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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

    © BrainMass Inc. brainmass.com December 24, 2021, 5:06 pm ad1c9bdddf
    https://brainmass.com/statistics/confidence-interval/test-hypothesis-confidence-interval-28462

    SOLUTION This solution is FREE courtesy of BrainMass!

    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!

    © BrainMass Inc. brainmass.com December 24, 2021, 5:06 pm ad1c9bdddf>
    https://brainmass.com/statistics/confidence-interval/test-hypothesis-confidence-interval-28462

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