A random sample of 29 lunch orders at Noodles & Company showed a mean bill of 9.60 with a standard deviation of 5.73.

Find the 95% confidence interval for the mean of all orders

The 95% confidence interval is from to

Given;

x^- = $9.60

σ= $5.73

n = 29

At the 95% confidence level the z is;

α = 1 - 95% = 1 - 0.95 = 0.05

α/2= 0.05 / 2 = 0.025

Zα/2=Z_0.025= 1.96

The margin error= E=Zα/2*(σ/√n)

= 1.96 * (5.73 /√29)

=  2.0855

At a 95% confidence interval estimate of the population mean is;

x^--E<μ<x^-<9.60+20855

7.5145<μ<11.6855

(7.5145 to 11.6855)

Step-by-step explanation

Given;

x^- = $9.60

σ= $5.73

n = 29

At the 95% confidence level the z is;

α = 1 - 95% = 1 - 0.95 = 0.05

α/2= 0.05 / 2 = 0.025

Zα/2=Z_0.025= 1.96

The margin error= E=Zα/2*(σ/√n)

= 1.96 * (5.73 /√29)

=  2.0855

At a 95% confidence interval estimate of the population mean is;

x^--E<μ<x^-<9.60+20855

7.5145<μ<11.6855

(7.5145 to 11.6855)

The 95% confidence interval is from 7.5145 to 11.6855.