)

\(n \ge 19.6^2\) \(=>\) \(n \ge 384.16\)

Round this value up, so the minimum sample size to have a margin of error of $10 is 385 observations.

Check:

Let us check on the normal distribution plot, use SE = 100/19.6 = 5.10

test

test

Indeed, the rejection region for the difference of the means is above ±10