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

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

Question
Comparing two groups
asked 2021-02-25

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

Answers (1)

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.

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\(\begin{array}{|c|c|} \hline & Housework Hours \\ \hline Gender & Sample\ Size & Mean & Standard\ Deviation \\ \hline Women & 473473 & 33.133.1 & 14.214.2 \\ \hline Men & 488488 & 18.618.6 & 15.715.7 \\ \end{array}\)

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b. Find the standard error for comparing the means. What factor causes the standard error to be small compared to the sample standard deviations for the two​ groups? The cause the standard error to be small compared to the sample standard deviations for the two groups.

c. Calculate the​ 95% confidence interval comparing the population means for women Interpret the result including the relevance of 0 being within the interval or not. The​ 95% confidence interval for ​\(\displaystyle{\left(\mu_{{W}}-\mu_{{M}}​\right)}\) is: (Round to two decimal places as​ needed.) The values in the​ 95% confidence interval are less than 0, are greater than 0, include 0, which implies that the population mean for women could be the same as is less than is greater than the population mean for men.

d. State the assumptions upon which the interval in part c is based. Upon which assumptions below is the interval​ based? Select all that apply.

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C.The samples from the two groups are independent.

D.The samples from the two groups are random.

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