Question

If the prevalence ratio includes the null value, then we cannot rule out the null hypothesis that there is no relationship between obesity and metabolic syndrome.

1. Use the table below to calculate the prevalence ratio and chi-square statistic and calculate the confidence interval for the prevalence ratio.

2. Does the confidence interval include the null value?

Caucasians Obese Non-obese Metabolic Syndrome Yes No 17 26 228 41 254 24 43 252 295 1 1 1 SE{In (RR)} + 1 a+b a с c+d 95% CI = exp ( In(RR) – 1.96 x SE{In(RR)}) to exp ( 1n(RR) +1.96 x SE{In(RR)})

Answer 1

Given:

Metabolic Syndrome

	Yes	No	Row Total
Obese	17	26	43
Non-obese	24	254	252
Column Total	41	254	N = 295

1) Prevalence ratio (RR)

a* d. 17 * 228 26 * 24 = 6.2115 b*c

2) 95% confidence interval

exp(ln(RR) = [1.96 * SE(In(RR))]) = expln(6.2115) + (1.96 * 0.3786])

Here,

$SE(ln(RR)) = \sqrt{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}$$= \sqrt{\frac{1}{17}+\frac{1}{26}+\frac{1}{24}+\frac{1}{228}}$  

3) Chi square test statistic

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

Here

O = observed frequencies [ given table values i.e; 17, 26, 24, 228]

E = expected frequencies [ E = (row total * column total) / N i.e; 5.98, 37.02, 35.02, 216.98]

$\chi^2= (20.33 + 3.28 +3.47 + 0.56) = 27.64$

Hence calculated $\chi^2$ = 27.64

For 5% level of significance, $\chi^2_{(df,N)} = \chi^2_{(1, 295)}$ = 3.8414

Calculated value > Tabulated value. Hence we reject the null hypothesis.

4)

It is given that, if the prevalence ratio includes the null values, we cannot rule out the null hypothesis. This means that we cannot reject the null hypothesis. But here we are rejecting the null hypothesis and hence we can conclude that the confidence interval does not include null values.