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# Test of hypothesis, Confidence Interval

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1) Recruit 25 volunteers who have received a medication and find they have a mean concentration of 7 with a standard deviation of 2. Assume normally distributed.

FIND: 95% CONFIDENCE INTERVAL for pop mean concentration
Find: 99% confidence interval for pop var of conc
HOW large a sample is needed to ensure the length of the CI in prob 1 is 0.5.

2.

100 hypertensive people are given z drug that is effective on 20 people. Effective=there blood pressure is lowered by at least 10.

6.54 What is the standard error of d?
6.55 What is the 95% confidence interval for the pop mean of d?
6.56 Can we make a statement about the effectiveness of this drug?
6.57 What does a 95% confidence interval mean in words

https://brainmass.com/statistics/confidence-interval/test-hypothesis-confidence-interval-28462

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Problem 1

Recruit 25 volunteers who have recieved a medication and find they have a mean concentration of 7 with a standard deviation of 2. Assume normall ditributed.

FIND: 95% CONFIDENCE INTERVAL for pop mean concentration
Find: 99% confidence interval for pop var of conc
HOW large a sample os needed to ensure the length of the CI in prob 1 is 0.5.

FIND: 95% CONFIDENCE INTERVAL for pop mean concentration

Mean=M = 7
Standard deviation =s= 2
sample size=n= 25
sx=standard error of mean=s/square root of n= 0.4 = ( 2 /square root of 25)
Confidence level= 95%
Therefore Significance level=alpha (a) = 5% =100% -95%
No of tails= 2

Since sample size= 25 < 30
and we are estimating population standard deviation from the sample use t distribution

t at the 0.05 level of significance and 24 degrees of freedom (=n-1) and 2 tail= 2.0639

Upper confidence limit= M +t*sx= 7.8256 =7+2.0639*0.4
Lower confidence limit= M -t*sx= 6.1744 =7-2.0639*0.4

Confidence interval is between 6.174 and 7.8256

Find: 99% confidence interval for pop var of conc

sample size=n= 25
standard deviation =s= 2
Variance =s 2 = 4 =2 ^2

We use the chi square distribution for the confidence interval for the variance

degrees of freedom=n-1= 24 =25-1
Confidence interval of 99%
Corresponding to this confidence interval and degrees of freedom the chi square values are
X L^ 2 = 9.8862
X U ^ 2 = 45.5584

sL 2 =(n-1) s 2 / X U = 2.1072 =24*4 / 45.5584
or sL = 1.45 =square root of 2.1072

sU 2 =(n-1) s 2 / X L = 9.7105 =24*4 / 9.8862
or sU = 3.12 =square root of 9.7105

Answer Upper confidence limit for variance= 9.71
Lower confidence limit for variance= 2.11

Upper confidence limit for standard deviation= 3.12
Upper confidence limit for standard deviation= 1.45

HOW large a sample is needed to ensure the length of the CI in prob 1 is 0.5.

Here the sample size would be greater than 30 as the CI = 0.5
as against 1.651 =7.8256-6.1744 in part 1
Therefore we will use z distribution

Confidence level= 95%
Therefore Significance level=alpha (a) = 5% =100% -95%
Standard deviation =s= 2

Z at the 0.05 level of significance 2 tail = 1.96

Since the confidence interval = 0.5

Upper confidence limit= M +z*sx= 7.25
Lower confidence limit= M -z*sx= 6.75
difference= 0.5

Thus sx=(7.25-7)/z
Since Z= 1.96
sx=(7.25-7)/1.96= 0.1276

Standard deviation =s= 2
sx=standard error of mean=s/square root of n=2/square root of n= 0.1276 (calculated above)
Thus the only unknown is n or the sample size
square root of n= 2/0.1276
or n= 245.6736864 or rounding it up= 246

Answer : the sample size should be equal to 246

We can check this

Mean=M = 7
Standard deviation =s= 2
sample size=n= 245.6736864
sx=standard error of mean=s/square root of n= 0.1276 = ( 2 /square root of 245.673686382799)
Confidence level= 95%
Therefore Significance level=alpha (a) = 5% =100% -95%
No of tails= 2

Since sample size= 245.6736864 >= 30
use normal distribution

Z at the 0.05 level of significance 2 tailed test = 1.96

Upper confidence limit= M +z*sx= 7.2501 =7+1.96*0.1276
Lower confidence limit= M -z*sx= 6.7499 =7-1.96*0.1276

Confidence interval is between 6.7499 and 7.2501

( The minor difference is because we have rounded up the sample size to a whole number)

Problem 2
100 hypertensive people are given z drug that is effective on 20 people. Effective=there blood pressure is lowered by at least 10.

6.54 What is the standard error of d?
6.55What is the 95% confidence interval for the pop mean of d?
6.56Can we make a statement about the effectiveness of this drug?
6.57What does a 95% confidence interval mean in words

6.54 What is the standard error of d?

Let p= proportion of people on which the drug is effective

p= 20.00% =20/100
q=1-p= 80.00%
n=sample size= 100
sp=standard error of proportion=square root of (pq/n)= 4.00% =square root of ( 20.% * 80.% / 100)

standard error= 4.00% or 0.04

6.55What is the 95% confidence interval for the pop mean of d?

Confidence level= 95%
Significance level=alpha (a) = 5% =100% -95%
No of tails= 2
This is 2 tails because we are calculating upper and lower confidence level

Since sample size= 100 >= 30 use normal distribution
Z at the 0.05 level of significance 2 tailed test = 1.96

Upper confidence limit= p+z*sp= 27.84% =20.%+1.96*4.%
Lower confidence limit= p-z*sp= 12.16% =20.%-1.96*4.%

The 95 % confidence interval is between 12.16% and 27.84%

6.56Can we make a statement about the effectiveness of this drug?

Yes we can make a statement about the effectiveness of the drug.

From the sample results we can say that the drug is effective in 20% of the cases.

6.57What does a 95% confidence interval mean in words

It means that we can be 95% confident that if we take a number of samples of hypertensive people who are given the drug, the proportion of people on whom the drug is effective will be between 12.16% and 27.84%

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!