A study utilizes ANOVA to examine the average distribution rates of an antibiotic in infants, toddlers...
A study utilizes ANOVA to examine the average distribution rates of an antibiotic in infants, toddlers and school age children. The results include an F statistic that is 1.04. You know this means:
Homework Answers
This means that the result is not significant at 0.01, 0.05 and 0.10 significance levels.
p-value for F =1.04 and at df(numerator) =number of factors - 1 =3 - 1 =2 and df(denominator) =overall sample size - number of factors =N - 3 (N is not given) is around 0.3
So, p-value of about 0.3 >0.01; >0.05 and >0.10 significance levels. Hence, we failed to reject the null hypothesis(H0) that says "no two population means differ significantly".
Thus, there is no sufficient evidence to claim that at least two population means differ significantly(H1: alternative hypothesis).
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A study utilizes ANOVA to examine the average distribution rates
of an antibiotic in infants, toddlers...
A study utilizes ANOVA to examine the average distribution rates of an antibiotic in infants, toddlers and school age ch...
A study utilizes ANOVA to examine the average distribution rates of an antibiotic in infants, toddlers and school age children. The results include an F statistic that is 1.04. You know this means:
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A study is designed to examine whether there is a difference in
mean daily calcium intake among three groups of adults with normal
bone density (Norm), adults with osteopenia (OstPNia) (a low bone
density which may lead to osteoporosis) and adults with
osteoporosis (OstPSis). A total of twenty-one adults at age 60 was
recruited in the study (7 adults in each group). Each participant's
daily calcium intake was measured based on reported food intake and
supplements in milligrams. We...
Imagine that you are doing an exhaustive study on the children in all of the elementary schools in your school district. You are particularly interested in how much time children spend doing active play on weekends. You find that for this population of 2,431 children, the average number of minutes spent doing active play on weekends is μ = 87.85, with a standard deviation of ơ-118.1. You select a random sample of 25 children of elementary school age in this...
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