22. How profi question is to examine profit as a percentage of stockholder 30 re

Question

22. How profi question is to examine profit as a percentage of stockholder 30 retail stocks was studied for z1. profit as a percentage o result was13.7. A rand percentage of stockholder equity. The result was = 9.5. Assume ?,-3.1 and table are different sectors of the stock market? One way to answer such a equity. A random sample of om sample of 35 utility stocks was studied for 2 profit as a ?2 2.8. (4 pt. ea.) 2SampTInt or 2-PropZlnt? Explain why aitWhich confidence interval procedure is appropriate in this case - 2SampZImt, b) Find the appropriate 95% confidence interval for the difference between the two populations. c) Interpret your answer. Does it appear at the 95% confidence level that the profit as a percentage of stockholder equity for retail stocks is higher than that for utility stocks? Explain. 3. Let p and p be the respective proportions of women with nutritional anemia in f two developing countries. If a random sample of 380 women from the first count ielded 254 women with nutritional anemia, and an independently chosen, random mple of 365 women from the second country yielded 201 women with nutritional emia, find a 99% confidence interval for P1-P2, (4 pt.)

Explanation / Answer

Q22.

TRADITIONAL METHOD
given that,
mean(x)=13.7
standard deviation , ?1 =3.1
population size(n1)=30
y(mean)=9.5
standard deviation, ?2 =2.8
population size(n2)=35
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((9.61/30)+(7.84/35))
= 0.7378
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, ? = 0.05
from standard normal table, two tailed z ?/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.7378
= 1.4461
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (13.7-9.5) ± 1.4461 ]
= [2.7539 , 5.6461]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=13.7
standard deviation , ?1 =3.1
number(n1)=30
y(mean)=9.5
standard deviation, ?2 =2.8
number(n2)=35
CI = x1 - x2 ± Z 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)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 13.7-9.5) ±Z a/2 * Sqrt( 9.61/30+7.84/35)]
= [ (4.2) ± Z a/2 * Sqrt( 0.5443) ]
= [ (4.2) ± 1.96 * Sqrt( 0.5443) ]
= [2.7539 , 5.6461]
-----------------------------------------------------------------------------------------------

a.
2 sample Z int

b.
[2.7539 , 5.6461]

c.
interpretations:
1. we are 95% sure that the interval [2.7539 , 5.6461] contains the difference between
true population mean U1 - U2
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 difference between
true population mean U1 - U2
3. Since this Cl does contain a zero we can conclude at 0.05 true mean
difference is zero

and it appears that the profit as a percentage of stock holder equity for retail stocks
is higher than that for uility stocks

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