Question

The mean serum- creatinine level measured in 12 patients 24 hours after they received a newly proposed antibiotics was 1.2 mg/ dL. Show your work

a) If the mean and standard deviation of serum creatinine in the general population are 1.0 and 0.4 mg/dL respectively, then using a significance level of 0.05, test whether the mean serum- creatinine level in this group is different from that of the general population.

b) Suppose the standard deviation of serum creatinine is 0.6mg/dl. Assume now that the standard deviation of the serum creatinine in the general population is not known; perform a hypothesis testing to see whether the mean serum- creatinine level in this group is different from that of the general population.

Answer 1

To test whether the mean serum-creatinine level in a group of patients is different from that of the general population, we use a z-test when the population standard deviation is known, and a t-test when it is unknown. Calculations involve comparing a calculated z-score or t-score to critical values at the 0.05 significance level to decide whether to reject the null hypothesis.

Step-by-step explanation:

To answer the student's question, we must conduct hypothesis tests to compare the sample mean to the population mean under different circumstances. In part a), we will assume the population standard deviation is known, and in part b), we will assume it is unknown and use a t-test instead. Here is the step-by-step process:

Part a)

First, we state the null hypothesis (H0: μ = 1.0) and alternative hypothesis (H1: μ ≠ 1.0), where μ is the population mean of serum-creatinine levels.

We calculate the z-score using the formula Z = (Xbar - μ0) / (σ / sqrt(n)), where Xbar = 1.2 mg/dL, μ0 = 1.0 mg/dL, σ = 0.4 mg/dL, and n = 12.

Upon calculating, we compare the z-score to the critical value at the 0.05 significance level.

If the z-score is outside the range of critical values, we reject H0; otherwise, we fail to reject H0.

Part b)

1. Since the population standard deviation is unknown, we use the sample standard deviation (s = 0.6 mg/dL).
2. The t-score is calculated using the formula T = (Xbar - μ0) / (s / sqrt(n)), and degrees of freedom (df) are n-1.
3. The critical t-value is found using a t-distribution table at the 0.05 significance level and df = 11.
4. As before, if the t-score exceeds the critical value, we reject H0; otherwise, we fail to reject H0.