# Confidence Intervals Essay

## Confidence Intervals

As hypothesized, high cholesterol levels in children can lead to their children being affected with hyperlipidemia. A study is conducted to estimate the mean cholesterol in children between the ages of 2 – 6 years of age. It also attempted to establish a correlation as to the effect family history has on the onset of the disease. From data collected as shown in Table 1 of the spreadsheet attached, a sample size of 9 (n=9) participants enrolled in the study. Total cholesterol levels measured in children between ages 2 – 6 years was summarized at 1,765. The sample mean (X) and standard deviation (S) computed as (1765/90) =196.1 and square root summation (X-X) square / n-1 =29.0 respectively. Now to generate a 95% confidence interval for the true mean total cholesterol levels in children from data collected, we used the z value for 95% as (z= 1.96).

From sample statistics the confidence interval for 95% computed from the formula is (196.1 +/- 1.96 X 29/3) we now have 196.1 +/-19.0. Now, by adding and subtracting the margin of error, we have (215.1, 177.1) respectively. A point estimate for the true mean cholesterol levels in the population is 196.1 and that we are 95% confident that the true mean is between 215.1 and 177.1. The margin or is large because of the small sample size. A pilot study with 10 participants was conducted to assess how systolic blood pressure changes overtime if left. Clinical trial compared experimental medication designed to lower to that of a placebo. The sample mean (X) for the difference in blood pressures over a four weeks period is computed as (9/10) =0.09 or 90%. Also, the standard deviation for the difference over the same four weeks period computed as [(X) 2/square root n-1] = 183/20.33= 4.5, and. Z value =1.96. The confident interval (CI) for 95% difference in blood pressures from sample data collected as seen in table 2 on the spreadsheet attached is computed as [(0.9 +/- 1.96 (4.5)/3.162 = 0.9 +/- 2.78.

By adding and subtracting the margin of error, we have (3.68, -1.88) respectively. The point estimate for the true difference in blood pressures over a four weeks period in the population is 0.9 and we are 95% confident that the true mean for the difference in blood pressures over four weeks is between 3.68 and -1.88. Now a main trial is conducted with 200 patients enrolled and randomly assigned to either one of the two groups: experimental medication group or placebo group. At the end of the six weeks treatment period the data displayed both groups as assigned in table 3 on the spreadsheet attached. The two comparison groups physically separated where 100 patients were assigned to receive the experimental drug and another 100 patients to receive the placebo.

We first compute the descriptive statistics on each of the of the two samples i.e. sample size, mean, and standard deviation denoted as n1, X1, s1 for sample 1 and n2, X2, s2 for sample 2 respectively. The point estimate for the difference in the population is the difference in the sample means indicated as (X1-X2) and square root of 1/n1 + 1/n2 as the pooled estimate of the common standard deviation (Sp). Sp = square root (n1-1) s1 square + (n2-1) square / n1 + n2 -2; thus Sp = 17.07. Given that Sp (17.07), sample deviation (-11.2) and z value at 95% (1.96). we now compute the 95% confident interval (CI) for the difference in mean systolic blood pressure between groups is (-6.52, -15.8). The CI is interpreted as follows: We are 95% confident that that the difference in the mean systolic blood pressures between group one and group two is between -6.52 and -15.8 respectively.

References

Sullivan, L. (2007). Essential of biosatistics in public health. Sudbury, MA: Jones and Barlett learning. http://www.jblearning.com