## Iron deficiency anemia is an important nutritional health problem in the U.S. A dietary assessment was performed on 51 boys 9-11-years old w

Iron deficiency anemia is an important nutritional health problem in the U.S. A dietary assessment was performed on 51 boys 9-11-years old whose families were below the poverty level. The mean daily iron intake among these boys was found to be 12.50 mg with standard deviation 4.75 mg. Suppose the mean daily iron intake among a large population of 9-11-years old boys from all income strata is 14.44 mg. We want to test if the mean iron intake among the low-income group is different from that of the general population.

a. State the hypothesis that you can use to consider this question.

b. Carry out the hypothesis test using the critical-value method with an alpha= 0.05

c. Construct a 95% confidence interval estimate for the true mean of iron intake of the general population.

## Answers ( )

Answer:a) Null hypothesis:

Alternative hypothesis:

b)

The degrees of freedom are given by:

Now we can calculate the p value taking in count the alternative hypothesis

Since the p value is lower than the significance we have enough evidence to reject the null hypothesis and the true mean is significantly different from 14.44

c)

Step-by-step explanation:Information givenrepresent the mean for the daily iron intake

represent the sample deviation

sample size

represent the reference value

represent the significance level for the hypothesis test.

t would represent the statistic

represent the p value for the test

Part aWe want to test if the mean iron intake among the low-income group is different from that of the general population, the system of hypothesis would be:

Null hypothesis:

Alternative hypothesis:

Part bSince we don’t know the population deviation the statistic would be given by:

(1)

Replacing the info we got

The degrees of freedom are given by:

Now we can calculate the p value taking in count the alternative hypothesis

Since the p value is lower than the significance we have enough evidence to reject the null hypothesis and the true mean is significantly different from 14.44

Part cThe confidence interval would be given by:

For the 95% confidence interval we can find the critical value in a t distribution with 50 degrees of freedom and we got:

And replacing we got: