Question

In: Statistics and Probability

We interviewed two groups of 50 college students respectively from UIC and DePaul to know if...

We interviewed two groups of 50 college students respectively from UIC and DePaul to know if they rather watch NFL Football vs. some other sport on Sunday.

Actual Data

UIC

DePaul

Row Total

Football

15

25

40

Other Sport

35

25

60

Total

50

50

100

Based on the statistics above, we need to determine if there is a relationship between the university of a student and watching Football. Answer to the following questions to arrive at the conclusion.

  • What are Ho and Ha?
  • What are the chi-square values?
  • What is the degrees of freedom?
  • What is the p-value?
  • For alpha=0.05, what would be our statistical conclusion?
  • Write out the full conclusion in English

Please show work

Solutions

Expert Solution

Solution:

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H0: There is no relationship between the University of a Student and watching Football.

Alternative hypothesis: Ha: There is a relationship between the university of a student and watching Football.

We assume/given level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

We are given

Number of rows = r = 2

Number of columns = c = 2

Degrees of freedom = df = (r – 1)*(c – 1) = 1*1 = 1

α = 0.05

Critical value = 3.841459

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies

Column variable

Row variable

UIC

DePaul

Total

Football

15

25

40

Other Sport

35

25

60

Total

50

50

100

Expected Frequencies

Column variable

Row variable

UIC

DePaul

Total

Football

20

20

40

Other Sport

30

30

60

Total

50

50

100

Calculations

(O - E)

-5

5

5

-5

(O - E)^2/E

1.25

1.25

0.833333

0.833333

Test Statistic = Chi square = ∑[(O – E)^2/E] = 4.166667

χ2 statistic = 4.166667

P-value = 0.041227

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that there is a relationship between the University of a Student and watching Football.


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