In: Statistics and Probability
Adults aged 18 years old and older were randomly selected for a survey on obesity. Adults are considered obese if their body mass index (BMI) is at least 30. The researchers wanted to determine if the proportion of women in the south who are obese is less than the proportion of southern men who are obese. The results are shown in the table below.
# who are obese | sample size | |
Men | 42,769 | 155,525 |
Women | 67,169 | 248,775 |
At the1% level of significance, is there evidence to support the claim? Explain.
work must include
1. Clear statement of hypotheses, with the correct
parameter(s)
2. An indication of the test used
3. The test statistic and p-value
4. An indication of the statistical decision (i.e. whether or not
to reject Ho)
along with an explanation.
5. An interpretation of the statistical decision in the context of
the problem.
1)
Ho: proportion of women in the south who are obese is equal to
the proportion of southern men who are obese
Ha: proportion of women in the south who are obese is less than the
proportion of southern men who are obese
Ho: p1 - p2 = 0
Ha: p1 - p2 > 0
2)
Two proportion z test
3)
sample #1 -----> Men
first sample size, n1=
155525
number of successes, sample 1 = x1=
42769
proportion success of sample 1 , p̂1=
x1/n1= 0.2750
sample #2 -----> Women
second sample size, n2 =
248775
number of successes, sample 2 = x2 =
67169
proportion success of sample 1 , p̂ 2= x2/n2 =
0.270
difference in sample proportions, p̂1 - p̂2 =
0.2750 - 0.2700 =
0.0050
pooled proportion , p = (x1+x2)/(n1+n2)=
0.2719
std error ,SE = =SQRT(p*(1-p)*(1/n1+
1/n2)= 0.0014
Z-statistic = (p̂1 - p̂2)/SE = ( 0.005
/ 0.0014 ) =
3.4753
p-value =
0.0003 [excel function
=NORMSDIST(-z)]
4)
decision : p-value<α,Reject null hypothesis
5)
Conclusion: There is enough evidence to support that the proportion of women in the south who are obese is less than the proportion of southern men who are obese.
Please revert back in case of any doubt.
Please upvote. Thanks in advance.