Question

Are births really evenly distributed across the days of a week? Here are data on 700 births in a hospital:

Day	Sun.	Mon.	Tue.	Wed.	Thu.	Fri.	Sat.
Observed births	89	110	116	104	94	106	81

<Step 1>

Null hypothesis: the births are evenly distributed across the days of the week

Research hypothesis: the births are not equally probable on all days of the week?

<Step 2> Choose α = 5%

<Step 3> Test statistic used: χ2 =

Decision : Reject Ho if χ2 is too big.

<Step 4> Calculations and Conclusion

1˚ Arrange the data in the form of a frequency distribution (See the table above).

2˚ Obtain the expected frequency for each day.

Day	Sun.	Mon.	Tue.	Wed.	Thu.	Fri.	Sat.
Expected births

3˚ Setup a summary table to calculate the Chi-square value.

4˚ Find the degree of freedom.

5˚ Compare the calculated Chi-square value with the appropriate value from the χ2 Table.

The calculated χ2 value is

1. 7.15

2. 8.27    

3. 9.26   

4. 10.10   

5. 11.92

6. 12.76

7. 13.68

8. 16.42

9. 19.12   

10. 23.86.

Q11: (This continues Q10: 2 marks) Find the p-value of the test.

1. Less than 0.25%

2. Between 0.25% and 0.5%

3. Between 0.5% and 1%

4. Between 1% and 2.5%

5. Between 2.5% and 5%

6. Between 5% and 10%

7. Between 10% and 15%

8. Between 15% and 20%

9. Between 15% and 20%

10. Bigger than 20%.

Answer 1

1)

The observed values are

Day	Observed values
Sun	89
Mon	110
Tue	116
Wed	104
Thu	94
Fri	106
Sat	81
Total	700

2)

The expected values are,

Day	Expected values
Sun	700/7=100
Mon	700/7=100
Tue	700/7=100
Wed	700/7=100
Thu	700/7=100
Fri	700/7=100
Sat	700/7=100

3)

The Chi-Square statistic is obtained using the formula,

Day	Observed values, Oi	Expected values, Ei
Sun	89	100	1.21
Mon	110	100	1
Tue	116	100	2.56
Wed	104	100	0.16
Thu	94	100	0.36
Fri	106	100	0.36
Sat	81	100	3.61
		Total	9.26

4)

The degree of freedom for the chi-square distribution = k - 1 = 7 - 1 = 6

where k = number of days

5)

The chi-square critical value is obtained from the chi-square critical value table for significance level = 0.05 and the degree of freedom = 6

Since,

at a 5% significance level. it can be concluded that the null hypothesis is not rejected.

11)

The P-value for the chi-square = 9.26 is obtained from the chi-square distribution table for the degree of freedom = 6

THe p-value si between 15% and 20%