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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 α level of .05, and summarize your findings.

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Answer:

Step-by-step explanation:

A. the hypothesis will be studied:

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

where mu is the population mean.

n = number of children from 9 to 11 years old = 51

Xbar = average daily iron intake among these children = 12.50 mg

standard deviation (sd) = 4.75 mg

alpha = significance level = 0.05

B. Here we use the t-test because the 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

C. P value and critical value that we can calculate using EXCEL.

the syntax is

= TDIST (x, freedom_of_degrees, tails)

= TINV (probability, freedom of degrees)

where, x is the absolute value of the test statistic.

freedom_degrees = 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 the value p <alpha and

The | t | > critical value

Therefore, reject H0 with a significance level of 5%.

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

D. The standard deviation of daily iron intake in the largest population of children ages 9 to 11 was 5.56 mg. We want to test whether the standard deviation of the low-income group is comparable to that of the general population. Please indicate the hypotheses that we can use to answer this question.

This is where we are given that = ɕ = 5.56 mg

The hypothesis for the test is,

H0: ɕ2 = 5.562 Vs H1: ɕ2 ≠ 5.562

Alpha = significance level = 0.05

The test statistic is,

Χ2 = ((n-1) * sd2) / ɕ2

Χ2 = ((51-1) * 4,752) / 5,562

Χ2 = 36.4928

The p value syntax in EXCEL is,

= CHIDIST (x, deg_freedom)

where, x is the statistical test value.

freedom_degrees = n - 1 = 51 - 1 = 50

p value = 0.9233

p value> alpha

H0 cannot be rejected with a significance level of 5%.

Conclusion: The standard deviation of daily iron intake in the largest population of children aged 9 to 11 years was 5.56 mg.

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