Question

n 10. A study of iron deficiency among infants compared samples of infants following different feeding regimens. One group contained breast-fed infants and the other group was fed a baby formula with iron supplements. Here are the summary results on blood hemoglobin levels at 12 months of age. Group Sample mean Sample St. Dev. Breast-fed 23 13.3 1.7 Formula 19 12.4 1.8 Does the data provide evidence at the 10% level that the hemoglobin levels differ for these two populations of infants? YOU NEED ONLY STATE YOUR HYPOTHESES FOR THIS QUESTION (and define your parameter(s).) 11. A significance test is done to determine if there is a positive correlation between H:p=0 college GPA and salary for first job. The hypotheses were H:p>0. If the p-value for the test was p = 0.001, give the conclusion for the significance test. 12. What is ANOVA used for?

Answer 1

10)

Answer:

$\text{parameters}:\mu_1,\ \mu_2$

$\text{Hypothesis}$

Ho: M1 = 42

$H_a:\mu_1\neq\mu_2$

Explanation:

Let the random variable, X = blood hemoglobin levels in the infants

Parameters:

A parameter is defined as the statistical number that represents the whole population. Hence the mean value is used to represents the blood hemoglobin levels in the infants.

Hypothesis:

The null hypothesis is defined as the mean blood hemoglobin levels between breast-fed group and the iron supplements groups is equal and the alternative hypothesis tests the claim that the mean blood hemoglobin levels between breast-fed group and the iron supplements groups is not equal.

This is a two-tailed test.

11)

Answer:

The p-value for the hypothesis test is very small such that the probability of incorrectly rejecting the null hypothesis is very small hence there is statistically significant evidence to conclude that the GPA and salary are positively correlated.

Explanation:

The p-value is defined as the propbability of rejecting the hypothesis when it is true such that it is the probability of incorrectly rejecting the null hypothesis. (The p-value is also known as the probability of making a type I error)

The p-value for the hypothesis test = 0.001 such that the probability of making a type I error is very small. Hence there is strong evidence to reject the null hypothesis.

12)

When we want to compare more than two group means, an ANOVA test is used to test whether all the sample groups are from the same population or not i.e. whether all the group means are equal or not. The ANOVA test uses the F distribution to test this hypothesis using the following formula,

$F=\frac{\text{Between group variance}}{\text{Within group variance}}$

Now, based on the above formula we can see that if the ratio between these two variances is not large (less than predefined critical value), we can say that the samples are from the same population.