Question

data were collected on the amount spent by 64 consumers for lunch at a major Houston restaurant. these data are contained in the file named Houston. based upon past studies the population standard deviation is known with $9.

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

Answer 1

For the given data mean is

Create the following table.

data	data-mean	(data - mean)2
20.50	-1.02	1.0404
14.63	-6.89	47.4721
23.77	2.25	5.0625
29.96	8.44	71.2336
29.49	7.97	63.5209
32.70	11.18	124.9924
9.20	-12.32	151.7824
20.89	-0.63	0.3969
28.87	7.35	54.0225
15.78	-5.74	32.9476
18.16	-3.36	11.2896
12.16	-9.36	87.6096
11.22	-10.3	106.09
16.43	-5.09	25.9081
17.66	-3.86	14.8996
9.59	-11.93	142.3249
18.89	-2.63	6.9169
19.88	-1.64	2.6896
23.11	1.59	2.5281
20.11	-1.41	1.9881
20.34	-1.18	1.3924
20.08	-1.44	2.0736
30.36	8.84	78.1456
21.79	0.27	0.0729
21.18	-0.34	0.1156
19.22	-2.3	5.29
34.13	12.61	159.0121
27.49	5.97	35.6409
36.55	15.03	225.9009
18.37	-3.15	9.9225
32.27	10.75	115.5625
12.63	-8.89	79.0321
25.53	4.01	16.0801
27.71	6.19	38.3161
33.81	12.29	151.0441
21.79	0.27	0.0729
19.16	-2.36	5.5696
26.35	4.83	23.3289
20.01	-1.51	2.2801
26.85	5.33	28.4089
13.63	-7.89	62.2521
17.22	-4.3	18.49
13.17	-8.35	69.7225
20.12	-1.4	1.96
22.11	0.59	0.3481
22.47	0.95	0.9025
20.36	-1.16	1.3456
35.47	13.95	194.6025
11.85	-9.67	93.5089
17.88	-3.64	13.2496

Find the sum of numbers in the last column to get.

So

For 99% CI, z value is 2.58

a. So Margin of Error is

b. Hence CI is