A study that looked at beverage consumption used sample sizes that were much sma

Question

A study that looked at beverage consumption used sample sizes that were much smaller than previous national surveys. One part of this study compared 20 children who were 7 to 10 years old with 5 who were 11 to 13. The younger children consumed an average of 8.3 oz of sweetened drinks per day while the older ones averaged 14.5 oz. The standard deviations were 10.7 oz and 8.3 oz respectively.Use younger children as population 1 (a) Do you think that it is reasonable to assume that these data are Normally distributed? Explain why or why not. (Hint: Think about the 68-95-99.7 rule.) The distributions appear to be Normal according to the 68-95-99.7 rule. We can see that the groups have means and standard deviations, two characteristics of Normal The 68-95-99.7 rule suggests that the distributions are not Normal. If they were, we would expect some of the children to drink negative amounts of sweetened drinks, which does not make sense (b) Using the methods in this section, test the null hypothesis that the two groups of children consume equal amounts of sweetened drinks versus the two-sided alternative. Report all details of the significance-testing procedure with your conclusion. (Round your answers to three decimal places.) SED P-value c) Give a 95% confidence interval for the difference in means Round your answers to four decimal places. (d) Do you think that the analyses performed in parts (b) and (c) are appropriate for these data? Explain why or why not. O The t procedures are appropriate for these data because the distributions are Normal and the samples are large enough O The t procedures are questionable for these data because the distributions are not Normal and the samples are small

Explanation / Answer

a)

Option B.

b)

SE = sqrt[ (s12/n1) + (s22/n2) ]

(s12/n1) = 5.7245

(s22/n2) = 13.7780

SE = 4.4162

Test Statistics

t = [ (x1 - x2) - d ] / SE

t = -1.4039

df = 8

p-value = 0.099

c)

x1bar - x2bar = -6.20

SE = 4.4162

CI = 95%

DF = 8

t-value = 2.3060

ME = t*SE = 10.1837

Confidence Interval is (x1bar - x2bar) +/- ME

Lower bound = -16.38369

Upper bound = 3.98369

Confidence Interval is (-16.3837 , 3.9837 )

d)

Option B

e)

Option B

x1(bar) 8.30 x2(bar) 14.50 s1 10.70 s2 8.30 n1 20 n2 5 SE = sqrt[ (s12/n1) + (s22/n2) ] (s12/n1) 5.7245 (s22/n2) 13.7780 SE 4.4162 DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] } [ (s12 / n1)2 / (n1 - 1) ] 1.725 [ (s22 / n2)2 / (n2 - 1) ] 47.46 (s12/n1 + s22/n2)2 380.35 DF = 8

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