Question

The concern of a study by Beynnon et al. (A-4) were nine subjects with chronic anterior cruciate ligamenttears. One of the variables of interest was the laxity of the anteroposterior, where higher values indicate more knee instability. The researchers found that among subjects with ACL-deficient knees, the mean laxity value was 17.4mm with a standard deviation of 4.3mm.

(a) What is the estimated standard error of the mean?
(b) Construct the 99 percent confidence interval for the mean of the population from which the nine subjects may be presumed to be a random sample.
(c) What is the precision of the estimate?
(d) What assumptions are necessary for the validity of the conidence interval you constructed?

Answer 1

a) Standard error of the mean = 1.433 mm

b) 99% confidence interval = (12.6, 22.2)

c) The precision of the estimate = 4 816

d) The assumptions that are necessary for the validity of the conidence interval constructed include

- The sample must be a random sample extracted from the population, with each variable in the sample independent from one another.

- The sample must be a normal distribution sample or approximate a normal distribution and the best way to establish this is when the population distribution where the sample was extracted from is normal or approximately normal.

Explanation:

a) Standard error of the mean is given as

σₓ = (σ/√n)

σ = Sample standard deviation = 4.3 mm

n = sample size = 9

σₓ = (4.3/√9) = 1.433 mm

b) Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = 17.4 mm

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 9 - 1 = 8.

Significance level for 99% confidence interval

(100% - 99%)/2 = 0.5% = 0.005

t (0.005, 8) = 3.36 (from the t-tables)

Standard error of the mean = 1.433 mm

99% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 17.4 ± (3.36 × 1.433)

CI = 17.4 ± 4.816

99% CI = (12.584, 22.216)

One crate orė

99% Confidence interval = (12.584, 22.216)

c) The precision of the estimate is gven as the length of the, margin of error of the confidence interval. The precision of the estimate = 4.816

d) They include

- The sample must be a random sample extracted from the population, with each variable in the sample independent from one another.

- The sample must be a normal distribution sample or approximate a normal distribution and the best way to establish this is when the population distribution where the sample was extracted from is normal or approximately normal.

Hope this Helps!!!