A recent study of undergraduates looked at gender differences in dieting trends.

Question

A recent study of undergraduates looked at gender differences in dieting trends. There were 188 women and 101 men who participated in the survey. The table below summarizes whether a student tried a low-fat diet or not by gender:

(a) Fill in the missing cells of the table.

(b) Summarize the data numerically. What percent of each gender has tried low-fat diets? (Round your answers to two decimal places.)

(c) Test that there is no association between gender and the likelihood of trying a low-fat diet. (Round your 2 to three decimal places, and round your P-value to four decimal places.)

     Gender Tried low-fat diet Women      Men Yes 38 8 No

Explanation / Answer

Given table data is as below

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calculation formula for E table matrix

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expected frequecies calculaated by applying E - table matrix formulae

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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above

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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, = 0.05
from standard normal table, chi square value at right tailed, ^2 /2 =3.8415
since our test is right tailed,reject Ho when ^2 o > 3.8415
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 7.4174
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 3.8415
we got | ^2| =7.4174 & | ^2 | =3.8415
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0.0065
ANSWERS
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null, Ho: there is no association between gender and the likelihood of trying a low-fat diet
alternative, H1: association between gender and the likelihood of trying a low-fat diet
test statistic: 7.4174
critical value: 3.8415
p-value:0.0065
decision: reject Ho                              

we claim that association between gender and the likelihood of trying a low-fat diet

col1 col2 row 1 38 8 46 row 2 150 93 243 TOTALS 188 101 N = 289

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