# 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

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
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Usamah Prosser

Let the standard deviation of the d

$=30.9136$
So, the population variance is,
Let the sample standard deviation of lower income group of 9-to-11 year's old boys be

So, the sample variance is,

Test whether the standard deviation from the low-income group is comparable to that of the general population
The null and alternative hypotheses of the test are,

Let the sample size be $n=51$
The degrees of freedom for the test is,

Find the critical values.
Using the chi-square distribution tables for 50 degrees of freedom, the lower and upper critical values are, Lower critical value ${\chi }_{0.025}^{2}$

Upper critical value $={\chi }_{0.975}^{2}$

The decision rule is to reject ${H}_{0}\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}{\chi }^{2}<32.3874\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}{\chi }^{2}>71.4202$
The chi-square test statistic is,

$\approx 36.493$
So, the shi-equated test statistic is
The p-value is,
$×CHISQ.DIST\left(36.4928,50,TRUE\right)$

The probability value of test is
The $95\mathrm{%}$ confidence interval for the variance of daily iron intake in the low-income group is calculated as follows:

Thus, with $95\mathrm{%}$ confidence the variance of daily iron intake in the low-income group lies between 15.8mg and 34.9 mg.
Conclusion:
The test statistic value does not falls in the rejection region. That is, the test statistic falls between the two critical values