Question # Iron is very important for babies' growth. A common belief is that breastfeeding will help the baby to get more iron than formula feeding.

Comparing two groups
ANSWERED Iron is very important for babies' growth. A common belief is that breastfeeding will help the baby to get more iron than formula feeding. To justify the belief, a study followed 2 groups of babies from born to 6 months. With one group babies are breast fed, and the other group are formula fed without iron supplements. Data below shows iron levels of those two groups of babies.

$$\begin{array}{|c|c|} \hline Group & Sample\ size & mean & Standard\ deviation \\ \hline Breast-fed & 23 & 13.3 & 1.7 \\ \hline Formula-fed & 23 & 12.4 & 1.8 \\ \hline DIFF = Breast-Formula & 23 & 0.9 & 1.4 \\ \hline \end{array}$$

(1) There are two groups we need to compare for the study: Breast-Fed and Formula- Fed. Are those two groups dependent or independent? Based on your answer, what inference procedure should we apply for this research?

(2) Please perform the inference you decided in (1), and make sure to follow the 5-step procedure for any hypothesis test.

(3) Based on your conclusion in (2), what kind of error could you make? Explain the type of error using the context words for this research 2021-02-26

1. The groups are independent so the independent samples t-test should be followed. 2. The hypothesis being tested is: $$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}}$$
$$\begin{array}{|c|c|} \hline 1 & 2 \\ \hline 13.3 & 12.4 & mean \\ \hline 1.7 & 1.8 & std.\ dev.\\ \hline 23 & 23 & n \\ \hline \end{array}$$
$$\begin{array}{|c|c|} \hline 44 & df \\ \hline 0.9000 & difference (1 - 2) \\ \hline 3.0650 & pooled\ variance\\ \hline 1.7507 & pooled\ std.\ dev. \\ \hline 0.5163 & standard\ error\ of\ difference \\ \hline 0.5163 & standard\ error\ of\ difference \\ \hline 1.743 & t \\ \hline .0441 & p-value (one-tailed, upper) \\ \hline \end{array}$$

The p-value is 0.0441. Since the p-value (0.0441) is less than the significance level (0.05), we can reject the null hypothesis. Therefore, we can conclude that $$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}}$$ 3. Type I error could have been made since the null hypothesis is rejected.