As part of the study on ongoing fright symptoms due to exposure to horror movies
ID: 3365981 • 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:
(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 2 statistic and the P-value. (Round your answers for 2 and the P-value to three decimal places.)
symptoms Bedtime symptoms Yes No Yes 36 34 No 33 16
Explanation / Answer
a.
percent of the students have lasting waking-life symptoms = 36/119 =0.3025=30.25%
b.
percent of the students have both waking-life and bedtime symptoms = 69/119 = 0.5798 = 57.98%
c.
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 40.588 29.412 row 2 28.412 20.588 ------------------------------------------------------------------
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 40.588 -4.588 21.05 0.519 34 29.412 4.588 21.05 0.716 33 28.412 4.588 21.05 0.741 16 20.588 -4.588 21.05 1.022 ^2 o = 2.998 ------------------------------------------------------------------
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, = 0.05
from standard normal table, chi square value at right tailed, ^2 /2 =3.841
since our test is right tailed,reject Ho when ^2 o > 3.841
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 2.998
critical value
the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 2 - 1 ) = 1 * 1 = 1 is 3.841
we got | ^2| =2.998 & | ^2 | =3.841
make decision
hence value of | ^2 o | < | ^2 | and here we do not reject Ho
^2 p_value =0.083
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: 2.998
critical value: 3.841
p-value:0.083
decision: do not reject Ho
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.