Homework Assignment Number 9 Due Friday by 5pm Points 0 Submitting on paper Avai
ID: 3340285 • Letter: H
Question
Homework Assignment Number 9 Due Friday by 5pm Points 0 Submitting on paper Available Nov 1 at 10am - Dec 15 at 5pm about I month Statistics 2070 Assignment Number 9.Due Friday November 3 Use the following scenario for problem L through 14. What is your favorite restaurant in downtown Laramie? I randomly selected 160 students and asked them the following question, "what is your favorite restaurant in downtown Laramie: Crowbar (C), Born in a Ban (BB) Sweet Melissa's (SW), Grand Avenue Pizza (GAP)?" Results: Crowbar (f-40), Born in a Barn (f 45), Sweet Melissa's (f-10), Grand Avenue Pizza (f-65) 1. What is the variable in this scenario? 2. What are the possible outcomes? 3. What is its level of measurement? 4. What is the appropriate test situation? 5, p = ? 6. Interpret p. 7. HO in English 8. HA in EnglishExplanation / Answer
1)
Variable in this scenario is the favourite restaurant.
2)
There are four possible outcomes which are given as C, BB, SW, and GAP.
3)
The level of measurement for the variable favourite restaurant is nominal.
4)
Here, we have to use Chi square test for checking whether the given four restaurants are equally likely or not.
5)
P-value = 0.0000
6)
P-value = 0.0000 < = 0.01
So, we reject the null hypothesis
7)
Null hypothesis: H0: Given four restaurants are equally likely.
8)
Alternative hypothesis: Ha: Given four restaurants are not equally likely.
9)
The numerical value for E in this problem is 40.
10)
E is the expected frequency of each restaurant.
11)
Null hypothesis: H0: Given four restaurants are equally likely.
Alternative hypothesis: Ha: Given four restaurants are not equally likely.
Test statistic = Chi square = [(O – E)^2/E]
Where, O is observed frequencies and E is expected frequencies.
O
E
(O - E)
(O - E)^2
(O - E)^2/E
C
40
40
0
0
0
BB
45
40
5
25
0.625
SW
10
40
-30
900
22.5
GAP
65
40
25
625
15.625
Total
160
160
38.75
Test statistic = Chi square = [(O – E)^2/E] = 38.75
12)
Critical value = 7.814727764
(by using Chi square table or excel)
13)
Degrees of freedom = n – 1 = 4 – 1 = 3
P-value = 0.0000 (by using Chi square table or excel)
P-value < = 0.01
So, we reject the null hypothesis that given four restaurants are equally likely.
14)
There is insufficient evidence to conclude that given four restaurants are equally likely.
O
E
(O - E)
(O - E)^2
(O - E)^2/E
C
40
40
0
0
0
BB
45
40
5
25
0.625
SW
10
40
-30
900
22.5
GAP
65
40
25
625
15.625
Total
160
160
38.75
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