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Gallup recently conducted a survey about the proportion of Americans who use e-cigarettes (vape). The following...

Gallup recently conducted a survey about the proportion of Americans who use e-cigarettes (vape). The following example is based on hypothetical data that is meant to be similar to the data collected by Gallup.

The following table considers survey data about annual household income and whether or not a person vapes.

Problem 2. We now wish to decide if there “use of e-cigarettes” and “income category” are dependent. To assist with this process, the table from before has been augmented with most of the expected frequencies (listed in parentheses):

<$35,000

$35,000-$99,999

$100,000+

Total

Vape

47 (34)

57 (58)

19 (31)

123

Do Not Vape

381 (394)

659 (???)

362 (???)

1402

Total

428

716

381

1525

  1. Find the missing two expected frequencies, labeled as (???). For credit, be sure to show your work.
  2. From a brief analysis of the frequencies, do you believe that there is convincing evidence of a dependence relationship between income category and whether or not a person vapes? Explain.
  3. Based on the given information, the test statistic computes to be 10.47. Based on this, what is the conclusion of the hypothesis test at the 0.05 significance level? Be sure to justify your answer by using either the p-value or critical value method.

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Answer #1

2.

Given table data is as below
MATRIX col1 col2 col3 TOTALS
row 1 34 58 31 123
row 2 394 658 350 1402
TOTALS 428 716 381 N = 1525
------------------------------------------------------------------

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
------------------------------------------------------------------

expected frequencies calculated by applying E - table matrix formula
E-TABLE col1 col2 col3
row 1 34.521 57.75 30.73
row 2 393.479 658.25 350.27
------------------------------------------------------------------

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 34.521 -0.521 0.271 0.008
58 57.75 0.25 0.063 0.001
31 30.73 0.27 0.073 0.002
394 393.479 0.521 0.271 0.001
658 658.25 -0.25 0.063 0
350 350.27 -0.27 0.073 0
ᴪ^2 o = 0.012

------------------------------------------------------------------

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, α = 0.05
from standard normal table, chi square value at right tailed, ᴪ^2 α/2 =5.991
since our test is right tailed,reject Ho when ᴪ^2 o > 5.991
we use test statistic ᴪ^2 o = Σ(Oi-Ei)^2/Ei
from the table , ᴪ^2 o = 0.012
critical value
the value of |ᴪ^2 α| at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 3 - 1 ) = 1 * 2 = 2 is 5.991
we got | ᴪ^2| =0.012 & | ᴪ^2 α | =5.991
make decision
hence value of | ᴪ^2 o | < | ᴪ^2 α | and here we do not reject Ho
ᴪ^2 p_value =0.994


ANSWERS
---------------
a.

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
c.

test statistic: 0.012
critical value: 5.991
p-value:0.994
decision: do not reject Ho

b.

we do not have enough evidence to support the claim that there is convincing evidence of a dependence relationship between income category.

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