Question

The manager of The Cheesecake Factory in Boston reports that on six randomly selected weekdays, the number of customers served was 120, 130, 100, 205, 185, and 220. She believes that the number of customers served on weekdays follows a normal distribution.

Construct the lower bound of the 90% confidence interval for the average number of customers served on weekdays. (Round the sample standard deviation to 2 decimal places, the "t" value to 3 decimal places, and the final answer to 2 decimal places.)

Answer 1

Solution:

x	x2
120	14400
130	16900
100	10000
205	42025
185	34225
220	48400
---	---
∑x=960	∑x2=165950

Mean ˉx=∑xn

=120+130+100+205+185+220/6

=960/6

=160

The sample standard is S

  S =( x2 ) - (( x)2 / n ) n -1
=165950-(960)265

=165950-1536005

=123505

=√2470

=49.699

Degrees of freedom = df = n - 1 = 6 - 1 = 5

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t /2,df = t0.05,5 = 2.015

Margin of error = E = t/2,df * (s /n)

= 2.015* (49.70 / 6)

=40.88

Margin of error =40.88

The 99% confidence interval estimate of the population mean is,

- E < < + E

160- 40.88 < < 160 + 40.88

119.11 < < 200.88

(119.11, 200.88)