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Problem 3 [25 points]: Consider a new proposed antibiotic is taken by 16 patients, and their serum-creatinine level are measured 24 hours after. Assume that the serum-creatinine level is normally distributed, and the sample mean of serum-creatinine level was 1.80 mg/dL and sample variance of serumcreatinine level was 0.0625. Suppose that the standard deviation of serum-creatinine level in population is known at 0.28 ngdl. (a) [3 points] Construct a 95% Confidence Interval for the mean serum-creatinine level ,u. (b) [4 points] Instead of using the 20025 = -1.96 and 20.075 =1.96 to construct the 95% Confidence Interval for p in part (a), construct another 95% Confidence Interval for )1: based on 20.0; and 2007 (Le. 2"d and 97th percentile of N (0,1)). (c) [4 points] Compare part (b) to part (a), what's the difference between these two confidence intervals? Which confidence intervals provide a more accurate estimation? Why? [Why we commonly use Symmetry setting (zW2 and. 21.\") in confidence interval?1 (d) [4 points] if we want to control the width of 99% confidence interval for p is less than 0.05 (Margin of Error is less than 0.025), what's the minimum value of the sample size n we should collect? QWEW (e) [6 points] Construct a 90% Confidence Interval for the mean of serum-creatinine level and a 90% Confidence Interval for the Variance of serumcreatinine level 02 based on the sample mean and sample variance. (f) [4 points] What's the difference between t-distribution and normal distribution, and what's the difference between the confidence interval generated by normal distribution and the confidence interval generated by t-distribution