Question

Relaxing in the sauna Researchers followed a random sample of $2315$middle-aged men from eastern Finland for up to $30$years. They recorded how often each man went to a sauna and whether or not he suffered sudden cardiac death (SCD). The two-way table shows the data from the study.

a. State appropriate hypotheses for performing a chi-square test for independence in this setting.

b. Compute the expected counts assuming that $H_0$is true.

c. Calculate the chi-square test statistic, df, and P-value.

d. What conclusion would you draw?

Short answer

(a) The appropriate hypothesis is the null hypothesis.

(b) The expected count is,

	$1$or FEWER	$2-3$	$4$or MORE
YES	$49.33$	$124.18$	$16.50$
NO	$551.67$	$1388.82$	$184.50$

(c) The test statistics is ${\chi }^2=6.0323$,$df=2,P-value=0.025<P<0.05$.

(d) There are many convincing pieces of evidence for the association between SCD and weekly sauna frequency.

Step-by-step solution

Part (a) Step 1: Given information

We need to find out the appropriate hypothesis for performing a chi-square test for independence in this setting.

Part (a) Step 2: Explanation

We know that

The null hypothesis asserts that the variables are unrelated, whereas the alternative hypothesis asserts that they are.

$H_0$is there is no association between SCD and Weekly sauna frequency.

$H_{\alpha }$ is there is an association between SCD and Weekly sauna frequency.

Part (b) Step 1: Given information

We need to find the expected counts.

Part (b) Step 2: Explanation

From part (a)

We know that

Expected frequencies are a product of row and column total divided by table total. So,

ROW AND COLUMN NUMBER	EXPECTED FREQUENCY
$E_{11}$	$49.33$
$E_{12}$	$124.18$
$E_{13}$	$16.50$
$E_{21}$	$551.67$
$E_{22}$	$1388.82$
$E_{23}$	$184.50$

Part (c) Step 1: Given information

We need to find the test statistics, df, P-value.

Part (c) Step 2: Explanation

From parts (a) and (b)

We know that

The squared differences between the actual and predicted frequencies, divided by the expected frequency, make up the chi-square subtotals.

Therefore, test-statistics is,${\chi }^2=6.0323$

And

$$\begin{gathered} df=2 \\ P-value=0.025<P<0.05=0.048989 \end{gathered}$$

Part (d) Step 1: Given information

We need to find out the conclusion drawn.

Part (d) Step 2: Explanation

From the parts (a) ,(b) and (c)

We know that

There are many convincing pieces of evidence for the association between SCD and weekly sauna frequency.

And this is the conclusion we drew.