Mr. Acosta, a sociologist, is doing a study to see if there is a relationship between the age of a young adult (18 to 35 years old) and the type of movie preferred. A random sample of 93 adults revealed the following data. Test whether age and type of movie preferred are independent at the 0.05 level.
Person's Age | ||||
Movie | 18-23 yr | 24-29 yr | 30-35 yr | Row Total |
Drama | 7 | 16 | 11 | 34 |
Science Fiction | 11 | 12 | 7 | 30 |
Comedy | 9 | 7 | 13 | 29 |
Column Total | 27 | 35 | 31 | 93 |
(a) What is the level of significance?
State the null and alternate hypotheses.
H0: Age and movie preference are not
independent.
H1: Age and movie preference are
independent.H0: Age and movie preference are
independent.
H1: Age and movie preference are not
independent. H0: Age and
movie preference are not independent.
H1: Age and movie preference are not
independent.H0: Age and movie preference are
independent.
H1: Age and movie preference are
independent.
(b) Find the value of the chi-square statistic for the sample.
(Round the expected frequencies to at least three decimal places.
Round the test statistic to three decimal places.)
Are all the expected frequencies greater than 5?
YesNo
What sampling distribution will you use?
chi-squarenormal binomialStudent's tuniform
What are the degrees of freedom?
(c) Find or estimate the P-value of the sample test
statistic.
P-value > 0.1000.050 < P-value < 0.100 0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis of independence?
Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis. Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.
(e) Interpret your conclusion in the context of the
application.
At the 5% level of significance, there is sufficient evidence to conclude that age of young adult and movie preference are not independent.At the 5% level of significance, there is insufficient evidence to conclude that age of young adult and movie preference are not independent.
Given table data is as below
MATRIX | col1 | col2 | col3 | TOTALS |
row 1 | 7 | 16 | 11 | 34 |
row 2 | 11 | 12 | 7 | 30 |
row 3 | 9 | 7 | 13 | 29 |
TOTALS | 27 | 35 | 31 | N = 93 |
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calculation formula for E table matrix
E-TABLE | col1 | col2 | col3 |
row 1 | row1*col1/N | row1*col2/N | row1*col3/N |
row 2 | row2*col1/N | row2*col2/N | row2*col3/N |
row 3 | row3*col1/N | row3*col2/N | row3*col3/N |
------------------------------------------------------------------
expected frequecies calculated by applying E - table matrix
formulae
E-TABLE | col1 | col2 | col3 |
row 1 | 9.871 | 12.796 | 11.333 |
row 2 | 8.71 | 11.29 | 10 |
row 3 | 8.419 | 10.914 | 9.667 |
------------------------------------------------------------------
calculate chisquare test statistic using given observed
frequencies, calculated expected frequencies from above
Oi | Ei | Oi-Ei | (Oi-Ei)^2 | (Oi-Ei)^2/Ei |
7 | 9.871 | -2.871 | 8.243 | 0.835 |
16 | 12.796 | 3.204 | 10.266 | 0.802 |
11 | 11.333 | -0.333 | 0.111 | 0.01 |
11 | 8.71 | 2.29 | 5.244 | 0.602 |
12 | 11.29 | 0.71 | 0.504 | 0.045 |
7 | 10 | -3 | 9 | 0.9 |
9 | 8.419 | 0.581 | 0.338 | 0.04 |
7 | 10.914 | -3.914 | 15.319 | 1.404 |
13 | 9.667 | 3.333 | 11.109 | 1.149 |
chisqr^2 o = 5.787 |
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set up null vs alternative as
null, Ho: age and movie preference are independent.
alternative, H1: age and movie preference are not
independent.
level of significance, alpha = 0.05
from standard normal table, chi square value at right tailed,
chisqr^2 alpha/2 =9.488
since our test is right tailed,reject Ho when chisqr^2 o >
9.488
we use test statistic chisqr^2 o = Σ(Oi-Ei)^2/Ei
from the table , chisqr^2 o = 5.787
critical value
the value of |chisqr^2 alpha| at los 0.05 with d.f (r-1)(c-1)= ( 3
-1 ) * ( 3 - 1 ) = 2 * 2 = 4 is 9.488
we got | chisqr^2| =5.787 & | chisqr^2 alpha | =9.488
make decision
hence value of | chisqr^2 o | < | chisqr^2 alpha | and here we
do not reject Ho
chisqr^2 p_value =0.216
------------------------------------------------------------------------------
(a) null, Ho: age and movie preference are independent.
alternative, H1: age and movie preference are not
independent.
(b) test statistic: 5.787
yes
chi-square
(c) p-value:0.216, P-value > 0.100
critical value: 9.488
(d) since the P-value > α, we fail to reject the null
hypothesis
(e) at the 5% level of significance, there is insufficient evidence
to conclude that age of young adult and movie preference are not
independent
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