Chapter 11.2, Problem 41E

a)

To determine

To state the null and alternative hypotheses.

Answer to Problem 41E

  $H_0$ : There is no association between Weekly Sauna frequency and SCD.

  $H_a$ : There is association between Weekly Sauna frequency and SCD.

Explanation of Solution

Given:

		Weekly sauna frequency
		1 or fewer	2-3	4 or more
SCD	Yes	61	119	10
	No	540	1394	191

Calculation:

The null and alternative hypotheses:

  $H_0$ : There is no association between Weekly Sauna frequency and SCD.

  $H_a$ : There is association between Weekly Sauna frequency and SCD.

b)

To determine

To calculate expected counts.

Explanation of Solution

Given:

		Weekly sauna frequency
		1 or fewer	2-3	4 or more
SCD	Yes	61	119	10
	No	540	1394	191

Formula:

  $E=\frac{r\times c}{n}$

Where, r = row total, c = column total and n = grand total.

Calculation:

Therefore, expected count is,

  $\begin{array}{l}{E}_{11}=\frac{r_1\times c_1}{n}=\frac{190\times 601}{2315}=49.33\\ .\\ .\\ .\\ E_{23}=\frac{r_2\times c_3}{n}=\frac{2125\times 201}{2315}=184.50\end{array}$

Therefore, table of expected count is,

	1 or fewer	2-3	4 or more
Yes	49.33	124.18	16.5
No	551.67	1388.82	184.5

c)

To determine

To calculate value of chi square test statistic, df and p-value.

Answer to Problem 41E

The chi-square test statistic is 6.03. The degrees of freedom = df = 2 and p-value = 0.0490.

Explanation of Solution

Given:

		Weekly sauna frequency
		1 or fewer	2-3	4 or more
SCD	Yes	61	119	10
	No	540	1394	191

Expected count is

	1 or fewer	2-3	4 or more
Yes	49.33	124.18	16.5
No	551.67	1388.82	184.5

Formula:

  ${\chi }^2={\displaystyle \sum \frac{{(O-E)}^2}{E}}$

Calculation:

		1 or fewer	3-Feb	4 or more	Total
Yes	Observed	61	119	10	190
	Expected	49.33	124.18	16.50	190.00
	(O - E)² / E	2.76	0.22	2.56	5.54
No	Observed	540	1394	191	2125
	Expected	551.67	1388.82	184.50	2125.00
	(O - E)² / E	0.25	0.02	0.23	0.50
Total	Observed	601	1513	201	2315
	Expected	601.00	1513.00	201.00	2315.00
	(O - E)² / E	3.01	0.24	2.79	6.03
		6.03	chi-square
		2	Df
		.0490	p-value

Therefore, chi-square test statistic is 6.03. The degrees of freedom = df = 2 and p-value = 0.0490.

d)

To determine

To state the conclusion.

Answer to Problem 41E

There is convincing evidence that there is an association between Weekly sauna frequency and SCD.

Explanation of Solution

Given:

The chi-square test statistic is 6.03. The degrees of freedom = df = 2 and p-value = 0.0490.

Calculation:

Decision: P-value < 0.05, reject H0.

Conclusion: There is convincing evidence that there is an association between Weekly sauna frequency and SCD.