Expert: Scott A Wilson - 3/28/2009

Question
Sir,I am MBA student. Please give the answers of my questions. Thank you.
1.Nonstandard dice can produce interesting distributions of outcomes. You have two balanced, six-sided dice. One is a standard dice,with faces having 1,2,3,4,5 and 6 spots.The other die has three faces with 0 spots and three faces with 6 spots.Find the probability distribution for the total number of spots Y on the up-faces when you roll these two dice.
2.What is a chi-square test? How do you find the degrees of freedom in a chi-square distribution?Discuss chi-square test as a test for goodness of fit and as a test of independence.
3.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 other 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      Mean    s
Breast-fed   23     13.3    1.7
Formula      19     12.4    1.8
a. Is there significant evidence that the mean hemoglobin level is higher among breast-fed babies? State null hypothesis and alternate hypothesis and conduct a t-test.
b. Give a 95% confidence interval for the mean difference in hemoglobin level between the two populations of infants.
4.What do you mean by a seasonal index? Expain ratio to link relatives method of measuring seasonal variations.

Answer
1.Nonstandard dice can produce interesting distributions of outcomes. You have two balanced, six-sided dice. One is a standard dice,with faces having 1,2,3,4,5 and 6 spots.The other die has three faces with 0 spots and three faces with 6 spots.Find the probability distribution for the total number of spots Y on the up-faces when you roll these two dice.

One of the dice could come up with 1 through 6, all with 1/6 chance.
The other die could come up with a 0 or a 6, with ½ chance for each.
If the second die is a 0, the sum is 1 through 6.
If the second die is a 6, the sum is 7-12.
This means all occurrences are equally likely.

The probability distribution is

1   1/12
2   1/12
3   1/12
4   1/12
5   1/12
6   1/12
7   1/12
8   1/12
9   1/12
10   1/12
11   1/12
12   1/12


2.What is a chi-square test? How do you find the degrees of freedom in a chi-square distribution?
Discuss chi-square test as a test for goodness of fit and as a test of independence.

A paper describing the chi-square distribution is http://en.wikipedia.org/wiki/Chi-square_distribution
A good sample of it is found in http://www.colby.edu/biology/BI17x/freq.html
Basically, it is the test between the expected results and the measured results.

The degrees of freedom of a chi-square test are the number of parameters minus the number of averages looked at.  In most cases, it is 1.  If you looked at who consumed oranges, the number of degrees of freedom would be the number - 1.  If you also looked at whether they were male or female and how many oranges they ate, the degrees of freedom would be the number – 2.  If, in addition to how many oranges and whether they were male or female, you looked into whether they were eaten in the daylight or not, the degrees of freedom would be the number of observations – 3.


3.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 other 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      Mean    s
Breast-fed   23     13.3    1.7
Formula      19     12.4    1.8
a. Is there significant evidence that the mean hemoglobin level is higher among breast-fed babies? State null hypothesis and alternate hypothesis and conduct a t-test.
b. Give a 95% confidence interval for the mean difference in hemoglobin level between the two populations of infants.

I found a papper that explains the difference between two means with differing standard deviations.  It is
//stattrek.com/AP-Statistics-4/Unpaired-Means.aspx?Tutorial=AP

Let the SE = sqrt(Sqr(s1)/n1 + sqr(s2)/n2)).
The t statistic is (avg1-avg2)/SE.

A t table will give you whether or not this is significant.
A 95% confindecne interval can be found there as well.
It could be the overall average { (n1*mean1 + n2*mean2)/(n1+n1) }
plus/minu the 95% t statistic with n1 + n2 - 2 degrees of freedom.


4. A seasonal index is meant by noting any difference in which season it is.  For example, you might consider measurements throughout the year and note and statistical difference in them that implied it varied by season.  I would lump the data into four groups by season if it was available