1. A nutrition researcher wishes to compare two methods of obtaining dietary inf
ID: 3315282 • Letter: 1
Question
1. A nutrition researcher wishes to compare two methods of obtaining dietary information for the purpose of estimating caloric intake. Method 1 is a 24-hr recall (respondents record all they ate during last 24 hours). Method 2 is a dietary survey (respondents are asked questions about what they typically eat). A group of college women are randomly selected, and they provide information using both methods. The researcher estimates the daily caloric intake for each method and sees whether they differ Caloric Intake Measured by 24-Hour Recall and Dietary Survey Student 2 3 4 5 24-Hour 1530 2130 2940 1960 2270 Recall Survey Difference 240 -120 510 1290 2250 2430 1900 2120 60 150 Determine the appropriate hypotheses and perform a nonparametric test to answer the researchExplanation / Answer
Given that,
mean(x)=2166
standard deviation , s.d1=514.3248
number(n1)=5
y(mean)=1998
standard deviation, s.d2 =440.4203
number(n2)=5
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.776
since our test is two-tailed
reject Ho, if to < -2.776 OR if to > 2.776
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =2166-1998/sqrt((264529.9999/5)+(193970.04065/5))
to =0.5548
| to | =0.5548
critical value
the value of |t | with min (n1-1, n2-1) i.e 4 d.f is 2.776
we got |to| = 0.55478 & | t | = 2.776
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.5548 ) = 0.609
hence value of p0.05 < 0.609,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.5548
critical value: -2.776 , 2.776
decision: do not reject Ho
p-value: 0.609
we don't have proper evidence to support the claim
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