Data were collected on the amount spent by 64 customers for lunch at a major Houston restaurant. These data are contained in the

Question

3 years ago

6

Data were collected on the amount spent by 64 customers for lunch at a major Houston restaurant. These data are contained in the

WEBfile named Houston. Based upon past studies the population standard deviation is known with = $6.
Click on the webfile logo to reference the data.

Round your answers to two decimal places.

a. At 99% confidence, what is the margin of error?

b. Develop a 99% confidence interval estimate of the mean amount spent for lunch.

____ to ____

Amount
20.50
14.63
23.77
29.96
29.49
32.70
9.20
20.89
28.87
15.78
18.16
12.16
11.22
16.43
17.66
9.59
18.89
19.88
23.11
20.11
20.34
20.08
30.36
21.79
21.18
19.22
34.13
27.49
36.55
18.37
32.27
12.63
25.53
27.71
33.81
21.79
19.16
26.35
20.01
26.85
13.63
17.22
13.17
20.12
22.11
22.47
20.36
35.47
11.85
17.88
6.83
30.99
14.62
18.38
26.85
25.10
27.55
25.87
14.37
15.61
26.46
24.24
16.66
20.85

Mathematics

1 answer:

galina1969 [7] · Answer 1 · 2022-04-10T00:43:24+03:00

Answer:

a. The margin of error is 2.29.

b. 19.23 to 23.81

Step-by-step explanation:

The sample size is n=64.

We start by calculating the sample mean and standard deviation with the following formulas:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}$

The sample mean is M=21.52.

The sample standard deviation is s=6.89.

We have to calculate a 99% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{6.89}{\sqrt{64}}=\dfrac{6.89}{8}=0.861$