Question

Find the p-value of the test.

A) Less than 0.25%

B) Between 0.25% and 0.5%

C) Between 0.5% and 1%

D) Between 1% and 2.5%

E) Between 2.5% and 5%

F) Between 5% and 10%

G) Between 10% and 15%

H) Between 15% and 20%

I) Between 15% and 20%

J) Bigger than 20%.

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

A) 7.15

B) 8.27

C) 9.26

D) 10.10

E) 11.92

F) 12.76

G) 13.68

H) 16.42

I) 19.12

J) 23.86.

Answer 1

<Step 1>

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

Research hypothesis: Alternate Hypothesis : H1 : the births are not equally probable on all days of the week

<Step 2>

Choose α = 5%

1˚ Arrange the data in the form of a frequency distribution

2˚ Obtain the expected frequency for each day.

Day	O: Observed births	E:Expected Frequency
Mon	89	700/7=100
Tue	110	700/7=100
Wed	116	700/7=100
Thu	104	700/7=100
Fri	94	700/7=100
Sat	106	700/7=100
Sun	81	700/7=100
Total	700	700

O : Observed Frequency
E: Expected Frequency

O	E	O-E	(O-E)2	(O-E)2/E
89	100	-11	121	1.21
110	100	10	100	1
116	100	16	256	2.56
104	100	4	16	0.16
94	100	-6	36	0.36
106	100	6	36	0.36
81	100	-19	361	3.61
			Total	9.26

The calculated value is

C) 9.26

4. Degrees of freedom = Number of days - 1 =7-1 =6

Degrees of freedom = 6

For = 0.05 ; Critical value of for 6 degrees of freedom = 12.59

As calculated value : 9.26 < 12.59 : Fail to reject the null hypothesis.

for 6 degrees of freedom,

i.e

p-value of the test.

H) Between 15% and 20%

I) Between 15% and 20%