Question

Iron-deficiency anemia is an important nutritional health problem in the United States. A dietary assessment was performed on 51 boys 9 to 11 years of age 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- to 11-year-old boys from all income strata is 14.44 mg. We want to test whether the mean iron intake among the low-income group is lower than that of the general population.

(a) State the hypotheses that we can use to consider this question.
(b) Carry out the hypothesis test in (a) using the critical-value method with a significance level of .05, and summarize your findings.
(c) What is the p-value for the test conducted in (b)?
(d) The standard deviation of daily iron intake in the larger population of 9- to 11-year-old boys was 5.56 mg. We want to test whether the standard deviation from the low-income group is lower than that of the general population. State the hypotheses that we can use to answer this question.
(e) Carry out the test in (d) and report the p-value with an α level of .05, and summarize your findings.

Answer 1

The null hypothesis is that the mean daily iron intake of the low-income group is equal to or greater than the general population, and the alternative hypothesis is that the mean daily iron intake of the low-income group is less than that of the general population.

H0: μ >= 14.44

Ha: μ < 14.44

(b) We can use a one-tailed t-test to test this hypothesis. With a sample size of 51, degrees of freedom of 50, a sample mean of 12.50, a population mean of 14.44, and a standard deviation of 4.75, we can calculate the t-statistic as follows:

t = (12.50 - 14.44) / (4.75 / sqrt(51)) = -3.16

Using a t-distribution table with 50 degrees of freedom and a significance level of .05, the critical value is -1.677. Since the calculated t-statistic is less than the critical value, we reject the null hypothesis.

Therefore, we can conclude that the mean daily iron intake among the low-income group is significantly lower than that of the general population at a significance level of .05.

(c) The p-value for this test is the probability of obtaining a t-value of -3.16 or more extreme assuming the null hypothesis is true. Using a t-distribution table with 50 degrees of freedom, we find the p-value to be less than .005.

(d) The null hypothesis is that the standard deviation of the low-income group is equal to or greater than that of the general population, and the alternative hypothesis is that the standard deviation of the low-income group is less than that of the general population.

H0: σ >= 5.56

Ha: σ < 5.56

(e) We can use a chi-square test to test this hypothesis. With a sample size of 51, degrees of freedom of 50, and a sample standard deviation of 4.75, we can calculate the chi-square statistic as follows:

chi-square = (n - 1) * s^2 / σ^2 = 50 * 4.75^2 / 5.56^2 = 39.70

Using a chi-square distribution table with 50 degrees of freedom and a significance level of .05, the critical value is 68.67. Since the calculated chi-square statistic is less than the critical value, we fail to reject the null hypothesis.

Therefore, we do not have sufficient evidence to conclude that the standard deviation of daily iron intake among the low-income group is significantly lower than that of the general population at a significance level of .05.