Question

Iron-deficiency anemia is an important nutritional health problem in the United States. A dietary assessment was performed on 86 boys 9 to 11 years of age whose family incomes were below the poverty level. The mean daily iron intake among these boys was found to be 13.31 mg with standard deviation 4.46 mg. Previous studies indicate the mean daily iron intake in the generalpopulation of 9- to 11- year- old boys from all income strata is 14.58 mg and s= 5.43 mg. The researchers want to test whether the mean iron intake among the low-income population is differentfrom that of the general population at the .05 (5%) level of significance.

A) This study is designed to test hypothesizedequivalence of [circle correctanswer(s)]:

(a) a sample mean and a reference value

(b) a population mean and a reference value

(c) a population proportion and a reference value

B) Which hypothesis testing procedure would be most appropriate in this scenario?

C) A Type I error would occur if, in fact, mean daily iron intake is 14.58 mg in the low-income population, but the results of the hypothesis test lead the researchers to _______________________________________.  The long-term probability of this occurring that is given for this example is ________. One reason why Type I (and Type II) errors occur is _________________________________________________.

D) Assume the required conditions for the test are met and calculate the value of the test statistic.

E)

(1) What is the p-value for the test statistic?

(2) Does the p-value suggest that µ is lowerthan 14.58 mg/day for the population of low-income boys, ages 9 to 11 years?  Explain why or why not.

Answer 1

A)
option b) a population mean and a reference value

B)
Two tailed, z-test, one sample

C)
A Type I error would occur if, in fact, mean daily iron intake is 14.58 mg in the low-income population, but the results of the hypothesis test lead the researchers to reject the null hypothesis.
The long-term probability of this occurring that is given for this example is 0.05. One reason why Type I (and Type II) errors occur is sample size.

D)
H0: mu = 14.58
Ha: mu not equals to 14.58

test statistic, z = (13.31 - 14.58)/(4.46/sqrt(86))
z = -2.6407

p-value = 0.0083
As p-value is less than the significance level, we reject the null hypothesis.

p-value suggess mu is lower than 15.58mg/day for the population of low income boys.