In: Math
Use the following info for Questions 6-10.
Classic Golf, Inc., manages five golf courses in Florida. The director wishes to compare the rounds of golf played at the five courses, so he counted the rounds for a sample week. At the .05 significance level, is there an actual difference in the numbers of rounds playd at the five courses?
(Part of the data analysis is provided for you.)
Course #Rounds played (fe) (fo - fe) (fo - fe)2 (fo -fe)2/fe
Dogwood 124 104 20 400 3.846
Forsythia 74 104 -30 900 8.654
Starburst 104
Azalia 98
Sunflower 120
Golf Course Question #6: how was (fe) calculated?
A. |
It's calculated from the #rounds played at Starburst. |
|
B. |
1040/10 |
|
C. |
520/5 |
|
D. |
20.8 x 5 |
QUESTION 7
Golf Course Question #7: Calculate ∑( fo - fe) 2
A. |
1590 |
|
B. |
1592 |
|
C. |
1594 |
|
D. |
1596 |
QUESTION 8
Golf Course Question #8: What is the critical value of Chi-Square?
A. |
9.263 |
|
B. |
9.488 |
|
C. |
11.070 |
|
D. |
15.211 |
Golf Course Question #9: What is the computed value of Chi-Square?
A. |
15.308 |
|
B. |
16.517 |
|
C. |
17.342 |
|
D. |
18.166 |
QUESTION 10
Golf Course Question #10: What is your decision?
A. |
#Rounds played is the same at all 5 |
|
B. |
#Rounds played is not the same at all 5 |
|
C. |
Cannot be determined |
6)
total frequency= 520
expected frequncy,E = expected proportions*total frequency
=520/5 = 104
answer: 520/5
7)
observed frequencey, fo | expected proportion | expected frequency,fe | (fo-fe)^2 | (fO-fE)²/fE | |
124 | 0.200 | 104.00 | 400.00 | 3.846 | |
74 | 0.200 | 104.00 | 900.00 | 8.654 | |
104 | 0.200 | 104.00 | 0.00 | 0.000 | |
98 | 0.200 | 104.00 | 36.00 | 0.346 | |
120 | 0.200 | 104.00 | 256.00 | 2.462 |
answer: Σ(fo-fe)^2 = 1592
8)
level of significance, α= 0.05
Degree of freedom=k-1= 5 -
1 = 4
Critical value = 9.488 [ Excel
function: =chisq.inv.rt(α,df) ]
9)
chi square test statistic,X² = Σ(fo-fe)²/fe =
15.308
10)
since, test stat> criitcal value, reject Ho
answer: B.
#Rounds played is not the same at all 5