Question

You have measured the blood hemoglobin concentrations in a random sample of 12 males aged 20-29 years and have obtained the following values in mg/dL: [ 14.7, 15.22, 15.28, 16.58, 15.1, 15.66, 15.91, 14.41, 14.73, 15.09, 15.62, 14.92] Calculate the following from the above sample: 1.95% confidence interval for the mean hemoglobin concentration in the population of 20-29 year old males. 2. 99% confidence interval for the mean hemoglobin concentration in the same population 3. 95% confidence interval for the variance and standard deviation of the hemoglobin concentration in this population 4.99% confidence interval for the variance and standard deviation of the hemoglobin concentration in this population Perform a one-sample t-test to check if the mean hemoglobin concentration in the population of 20-29 year old males is 16.0 mg/dL. The test questions will ask you about the final results of the above calculations as well as intermediate results that led to them, so save your work before answering the questions.

Answer 1

14.7 15.22 15.28 16.58 15.1 15.66 15.91 14.41 14.73 15.09 15.62 14.92 15.26833 0.598693 MEAN S.D

SOLUTION 1:SAMPLE MEAN = 15.27
t CRITICAL = 2.2
s_M = √(0.599²/12) = 0.17

μ = M ± t(s_M)
μ = 15.27 ± 2.2*0.17
μ = 15.27 ± 0.374

Result

M = 15.27, 95% CI [14.89, 15.64].

You can be 95% confident that the population mean (μ) falls between 14.89 and 15.64.

SOLUTION 2: SAMPLE MEAN = 15.27
t CRITICAL VALUE= 3.11
s_M = √(0.599²/12) = 0.17
μ = M ± t(s_M)
μ = 15.27 ± 3.11*0.17
μ = 15.27 ± 0.53

99% CI [14.74, 15.80].

You can be 99% confident that the population mean (μ) falls between 14.74 and 15.80.

SOLUTION 3: The critical values for α=0.05 and df=11 degrees of freedom are χL2​=χ1−α/2,n−1/2​=3.8157 and χU2​=χα/2,n−1/2​=21.92. The corresponding confidence interval is computed as shown below:

CI(Variance) = (n-1) s x 2,1-1 (n-1)s xi-a 2,n-1 - (0222015 (1303585 (12 - 1) x 0.3584 (12-1) X 0.3584 3.8157 21.92 = (0.1799, 1.0332)

Now that we have the limits for the confidence interval, the limits for the 95% confidence interval for the population standard deviation are obtained by simply taking the squared root of the limits of the confidence interval for the variance, so then:

CI(Standard Deviation) = (0.1799, V1.0332) = (0.4241, 1.0165)

SOLUTION 4: The critical values for α=0.01 and df=11 degrees of freedom are χL2​=χ1−α/2,n−1/2​=2.6032 and χU2​=χα/2,n−1/2​=26.7568. The corresponding confidence interval is computed as shown below

Now that we have the limits for the confidence interval, the limits for the 99% confidence interval for the population standard deviation are obtained by simply taking the squared root of the limits of the confidence interval for the variance, so then:

CI(Variance) = ((n − 1)s? (n − 1) s2 ( xa 2.1-1 'xi-a/2n-1 ((12-1) x 0.3584 (12-1) x 0.3584 26.7568 2.6032 = (0.1473, 1.5144)

AS PER THE GUIDELINES I HAVE DONE THE FIRST FOUR PLEASE RE POST THE REST. THANK YOU

CI(Standard Deviation) = (v0.1473, 71.5144) = (0.3839, 1.2306)