Question

As part of the study on ongoing fright symptoms due to exposure to horror movies at a young age, the following table was presented to describe the lasting impact these movies have had during bedtime and waking life:

		Waking symptoms
Bedtime symptoms		Yes		No
Yes		34		32
No		32		21

a) What percent of the students have lasting waking-life symptoms? (Round answer to two decimal places.)

b) What percent of the students have both waking-life and bedtime symptoms? (Round answer to two decimal places.)

c) Test whether there is an association between waking-life and bedtime symptoms. State the null and alternative hypotheses. (Use a=.01)

d) State the X² statistic and the P-value. (Round your answers for X² and the P-value to three decimal places.)

X²=

df=

P=

Answer 1

GIVEN:

The 2*2 contingency table that describe the lasting impact that horror movies have had during bedtime and waking life:

		Walking symptoms		Row Total
Bed time symptoms		Yes	No
	Yes	34	32	66
	No	32	21	53
Column Total		66	53	119

(a) PERCENT OF STUDENTS HAVING LASTING WALKING LIFE SYMPTOMS:

Total number of students

Number of students having walking life symptoms

The percent of students having walking life symptoms = Number of students having walking life symptoms / Total number of students

Thus the percent of students having walking life symptoms is 0.55.

(b) PERCENT OF STUDENTS HAVING BOTH WALKING LIFE AND BEDTIME SYMPTOMS:

Total number of students

Number of students having both walking life and bedtime symptoms

The percent of students having both walking life and bedtime symptoms = Number of students having both walking life and bedtime symptoms/ Total number of students

Thus the percent of students having both walking life and bedtime symptoms is 0.29.

(c) CHI SQUARE TEST FOR ASSOCIATION:

HYPOTHESIS:

There is no association between waking-life and bedtime symptoms.

There is association between waking-life and bedtime symptoms.

LEVEL OF SIGNIFICANCE:

TEST STATISTIC:

which follows chi square distribution with degrees of freedom

where

r is the number of rows

c is the number of columns

is the observed frequencies and

is the expected frequencies is given by,

CRITICAL VALUE:

The chi square critical value with degree of freedom at is .

CALCULATION:

OBSERVED FREQUENCIES:

The observed frequency table is given in the question as follows:

		Walking symptoms		Row Total
Bed time symptoms		Yes	No
	Yes	34	32	66
	No	32	21	53
Column Total		66	53	119

EXPECTED FREQUENCIES:

The expected frequencies is given by,

		Walking symptoms
Bed time symptoms		Yes	No
	Yes	(66*66)/119	(66*53)/119
	No	(66*66)/119	(66*53)/119

Thus the expected frequency table is given by:

		Walking symptoms
Bed time symptoms		Yes	No
	Yes	36.61	29.39
	No	36.61	29.39

The chi square test statistic is given by,

  

INFERENCE:

Since the calculated chi square test statistic (3.394) is less than the chi square critical value (6.6349) at , we fail to reject null hypothesis and conclude that there is no association between waking-life and bedtime symptoms.

P VALUE:

The p value for the chi square statistic (3.39) at is .

CONCLUSION ON P VALUE:

Since the calculated p value 0.066 is greater than , we fail to reject null hypothesis and conclude that there is no association between waking-life and bedtime symptoms.