1. Assuming that n1 = n2, find the sample sizes needed to estimate (p1 - p2) cor

Question

1.

Assuming that n1 = n2, find the sample sizes needed to estimate (p1 - p2) correct to within 0.07 with probability 0.90. Assume that there is no prior information available to obtain sample estimates of p1 and p2.

2.  

In order to compare the means of two populations, independent random samples are selected from each population, with the results shown in the table below. Use these data to construct a 98% confidence interval for the difference in the two population means.

3.  

A beef cattle nutritionist wants to compare the birth weights of calves from cows that receive two different diets during gestation. He therefore selects 16 pairs of cows, where the cows within each pair have similar characteristics. One cow within each pair is randomly assigned to diet 1, while the other cow in the pair is assigned to diet 2. He obtains the following results:

     Mean difference in birth weights of the pairs of calves = 10 lb

     Standard deviation of the difference in birth weights of the pairs = 8.0 lb

Construct a 95% confidence interval for the true mean difference in birth weights of the calves from cows receiving diet 1 vs. diet 2.

4.  

Frame score in beef cattle is based on height at the hips and is used as a measure of skeletal size. Frame scores range from 1 to 10 with a higher number indicating a taller animal. Independent random samples of frame scores were selected from the Angus and Simmental breeds of beef cattle with the following results:

In the Analysis of Variance table, the degrees of freedom for total, breeds, and error, respectively, are:

Explanation / Answer

QUESTION 2.

TRADITIONAL METHOD
given that,
mean(x)=5275
standard deviation , 1 =150
population size(n1)=400
y(mean)=5240
standard deviation, 2 =200
population size(n2)=400
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((22500/400)+(40000/400))
= 12.5
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.02
from standard normal table, two tailed z /2 =2.326
since our test is two-tailed
value of z table is 2.326
margin of error = 2.326 * 12.5
= 29.075
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (5275-5240) ± 29.075 ]
= [5.875 , 64.125]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=5275
standard deviation , 1 =150
number(n1)=400
y(mean)=5240
standard deviation, 2 =200
number(n2)=400
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 = [ ( 5275-5240) ±Z a/2 * Sqrt( 22500/400+40000/400)]
= [ (35) ± Z a/2 * Sqrt( 156.25) ]
= [ (35) ± 2.326 * Sqrt( 156.25) ]
= [5.875 , 64.125]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 98% sure that the interval [5.875 , 64.125] 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, 98% 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 0.02 true mean
difference is zero

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