Iron-deficiency anemia is an important nutritional health problem in the United

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 different from 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 comparable to 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 level of .05, and summarize your findings.

Explanation / Answer

Here we have to test the hypothesis that,

H0 : mu = 14.44 mg Vs H1 : mu 14.44 mg

where, mu is population mean.

n = number of  boys who are 9 to 11 years of age= 51

Xbar = mean daily iron intake among these boys = 12.50 mg

standard deviation (sd) = 4.75 mg

alpha = level of significance = 0.05

Here we use t-test because sample standard deviation is unknown.

The test statistic is,

t = (Xbar - mu) / (sd/sqrt(n))

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

t = -2.9167

p-value and critical value we can calculate by using EXCEL.

syntax is,

=TDIST(x, deg_freedom, tails)

=TINV(probability, deg_freedom)

where, x is absolute value of test statistic.

deg_freedom = n - 1 = 51 - 1 = 50

probability = alpha/2 = 0.05/2 = 0.025

tails = 2

p-value = 0.005

critical value = 2.3109

We see that p-value < alpha and

| t | > critical value

Therefore, reject H0 at 5% level of significance.

Conclusion : The mean iron intake among the low-income group is different from that of the general population.

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 comparable to that of the general population. State the hypotheses that we can use to answer this question.

That is here we are given that = = 5.56 mg

Hypothesis for testing is,

H0 : 2 = 5.562   Vs H1 : 2 5.562       

Alpha = level of significance = 0.05

The test statistic is,

2 = ( (n-1)*sd2 ) / 2

2 = ( (51-1)*4.752 ) / 5.562

2 = 36.4928

p-value syntax in EXCEL is,

=CHIDIST(x,deg_freedom)

where, x is test statistic value.

deg_freedom = n - 1 = 51 - 1 = 50

p-value = 0.9233

p-value > alpha

Fail to reject H0 at 5% level of significance.

Conclusion : The standard deviation of daily iron intake in the larger population of 9- to 11-year-old boys was 5.56 mg.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.