In: Statistics and Probability
1. At the emergency room at the one hospital, the nurses are convinced that the number of patients that they see during the midnight shift is affected by the phases of the moon. The doctors think this is an old wives’ tale. The nurses decided to test their theory at the 0.10 level of significance level by recording the number of patients they see during the midnight shift for each moon phase over the course of one lunar cycle. The results are in the table below. ( can do by hand or using JMP)
ER patients During the Midnight Shift |
|||||
New Moon |
1st Quarter |
Full Moon |
3 rd Quarter |
||
Number of patients |
85 |
66 |
97 |
68 |
|
2. A local fast-food restaurant serves buffalo wings. The restaurant’s managers notice that they normally sell the following proportions of flavors for their wings: 20% Spicy Garlic, 50% Classic Medium,10% Teriyaki, 10% Hot BBQ and 10% Asian Zing. After running a campaign to promote their nontraditional specialty wings, they want to know of the campaign has made an impact. The results after 10 days are listed in the following table. ( By hand or using JMP)
Buffalo Spicy Wings |
||
Number Sold |
||
Spicy Garlic |
251 |
|
Classic Medium |
630 |
|
Teriyaki |
115 |
|
Hot BBQ |
141 |
|
Asian ZIng |
121 |
3. According to an online survey b Harris Interactive for job site CareerBuilder.com more than half of IT workers say they have fallen asleep at work. The same is also true for government workers. Assume the following contingency table is representative of the survey results.
Job Category |
||
Slept of the Job? |
IT professional |
Government Professional |
Yes |
155 |
256 |
No |
145 |
144 |
Chi-square Goodness-of Fit Chi-square of Independence
H0:
HA:
Thank you! I'll like your answer!
1)
a)
Ho: two varaibles are as expected
H1:two variables are not as expected
b)
observed frequencey, O | expected proportion | expected frequency,E | (O-E)²/E | ||
85 | 0.250 | 79.00 | 0.456 | ||
66 | 0.250 | 79.00 | 2.139 | ||
97 | 0.250 | 79.00 | 4.101 | ||
68 | 0.250 | 79.00 | 1.532 |
c)
chi square test statistic,X² = Σ(O-E)²/E =
8.228
level of significance, α= 0.1
Degree of freedom=k-1= 4 -
1 = 3
P value = 0.0415 [ excel function:
=chisq.dist.rt(test-stat,df) ]
Decision: P value < α, Reject Ho
Conclusion: New moon effects the
results.
---------------------------
2)
observed frequencey, O | expected proportion | expected frequency,E | (O-E)²/E | ||
251 | 0.200 | 251.60 | 0.001 | ||
630 | 0.500 | 629.00 | 0.002 | ||
115 | 0.100 | 125.80 | 0.927 | ||
141 | 0.100 | 125.80 | 1.837 | ||
121 | 0.100 | 125.80 | 0.183 |
chi square test statistic,X² = Σ(O-E)²/E =
2.950
level of significance, α= 0.05
Degree of freedom=k-1= 5 -
1 = 4
P value = 0.5662 [ excel function:
=chisq.dist.rt(test-stat,df) ]
Decision: P value >α , Do not reject Ho
Conclusion: We can conclude that the promotional campaign has not made any difference in the proportions of flavors sold
Thanks in advance!
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