As part of the study on ongoing fright symptoms due to exposure to horror movies

ID: 3359845 • Letter: A

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 36, 33

No 33, 17

(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.) %

State the null and alternative hypotheses. (Use = 0.01.)

Null Hypothesis:

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

b)H0: Waking symptoms cause bedtime symptoms.

c)H0: Bedtime symptoms cause waking symptoms.

d)H0: There is a relationship between waking and bedtime symptoms.

Alternative Hypothesis:

a)Ha: Bedtime symptoms cause waking symptoms.

b)Ha: Waking symptoms cause bedtime symptoms.

c)Ha: There is a relationship between waking and bedtime symptoms.

d)Ha: There is no relationship between waking and bedtime symptoms. State the 2 statistic and the P-value.

(Round your answers for 2 and the P-value to three decimal places.)

2 =

df =

P =

Conclusion:

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

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

Explanation / Answer

a.
percent of the students have lasting waking-life symptoms = 33/119 = 0.2773 =27.73%

b.
percent of the students have both waking-life and bedtime symptoms = 69/119 =0.5798 = 57.98%

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set up null vs alternative as
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
level of significance, = 0.01
from standard normal table, chi square value at right tailed, ^2 /2 =6.6349
since our test is right tailed,reject Ho when ^2 o > 6.6349
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 2.2748
critical value
the value of |^2 | at los 0.01 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 2 - 1 ) = 1 * 1 = 1 is 6.6349
we got | ^2| =2.2748 & | ^2 | =6.6349
make decision
hence value of | ^2 o | < | ^2 | and here we do not reject Ho
^2 p_value =0.1315

ANSWERS
---------------
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
test statistic: 2.2748
critical value: 6.6349
p-value:0.1315
decision: do not reject Ho

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

Given table data is as below MATRIX col1 col2 TOTALS row 1 36 33 69 row 2 33 17 50 TOTALS 69 50 N = 119 ------------------------------------------------------------------

calculation formula for E table matrix E-TABLE col1 col2 row 1 row1*col1/N row1*col2/N row 2 row2*col1/N row2*col2/N ------------------------------------------------------------------

expected frequecies calculated by applying E - table matrix formulae E-TABLE col1 col2 row 1 40.0084 28.9916 row 2 28.9916 21.0084 ------------------------------------------------------------------

calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 36 40.0084 -4.0084 16.0673 0.4016 33 28.9916 4.0084 16.0673 0.5542 33 28.9916 4.0084 16.0673 0.5542 17 21.0084 -4.0084 16.0673 0.7648 ^2 o = 2.2748

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

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As part of the study on ongoing fright symptoms due to exposure to horror movies

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