2. At a certain college campus, students can choose to enter the cafeteria using
ID: 132502 • Letter: 2
Question
2. At a certain college campus, students can choose to enter the cafeteria using one of A custodian noticed that door #3 was always getting broken and suggested that more students use that door because it happens to have a hands-free opener. Science-minded stu counted the of people entering each door to see if door W3 is favored by students Door Number Number of Students who entered 60 56 84 Data showdoor #3 is used more often, but is the difference significant? a. Formulate a null hypothesis and an alternative hypothesis: Ho: ??: b. Formulate a prediction based on the null hypothesis: Conduct a Chi-squared test. Show your work for full credit d. Analyze your results using the chi-squared table at the end of the document, and write a conclusion (be sure to address whether the null hypothesis is supported or rejected).Explanation / Answer
For door experiment:
a) Formulation of null and alternative hypothesis
H0 = Door 3 is favored by students.
H1 = Door 3 is not favored by students.
b) Prediction based on null hypothesis
If all doors are equally favored, number of students entering through each door would be 33.33%. If null hypothesis is correct, Number of students entering from door 3 will be more than 33%
c) Chi square test:
Total students entering cafeteria = 200 students
Students entered through door 3 = 84 (observed, O)
Expected students to enter (more than 33.33% or at least 34%) = 34% of 200 = 0.34 x 200 = 68 (Expected, E)
Now, for chi square test, O-E = 84-68 = 16
(O-E)2 = (16)2 = 256
Chi square = (O-E)2/E = 256/68 = 3.764
d) Analysis of result on the basis of chi squared table:
The good fit is observed if the value is under 3.841 as indicated in the table,
The calculated chi square value is 3.764 which is less than 3.841
So, the null hypothesis is accepted at 0.05 level of significance
Conclusion: Door 3 is favored by students.
For kitten experiment:
a) Formulation of null and alternative hypothesis
H0 = The gene under consideration is recessive
H1 = The gene under consideration is not recessive.
b) Prediction based on null hypothesis
The kittens born with the birth defect will be approximately 25% of total population
c) Chi square test:
Total population =128 kittens
Kittens with birth defects = 44 (observed, O)
Expected kittens with birth defects = 25% of 128 = 0.25 x 128 = 32 (Expected, E)
Now, for chi square test, O-E = 44-32 = 12
(O-E)2 = (12)2 = 144
Chi square = (O-E)2/E = 144/32 = 4.5
d) Analysis of result on the basis of chi squared table:
The good fit is observed if the value is under 3.841 as indicated in the table,
The calculated chi square value is 4.5 which is more than 3.841
So, the null hypothesis is rejected at 0.05 level of significance
Conclusion: The gene under consideration is not recessive.
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