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		35		33
No		33		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: Bedtime symptoms cause waking symptoms.H₀: Waking symptoms cause bedtime symptoms.     H₀: There is a relationship between waking and bedtime symptoms.H₀: There is no relationship between waking and bedtime symptoms.

Alternative Hypothesis:

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.H_a: Waking symptoms cause 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:

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

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

Answer 1

	Column variable
Row variable	yes	no					Total
yes	35	33					68
no	33	18					51
Total	68	51					119

a)  
lasting walking lift symptoms = (35+33)/(35+33+33+18)=   57.14%
  
b)  
both walking life and bed time symptoms = 35/119=   29.41%

c)

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

H_a: There is a relationship between waking and bedtime symptoms

Chi-Square Test

	Expected frequency of a cell = sum of row*sum of column / total sum
	Expected Frequencies
	yes	no					Total
yes	38.8571	29.1429					68
no	29.1429	21.8571					51
Total	68	51					119
	(fo-fe)^2/fe
yes	0.3829	0.5105
no	0.5105	0.6807

Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =   2.085
  
Level of Significance =   0.01
Number of Rows =   2
Number of Columns =   2
Degrees of Freedom=(#row - 1)(#column -1) =   1
  
p-Value =   0.149
Decision:    p value > α , do not reject Ho

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