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 2021-01-30 Answered
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
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Usamah Prosser
Answered 2021-01-31 Author has 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

Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more