A study of iron deficiency among infants compared samples of infants following d

Question

A study of iron deficiency among infants compared samples of infants following different feeding regimens. One group contained breast-fed infants, while the infants in another group were fed a standard baby formula without any iron supplements. Here are summary results on blood hemoglobin levels at 12 months of age. Group n x s Breast-fed 22 13.1 1.7 Formula 18 12.6 1.8 (a) Is there significant evidence that the mean hemoglobin level is higher among breast-fed babies? State H0 and Ha. H0: breast-fed < formula; Ha: breast-fed = formula H0: breast-fed > formula; Ha: breast-fed = formula H0: breast-fed = formula; Ha: breast-fed > formula H0: breast-fed formula; Ha: breast-fed < formula Carry out a t test. Give the P-value. (Use = 0.01. Use breast-fed formula. Round your value for t to three decimal places, and round your P-value to four decimal places.) t = P-value = What is your conclusion? Fail to reject the null hypothesis. There is significant evidence that the mean hemoglobin level is higher among breast-fed babies. Fail to reject the null hypothesis. There is not significant evidence that the mean hemoglobin level is higher among breast-fed babies. Reject the null hypothesis. There is significant evidence that the mean hemoglobin level is higher among breast-fed babies. Reject the null hypothesis. There is not significant evidence that the mean hemoglobin level is higher among breast-fed babies. (b) Give a 95% confidence interval for the mean difference in hemoglobin level between the two populations of infants. (Round your answers to three decimal places.) , (c) State the assumptions that your procedures in (a) and (b) require in order to be valid. We need two independent SRSs from normal populations. We need the data to be from a skewed distribution. We need sample sizes greater than 40. We need two dependent SRSs from normal populations.

Explanation / Answer

PART A.
Given that,
mean(x)=13.1
standard deviation , s.d1=1.7
number(n1)=22
y(mean)=12.6
standard deviation, s.d2 =1.8
number(n2)=18
null, Ho: breast-fed = formula
alternate, H1:breast-fed > formula
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.74
since our test is right-tailed
reject Ho, if to > 1.74
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =13.1-12.6/sqrt((2.89/22)+(3.24/18))
to =0.896
| to | =0.896
critical value
the value of |t | with min (n1-1, n2-1) i.e 17 d.f is 1.74
we got |to| = 0.89606 & | t | = 1.74
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > 0.8961 ) = 0.19137
hence value of p0.05 < 0.19137,here we do not reject Ho
ANSWERS
---------------
H0: breast-fed = formula; Ha: breast-fed > formula
test statistic: 0.896
critical value: 1.74
decision: do not reject Ho
p-value: 0.19137
Fail to reject the null hypothesis. There is not significant evidence that
the mean hemoglobin level is higher among breast-fed babies

PART B.
TRADITIONAL METHOD
given that,
mean(x)=13.1
standard deviation , s.d1=1.7
number(n1)=22
y(mean)=12.6
standard deviation, s.d2 =1.8
number(n2)=18
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((2.89/22)+(3.24/18))
= 0.558
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 17 d.f is 2.11
margin of error = 2.11 * 0.558
= 1.177
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (13.1-12.6) ± 1.177 ]
= [-0.677 , 1.677]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=13.1
standard deviation , s.d1=1.7
sample size, n1=22
y(mean)=12.6
standard deviation, s.d2 =1.8
sample size,n2 =18
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 13.1-12.6) ± t a/2 * sqrt((2.89/22)+(3.24/18)]
= [ (0.5) ± t a/2 * 0.558]
= [-0.677 , 1.677]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-0.677 , 1.677] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

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