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		36		33
No		32		18

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


(b) What percent of the students have both waking-life and bedtime symptoms? (Round your 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 α = 0.01.)

Null Hypothesis:

H₀: There is no relationship between waking and bedtime symptoms.H₀: Waking symptoms cause bedtime symptoms.     H₀: There is a relationship between waking and bedtime symptoms.H₀: Bedtime symptoms cause waking symptoms.

Alternative Hypothesis:

H_a: Waking symptoms cause bedtime symptoms.H_a: Bedtime symptoms cause waking symptoms.     H_a: There is no relationship between waking and bedtime symptoms.H_a: There is a relationship between waking and bedtime symptoms.

State the χ² statistic and the P-value. (Round your answers for χ² and the P-value to three decimal places.)

χ2	=
df	=
P	=

Conclusion:

a. We do not have enough evidence to conclude that there is a relationship.

b. We have enough evidence to conclude that there is a relationship.    

Answer 1

a) Required percentage = (36 + 32)/(36 + 33+ 32 + 18) = 57.14%
b) Required percentage = 36/(36 + 34 + 32 + 17) = 30.25%

c)

H₀: There is no relationship between waking and bedtime symptoms.H₀: Waking symptoms cause bedtime symptoms.     

H_a: There is a relationship between waking and bedtime symptoms

Chi-Square Test
	Observed Frequencies
	column
row	yes	no					Total
Yes	36	33					69
No	32	18					50
Total	68	51					119
	Expected frequency of a cell = sum of row*sum of column / total sum
	Expected Frequencies
	yes	no					Total
Yes	39.4286	29.5714					69
No	28.5714	21.4286					50
Total	68	51					119
	(fo-fe)^2/fe
Yes	0.2981	0.3975
No	0.4114	0.5486

Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =   1.656  
      
Level of Significance =   0.01  
Number of Rows =   2  
Number of Columns =   2  
Degrees of Freedom=(#row - 1)(#column -1) =   1  
      
p-Value =   0.1982                                [Excel function:   =CHISQ.DIST.RT(χ²,df) ]
Decision:    p value > α ,   do not reject Ho  

a. We do not have enough evidence to conclude that there is a relationship.