Question

A transportation worker takes a simple random sample of 162 toll booth receipts for one day and finds that the mean is $7.38 with a standard deviation of $1.45. What is the 99% confidence interval for the mean amount of all the receipts that day?

Answer 1

Answer:

99% Confidence interval:  (7.09,7.67)

Step-by-step explanation:

We are given the following in the question:

Sample mean, $\bar{x}$ = $7.38

Sample size, n = 162

Alpha, α = 0.05

Sample standard deviation, σ = $1.45

99% Confidence interval:  

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,  

Degree of freedom = n - 1 = 161

$t_{critical}\text{ at degree of freedom 161 and}~\alpha_{0.01} = \pm 2.606$  

$7.38 \pm 2.606(\dfrac{1.45}{\sqrt{162}} )\\\\ = 7.38 \pm 0.2968 = (7.0832 ,7.6768)\\\approx (7.09,7.67)$  

is the required 99% confidence interval for the mean amount of all the receipts that day.