Question

A random sample of 50 cars in the drive-thru of a popular fast food restaurant revealed an average bill of $18.21 per car. The population standard deviation is $5.92. Round your answers to two decimal places.
(a) State the point estimate for the population mean cost of fast food bills at this restaurant $

(b) Calculate the 95% margin of error. $

(c) State the 95% confidence interval for the population mean cost of fast food bills at this restaurant. $ ≤ µ ≤ $

(d) What sample size is needed if the error must not exceed $1.00? n =

Answer 1

Answer:

a) $\bar{x} = 18.21$

b) 1.64

c) (16.57,19.85)

d) The sample size must be 135 or greater if the error must not exceed $1.00.

Step-by-step explanation:

We are given the following information in the question:

Sample size, n = 50

Sample mean =   $18.21

population standard deviation = $5.92

a)  Point estimate for the population mean cost

$\bar{x} = 18.21$

b) Margin of error =

$z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Margin of error =  $1.96\displaystyle\frac{5.92}{\sqrt{50}} = 1.64$

c) 95% Confidence interval

$\mu \pm z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$18.21 \pm 1.64) = (16.57,19.85)$

d) Marginal error less than $1.00

$1.96\displaystyle\frac{\sigma}{\sqrt{n}} \leq 1\\\\\sqrt{n} \geq 1.96\times 5.92\\n \geq (11.6)^2\\n \geq 134.56 \approx 135$

Thus, the sample size must be 135 or greater if the error must not exceed $1.00.