Question

A survey was conducted to study if parental smoking is associated with the incidence of smoking in children when they reach high school. Randomly chosen high school students were asked whether they smoked and whether at least one of their parents smoked.

The results are summarized in the following table:

Student Smoke Student Don’t

Parents Smoke 262 183

Parents Don’t 120 380

(a) For a randomly selected student in this study, find the conditional probability of smoking given his/her parents smoke.

(b) Suppose we are interested in testing whether parental smoking is independent of children smoking. Which statistical test would you consider for this problem?

(c) (4 points) Write down the R code to carry out that test. You first need to store the data into a matrix.

(d) Calculate the test statistic by yourself.

(e) Write down the R code to obtain the p-value based on your answer in
part(d).
(f) Suppose the p-value is 0.0001, what would be your really world conclusion? (You may use α = 0:05.)

Answer 1

A survey was conducted to study if parental smoking is associated with the incidence of smoking in children when they reach high school. Randomly chosen high school students were asked whether they smoked and whether at least one of their parents smoked.

The results are summarized in the following table:

	StudentSmoke	StudentDon’t	Total
ParentsSmoke	262	183	445
ParentsDon’t	120	380	500
Total	382	563	945

(a) For a randomly selected student in this study, find the conditional probability of smoking given his/her parents smoke.

P=262/445 =0.58876

(b) Suppose we are interested in testing whether parental smoking is independent of children smoking. Which statistical test would you consider for this problem?

Chi square test

(c) (4 points) Write down the R code to carry out that test. You first need to store the data into a matrix.

x<-c(262,183)

y<-c(120,380)

mydata <- data.frame(x,y)

chisq.test(mydata)

R output:

Pearson's Chi-squared test with Yates' continuity correction

data: mydata

X-squared = 117.48, df = 1, p-value < 2.2e-16

(d) Calculate the test statistic by yourself.

Chi-Square Test
Observed Frequencies
	Column variable				Calculations
	StudentSmoke	StudentDon’t	Total		fo-fe
ParentsSmoke	262	183	445		82.1164	-82.1164
ParentsDon’t	120	380	500		-82.1164	82.1164
Total	382	563	945
Expected Frequencies
	Column variable
	StudentSmoke	StudentDon’t	Total		(fo-fe)^2/fe
ParentsSmoke	179.8836	265.1164	445		37.4859	25.4345
ParentsDon’t	202.1164	297.8836	500		33.3625	22.6367
Total	382	563	945
Data
Level of Significance	0.05
Number of Rows	2
Number of Columns	2
Degrees of Freedom	1
Results
Critical Value	3.841
Chi-Square Test Statistic	118.92
p-Value	0.0000
Reject the null hypothesis

(e) Write down the R code to obtain the p-value based on your answer in
part(d).

pchisq(118.92, df=1, lower.tail=FALSE)

(f) Suppose the p-value is 0.0001, what would be your really world conclusion? (You may use α = 0:05.)

Since p value 0.0001 < 0.05 level of significance, Ho is rejected.

We conclude that parental smoking is not independent of children smoking