Question

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 percent confidence interval for the mean bill of all lunch orders. (Round your answers to 4 decimal places.) to The 95% confidence interval is from

Answer 1

Solution :

Given that,

$\bar x$ = $9.60

$\sigma$ = $5.73

n = 29

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.96

Margin of error = E = Z_$\alpha$/2* ($\sigma$ /$\sqrt$n)

= 1.96 * (5.73 / $\sqrt$29)

= 2.0855

At 95% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

9.60 - 2.0855 < $\mu$ < 9.60 + 2.0855

7.5145< $\mu$ < 11.6855

(7.5145 to 11.6855)