Question

1a. Construct a 99% confidence interval for proportion of obese people in the population the above sample was taken. (BMI>29.5 is counted as ob

1.b Is there any evidence to show there is difference between proportion of smokers between male and female? Apply appropriate test with alpha=0.05.

USE R AND SHOW CODES

Male=1,   Female=0

Smoker=1, Nonsmoker=0

DATA

sex	age	currentSmoker	BMI
1	39	0	26.97
0	46	0	28.73
1	48	1	25.34
0	61	1	28.58
0	46	1	23.1
0	43	0	30.3
0	63	0	33.11
0	45	1	21.68
1	52	0	26.36
1	43	1	23.61
0	50	0	22.91
0	43	0	27.64
1	46	1	26.31
0	41	0	31.31
0	39	1	22.35

ONLY ANSWER 1.b not 1.a & show how u got x values for p1 & p2

Answer 1

Result:

1.b Is there any evidence to show there is difference between proportion of smokers between male and female? Apply appropriate test with alpha=0.05.

R code:

sex <- c(1,0,1,0,0,0,0,0,1,1,0,0,1,0,0)

smoker <- c(0,0,1,1,1,0,0,1,0,1,0,0,1,0,1)

mydata <- data.frame(sex,smoker)

# to get frequencies

freq <-table(mydata$sex, mydata$smoker)

freq

R output:

freq

  

    0 1

0 6 4

1 2 3

Frequencies: out of 5 males, 3 are smokers. out of 10 females, 4 are smokers.

R code:

#proportion test

prop.test(c(3,4),c(5,10),correct=FALSE)

R output:

        2-sample test for equality of proportions without continuity

        correction

data: c(3, 4) out of c(5, 10)

X-squared = 0.53571, df = 1, p-value = 0.4642

alternative hypothesis: two.sided

95 percent confidence interval:

-0.3259135 0.7259135

sample estimates:

prop 1 prop 2

   0.6    0.4

Calculated chi square = 0.53571, P=0.4642 which is > 0.05 level of significance. Ho is not rejected.

There is no evidence to show that there is difference between proportion of smokers between male and female.