Question

A study is performed in a large southern town to determine whether the average amount spent on fod per four person family in the town is significantly different from the national average. Assume the national average amount spent on food for a four- person family is $150.

A what is the null and alternative hypothesis?

b. Is the sample evidence significant? significance level?

Family	Weekly food expense
1	$198.23
2	$143.53
3	$207.48
4	$134.55
5	$182.01
6	$189.84
7	$170.36
8	$163.72
9	$155.73
10	$203.73
11	$191.19
12	$172.66
13	$154.25
14	$179.03
15	$130.29
16	$170.73
17	$194.50
18	$171.14
19	$175.19
20	$177.25
21	$166.62
22	$135.54
23	$141.18
24	$158.48
25	$159.78
26	$157.42
27	$98.40
28	$181.63
29	$128.45
30	$190.84
31	$154.04
32	$190.22
33	$161.48
34	$113.42
35	$148.83
36	$197.68
37	$135.49
38	$146.72
39	$176.62
40	$154.60
41	$178.39
42	$186.32
43	$157.94
44	$116.35
45	$136.81
46	$195.58
47	$129.44
48	$146.84
49	$165.63
50	$158.97
51	$210.00
52	$175.46
53	$159.69
54	$154.56
55	$152.95
56	$177.30
57	$129.23
58	$127.40
59	$167.48
60	$183.83
61	$157.39
62	$163.24
63	$165.01
64	$137.43
65	$177.37
66	$142.68
67	$150.04
68	$161.44
69	$166.13
70	$190.96
71	$187.19
72	$116.63
73	$159.73
74	$159.64
75	$142.44
76	$153.03
77	$143.12
78	$156.35
79	$182.70
80	$129.03
81	$119.06
82	$137.99
83	$144.20
84	$183.51
85	$169.67
86	$134.66
87	$202.94
88	$143.43
89	$170.52
90	$139.53
91	$159.31
92	$134.77
93	$165.48
94	$127.20
95	$168.16
96	$125.39
97	$167.96
98	$178.64
99	$134.38
100	$111.87

Answer 1

A. To test

H₀: The average amount spent per week on food per four person family in the town is $150. Vs H_a : The average amount spent per week on food per four person family in the town is different from the national average, $150

i.e. H₀: Vs H_a: where = average amount spent per week on food per four person family.

B. To compare the sample mean to the population mean, the appropriate test statistic would be Z statistic given by,

Comparing the t score obtained with the critical value from the t tables for n-1 = 100-1 = 99 degrees of freedom, we arrive at a decision whether or not to reject the null hypothesis.

From the given data,

, , and n = 100.

Substituting in the test statistc,

Comparing t = 3.90 with the critical value t_0.05,99 = 1.984, (Using excel function TINV(0.05,99), we find that t = 3.90 > 1.984.It implies that there is not sufficient evidence to support the null hypothesis based on the given data.Hence H₀ may be rejected at 5% level of significance.i.e. the sample evidence is significant at a significance level of 5%.

( We may also compare t = 3.90 with critical value t_0.01,99 = 2.626 to claim the significance of the sample evidence at a significance level of 1%.We would reach at the same conclusion as above-the sample evidence is significant at a significance level of 1%.)

We may conclude that the average amount spent per week on food per four person family in the town is different from the national average, $150