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

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 Yes No

Yes 34 33

Bedtime symptom No 32 20

(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 = 0.01.) Null Hypothesis: H0: There is no relationship between waking and bedtime symptoms. H0: There is a relationship between waking and bedtime symptoms. H0: Bedtime symptoms cause waking symptoms. H0: Waking symptoms cause bedtime symptoms. Alternative Hypothesis: Ha: There is a relationship between waking and bedtime symptoms. Ha: There is no relationship between waking and bedtime symptoms. Ha: Bedtime symptoms cause waking symptoms. Ha: Waking symptoms cause 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: We do not have enough evidence to conclude that there is a relationship. We have enough evidence to conclude that there is a relationship.

Explanation / Answer

a.
percent of the students have lasting waking-life symptoms = 34/119 = 28.57%

b.
percent of the students have both waking-life and bedtime symptoms = 66/119 = 55.46%
c.

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 34 33 67 row 2 32 20 52 TOTALS 66 53 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 37.1597 29.8403 row 2 28.8403 23.1597 ------------------------------------------------------------------

calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 34 37.1597 -3.1597 9.9837 0.2687 33 29.8403 3.1597 9.9837 0.3346 32 28.8403 3.1597 9.9837 0.3462 20 23.1597 -3.1597 9.9837 0.4311 ^2 o = 1.3806 ------------------------------------------------------------------

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 = 1.3806
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| =1.3806 & | ^2 | =6.6349
make decision
hence value of | ^2 o | < | ^2 | and here we do not reject Ho
^2 p_value =0.24


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: 1.3806
critical value: 6.6349
p-value:0.240
decision: do not reject Ho

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