Question

A study is performed in a large southern town to determine whether the average amount spent on food per four-person family in the town is significantly different from the national average. A random sample of the weekly grocery bills of two-person families in this town is given in the file. Assume the national average amount spent on food for a four-person family is $150.
a. Identify the null and alternative hypotheses for this situation.
b. Is the sample evidence statistically significant? If so, at what significance levels can you reject the null hypothesis?
c. For which values of the sample mean (e.g., average weekly grocery bill) would you reject the null hypothesis at the 1% significance level? For which values of the sample mean would you reject the null hypothesis at the 10% 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

Step-by-step explanation:

H0: u = 150

H1: u ≠ 150

Total sum = 15933.24

N = 100

Mean = 15933.24/100

= 159.3324

σ² = 25954.03 - (159.3324)²/100

σ² = 556.213

σ = √556.213

σ = 23.8162

Testing hypothesis

t = (bar x - u)/ σ/√n

= 159.3324-150/23.8162/√100

= 3.91

We will have a p value of 0.02

0.0002 < 0.01

We reject null hypothesis at 1% level of significance

C. Mean = 159.3324

Se= 2.3936

Df = 100-1 = 99

Critical value at 0.01 = +-2.626

T = x-u/s.e

= -2.626 =( x -150)/2.3936

When we cross multiply and solve this

X = 143.714 for the lower tail

2.626 = (x-159)/2.3936

= 156.286 for upper tail.

We therefore reject H0 at

Bar X <= 143.71

Bar X >= 156.286

At 10%

Critical t = 1.660

-1.660 = (x - 150)/2.3936

Solving this ,

X =146.02 at the lower tail

1.660 = (x-150)/2.3936

X = 153.97

We reject H0 at

X<= 146.03

X>=153.97