Question

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.

Short answer

Part (a): We know $$\begin{gathered} H_0:p_1=p_2, \\ {H}_a:p_1\ne p_2, \\ \alpha =0.05, \\ {x}^2=0.13 \end{gathered}$$, P-value is $0.898$, where $H_0$rejects; at the $5\%$significance level, 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.

Part (b): We know $$\begin{gathered} H_0:p_1=p_2, \\ {H}_a:p_1\ne p_2, \\ \alpha =0.05, \\ {x}^2=0.016 \end{gathered}$$, P-value is $0.898$, where $H_0$rejects; at the $5\%$significance level, 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.

Part (c): The results of parts (a) and (b) are the same.

Part (d): The chi-square homogeneity test for comparing two population proportions and the two-tailed two proportions z-test are equivalent.

Step-by-step solution

Part (a) Step 1. Given information.

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$=14$

Part (a) Step 2. Consider the null and alternative hypotheses.

The test hypothesis are given below,

Null hypothesis is $H_0:p_1=p_2$, that is the data do not 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.

Alternative hypothesis is $H_a:p_1\ne p_2$, that is the data do 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,

Part (a) Step 3. Using the significance level.

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.

Part (b) Step 1. Consider the null and alternative hypotheses.

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,

Part (b) Step 2. Using the significance level.

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.

Part (c) Step 1. Consider parts (a) and (b).

Consider parts (a) and (b),

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

Part (d) Step 1. Consider the chi-square test.

The principle is being illustrated in the situation is that the chi-square test for homogeneity is same as the two proportions z-test for two tailed.

As there are only two populations are considered.