A consulting firm asks people in many different cultures how they felt about the

Question

A consulting firm asks people in many different cultures how they felt about the following statement: I try to avoid eating fast foods. In a random sample of 800 respondents, 408 people were 35 years old or younger, and, of those, 201 agreed (completely or somewhat) with the statement. Of the 392 people over 35 years old, 244 people agreed with the statement. Complete parts a and b below. a) Is there evidence that the percentage of people avoiding fast food is different in the two age groups? State the null and alternative hypotheses. Choose the correct answer below. A. Upper H 0H0: The proportions are different. Upper H Subscript Upper AHA: The proportions are the same. B. Upper H 0H0: The proportion of respondents 35 years old or younger is less than or equal to the proportion of respondents over 35 years old. Upper H Subscript Upper AHA: The proportion of respondents 35 years old or younger is greater than the proportion of respondents over 35 years old. C. Upper H 0H0: The proportions are the same. Upper H Subscript Upper AHA: The proportions are different. D. Upper H 0H0: The proportion of respondents 35 years old or younger is greater than or equal to the proportion of respondents over 35 years old. Upper H Subscript Upper AHA: The proportion of respondents 35 years old or younger is less than the proportion of respondents over 35 years old. Determine the chi squared2 statistic. chi squared2equals= nothing (Round to two decimal places as needed.) Find the P-value. P-valueequals= nothing (Round to three decimal places as needed.) Is there evidence that the percentage of people avoiding fast food is different in the two age groups? Reject Fail to reject Upper H 0H0 at the 0.05 level. There is insufficient sufficient evidence to conclude that the proportions of the two groups are the same. different. b) Give a 9090% confidence interval for the difference p1minusp2, where p1 and p2 are the proportions of respondents 35 years old or younger and respondents over 35 years old, respectively. The 90% confidence interval is left parenthesis nothing comma nothing right parenthesis , . (Round to four decimal places as needed. Use ascending order.)

Explanation / Answer

a.

i.

Given that,
sample one, x1 =201, n1 =408, p1= x1/n1=0.49
sample two, x2 =244, n2 =392, p2= x2/n2=0.62
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.493-0.62)/sqrt((0.556*0.44(1/408+1/392))
zo =-3.69
| zo | =3.69
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =3.694 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -3.6941 ) = 0.0002
hence value of p0.05 > 0.0002,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: -3.69
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.0002

we have enough evidence to support the claim that the percentage of people avoiding fast food is different in the two age groups

ii.

b.

TRADITIONAL METHOD
given that,
sample one, x1 =201, n1 =408, p1= x1/n1=0.4926
sample two, x2 =244, n2 =392, p2= x2/n2=0.6224
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.4926*0.5074/408) +(0.6224 * 0.3776/392))
=0.0348
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
margin of error = 1.645 * 0.0348
=0.0573
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.4926-0.6224) ±0.0573]
= [ -0.1871 , -0.0725]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =201, n1 =408, p1= x1/n1=0.4926
sample two, x2 =244, n2 =392, p2= x2/n2=0.6224
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.4926-0.6224) ± 1.645 * 0.0348]
= [ -0.1871 , -0.0725 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ -0.1871 , -0.0725] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the difference between
true population mean P1-P2

Given table data is as below MATRIX col1 col2 TOTALS row 1 408 201 609 row 2 392 244 636 TOTALS 800 445 N = 1245 ------------------------------------------------------------------

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 1 391.33 217.67 row 2 408.67 227.33 ------------------------------------------------------------------

calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 408 391.33 16.67 277.89 0.71 201 217.67 -16.67 277.89 1.28 392 408.67 -16.67 277.89 0.68 244 227.33 16.67 277.89 1.22 ^2 o = 3.89 ------------------------------------------------------------------

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 =3.84
since our test is right tailed,reject Ho when ^2 o > 3.84
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 3.89
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 3.84
we got | ^2| =3.89 & | ^2 | =3.84
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0.05


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: 3.89
critical value: 3.84
p-value:0.05
decision: reject Ho

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