P.Noakes et al. researched the effects of fatty acids found in oily fish on lowering the risk of allergic disease in the article "Increased Intake of Oily Fish in Pregnancy: Effects on Neonatal Immune Responses and on Clinical Outcomes in Infants at $6$Mo.". Pregnant women were randomly assigned to continue their habitual diet (control group), which was low in oily fish, or to consume two portions of salmon per week (treatment group). Their infants were clinically evaluated at $6$months of age and the frequency of many different symptoms was recorded. Of the $37$infants in the control group,$12$had symptoms of dry skin; and of the 45 infants in the experimental group, $14$had symptoms of dry skin. At thedata-custom-editor="chemistry" $5\%$significance level, do the data provide sufficient evidence to conclude that a difference exists in the proportions of infants who have symptoms of dry skin at$6$ months between those whose mothers continue their habitual diet and those whose mothers consume two portions of salmon per week?

Part (a): Use the two-portionsz-test to perform the required hypothesis test.

Part (b): Use the chi-square homogeneity test to perform the required hypothesis test.

Part (c): Compare your results in parts (a) and (b).

Part (d): Explain what principle is being illustrated.

Consider the given question,

Total no. of infants in control group$=37$

No. of infants with dry skin symptoms$=12$

Total no. of infants in experimental group$=45$

No. of infants with dry skin symptoms$=14$

Difference$=p\left(1\right)-p\left(2\right)$

Estimate for difference is $0.0132132$

$95\%$CI for difference is $\left(-0.189388,0.215815\right)$

Test for difference, role="math" localid="1651948088167" $x^2=0.13$

Fisher's exact test, $p-value=0.898$

Using the significance level, $\alpha =0.05$

Here, P-value is greater than the level of significance. That is $P-value\left(=0.898\right)>\alpha \left(=0.05\right)$.

Therefore, by the rejection rule, it can be concluded that there is no evidence to reject the null hypothesis $\left(H_0\right)$at $\alpha =0.05$.

Thus, the data do not provide sufficient evidence to dry skin at $6$months between those whose mothers proportions of infants who have symptoms of dry skin at $6$months between those whose mothers continue their habitual diet and those whose mothers consume two portions of salmon per week.

The test hypothesis are given below,

Null hypothesis: $H_0$, the data do not provide evidence to conclude that a difference exists in the proportions of infants who have symptoms of dry skin at $6$months between those whose mothers continue their habitual diet and those whose mothers consume two portions of salmon per week.

Alternative hypothesis: $H_1$, that is the data provide sufficient evidence to conclude that a difference exists in the proportions of infants who have symptoms of dry skin at $6$months between those whose mothers continue their habitual diet and those whose mothers consume two portions of salmon per week.

On finding the value of test statistics and P-value,

Consider the above table,

Chi-square$=0.016$,

$df=1$,

p-value$=0.898$

Using the significance level, $\alpha =0.05$

Here, P-value is greater than the level of significance. That is $P-value\left(=0.898\right)>\alpha \left(=0.05\right)$.

Therefore, by the rejection rule, it can be concluded that there is no evidence to reject the null hypothesis $\left(H_0\right)$at $\alpha =0.05$.

Thus, the data do not provide sufficient evidence to conclude a difference exists in the proportions of infants who have symptoms of dry skin at $6$months between those whose mothers continue their habitual diet and those whose mothers consume two portions of salmon per week.

Consider parts (a) and (b),

It is clear that the conclusion andP-value is exactly same for both the methods.