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

postillan4

postillan4

Answered question

2021-01-30

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

Answer & Explanation

Usamah Prosser

Usamah Prosser

Skilled2021-01-31Added 86 answers

Let the standard deviation of the d
s2 = 5.562
=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
s = 4.75mg
So, the sample variance is,
s2 = (4.75)2
= 22.5625
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,
H0: σ2 =30.9136mg
H1: σ2 =30.9136mg
Let the sample size be n=51
The degrees of freedom for the test is,
df = nl
= 511
= 50
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 χ0.0252
= 32.3574
Upper critical value =χ0.9752
= 71.4202
The decision rule is to reject H0ifχ2<32.3874orχ2>71.4202
The chi-square test statistic is,
χ2 = (n1)s2σ02χ(n1)2
= (511)4.7525.562
= 50×22.562530.9136
36.493
So, the shi-equated test statistic is χ2 = 36.493
The p-value is,
p = 2 ×CHISQ.DIST(36.4928,50,TRUE)
= 2 × 0.0767
= 0.1534
The probability value of test is p = 0.1534
The 95% confidence interval for the variance of daily iron intake in the low-income group is calculated as follows:
CI = ((511)4.752χ0.975,50  σ2  (511)4.752χ0.025,502)
= (50×22.5671.42  σ2  50×22.5632.36)
= (1128.12571.42  σ2  1128.12532.36)
= (15.8 σ2  34.9)
Thus, with 95% 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

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