A study of iron deficiency among infants compared samples of infants following d
ID: 3182218 • Letter: A
Question
A study of iron deficiency among infants compared samples of infants following different feeding regimens. One group contained breast-fed infants, while the children 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.
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.
Fail to reject the null hypothesis. There is 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. (which one is the correct answer)
We need sample sizes greater than 40.We need the data to be from a skewed distribution.
We need two independent SRSs from normal populations.
We need two dependent SRSs from normal populations.
Group n x s Breast-fed 23 13.2 1.7 Formula 19 12.5 1.8Carry 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? Which one is correct?
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.
Fail to reject the null hypothesis. There is 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. (which one is the correct answer)
We need sample sizes greater than 40.We need the data to be from a skewed distribution.
We need two independent SRSs from normal populations.
We need two dependent SRSs from normal populations.
Explanation / Answer
Given that,
mean(x)=13.2
standard deviation , s.d1=1.7
number(n1)=23
y(mean)=12.5
standard deviation, s.d2 =1.8
number(n2)=19
null, Ho: u1 < u2
alternate, H1: u1 > u2
level of significance, = 0.01
from standard normal table,right tailed t /2 =2.552
since our test is right-tailed
reject Ho, if to > 2.552
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =13.2-12.5/sqrt((2.89/23)+(3.24/19))
to =1.286
| to | =1.286
critical value
the value of |t | with min (n1-1, n2-1) i.e 18 d.f is 2.552
we got |to| = 1.28624 & | t | = 2.552
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > 1.2862 ) = 0.10733
hence value of p0.01 < 0.10733,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 < u2
alternate, H1: u1 > u2
test statistic: 1.286
critical value: 2.552
decision: do not reject Ho
p-value: 0.10733
Fail to reject the null hypothesis. There is not significant evidence that the mean hemoglobin level is higher among breast-fed babies.
c.
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=13.2
Standard deviation( sd1 )=1.7
Sample Size(n1)=23
Mean(x2)=12.5
Standard deviation( sd2 )=1.8
Sample Size(n2)=19
CI = [ ( 13.2-12.5) ±t a/2 * Sqrt( 2.89/23+3.24/19)]
= [ (0.7) ± t a/2 * Sqrt( 0.2962) ]
= [ (0.7) ± 2.101 * Sqrt( 0.2962) ]
= [-0.4434 , 1.8434]
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