91影视

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

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.

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

We need to find the expected counts.

From part (a)

We know that

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

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

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,${蠂}^2=6.0323$

And

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

We need to find out the conclusion drawn.

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.