Please help me and please give me the right answer please.. Thank you 3. [ 10 ma

Question

Please help me and please give me the right answer please.. Thank you

3. [ 10 marks ]
The file bmi.mtw contain the BMI for a sample of men and a sample of women. Two of the columns, OW_male and OW_female code the BMI values as:
0 - if BMI 25.4 (these are considered “not overweight”);
1 - if BMI 25.5 (these are considered “overweight”).

a. Test whether there is sufficient evidence to show that the proportion of overweight males (proportion of males who are overweight) is higher than the proportion of overweight females in the population. Use the critical value approach and the 5% level of significance. Perform the test manually after using Minitab to summarize the data (if you use descriptive statistics, the mean coded value in each sample is the sample proportion).

b. Explain how to find the p-value without the use of software (indicate what probability has to be found).

c. Finally calculate manually the 95% 1-sided confidence interval to estimate how much higher the proportion of overweight males is compared to the proportion of overweight females.

d. Explain how the results in parts b and c are consistent with your conclusion in part a.

Excel data below..

BMIfemale BMImale OWfemale OWmale 24.3 24.4 0 0 24.4 28.4 0 1 28.8 21.4 1 0 24.1 28.8 0 1 27.1 30.1 1 1 36.3 24.2 1 0 30.5 21.3 1 0 23.8 23.7 0 0 20.6 22.4 0 0 30.9 24.9 1 0 23.5 24.5 0 0 19.6 25.5 0 1 22.4 35.2 0 1 22.5 30.5 0 1 24.5 26.1 0 1 19.8 31.1 0 1 22.3 23.4 0 0 28.9 25.4 1 0 23.2 23 0 0 20.3 24.9 0 0 20.1 23.3 0 0 26.3 25.1 1 0 21.3 26.2 0 1 21.9 28.5 0 1 21.6 24.2 0 0 17.8 28.2 0 1 24.9 29.6 0 1 22.8 27.1 0 1 34.3 26.4 1 1 24.2 25.6 0 1 20.7 26.1 0 1 33.7 23.7 1 0 24.2 21.6 0 0 32.8 24.8 1 0 22.5 27.8 0 1 35.2 1 21.8 0 23.9 0 23.6 0 24.1 0 26.2 1 29.5 1 29.8 1 26.1 1 22.8 0 27 1

Explanation / Answer

for female nf=35 , overweight female=10 ,so proportion of overweight =pf=10/35=and

male nm=46, overweight male=23, so proportion of overweight male=pm=23/46

variance(pf)=pf(1-pf)/nf)

SE(pf)=sqrt(pf(1-pf)/nf)=sqrt((10/35)*(1-10/35)/35)=0.0764

SE(pm)=0.0737

SE(pm-pf)=sqrt(variance(pm)+variance(pf))=sqrt(0.0764*0.0764+0.0737*0.0737)=0.1062

(a) here here we use z-test with

null hypothesis H0:pm=pf

and alternate hypothesis H1: pm>pf

statistic z=(pm-pf)/SE(pm-pf)=(0.5-0.2857)/0.1062=2.0179

one tailed critical z(0.05)=1.6449 ( using ms-excel command=normsinv(0.05)) is less than calcualted z=2.0179, so we fail to accept H0 and conclude that proportion of overweight males (proportion of males who are overweight) is higher than the proportion of overweight females in the population.

(b) p-value=0.0218 ( using ms-excel=1-NORMSDIST(2.0179))

(c) 95% 1-sided confidence interval=(pm-pf)+z(0.05)*SE(pm-pf)=(0.5-0.2857)+1.6449*0.1062=0.389

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