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?

CI = x_bar +/- z_score*(s/sqrt(n))

where x_bar is the mean z_score is the confidence coefficient for 95% CI interval s is stantard deviation n is the population size

The margin of error is equal margin_error = z_score*(s/sqrt(n))

We know that for 95% CI interval the confidence coefficient is 1.96 We also have s = $100 We need to solve for n where z_score*(s/sqrt(n)) <= 10

s = 100
z_score = 1.96
margin_error = 10

Solving for n we get

n = (s*z_score/margin_error)^2
n
## [1] 384.16

So in order for us to get a margin of error less than $10, we would need a sample of at least 385 people (we can’t have .16 of a person).

Let’s double check

for (i in 384:400) {
  n = i
  z_score = 1.96
  s = 100
  margin_error = z_score*(s/sqrt(n))
  
  print(paste("n value: ",n, ", ME: ", margin_error))
}
## [1] "n value:  384 , ME:  10.0020831163646"
## [1] "n value:  385 , ME:  9.98908495217746"
## [1] "n value:  386 , ME:  9.97613733176204"
## [1] "n value:  387 , ME:  9.96323992839684"
## [1] "n value:  388 , ME:  9.95039241830947"
## [1] "n value:  389 , ME:  9.93759448064252"
## [1] "n value:  390 , ME:  9.92484579741993"
## [1] "n value:  391 , ME:  9.9121460535138"
## [1] "n value:  392 , ME:  9.89949493661167"
## [1] "n value:  393 , ME:  9.88689213718424"
## [1] "n value:  394 , ME:  9.87433734845362"
## [1] "n value:  395 , ME:  9.8618302663619"
## [1] "n value:  396 , ME:  9.84937058954028"
## [1] "n value:  397 , ME:  9.83695801927851"
## [1] "n value:  398 , ME:  9.82459225949487"
## [1] "n value:  399 , ME:  9.81227301670647"
## [1] "n value:  400 , ME:  9.8"

Notice 384 has a margin error slightly greater than $10.