Question

A medical statistician wants to estimate the average weight loss of those who have been on a new diet plan for 2 weeks. From a random sample of 100 participants in the diet plan, a mean weight change of -2.5 pounds and a standard deviation of 10 pounds are observed. In order to claim that the diet plan is effective, the whole part of the confidence interval for the mean weight change has to be below zero. In other words, the sample mean weight change has to be negative by more than the margin of error.

Hint : z0.1 = 1.28, z0.05 = 1.64, z0.025 = 1.96, z0.01 = 2.33 and z0.005 = 2.58.


1. Construct the 99% confidence interval. Can the statistician claim that the diet plan is effective?

2. Construct the 90% confidence interval. Can the statistician claim that the diet plan is effective? Is your answer different from that in part (a)?

Answer 1

a)

sample mean, xbar = -2.5
sample standard deviation, σ = 10
sample size, n = 100

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

ME = zc * σ/sqrt(n)
ME = 2.58 * 10/sqrt(100)
ME = 2.58

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (-2.5 - 2.58 * 10/sqrt(100) , -2.5 + 2.58 * 10/sqrt(100))
CI = (-5.08 , 0.08)

b)

sample mean, xbar = -2.5
sample standard deviation, σ = 10
sample size, n = 100

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

ME = zc * σ/sqrt(n)
ME = 1.64 * 10/sqrt(100)
ME = 1.64

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (-2.5 - 1.64 * 10/sqrt(100) , -2.5 + 1.64 * 10/sqrt(100))
CI = (-4.14 , -0.86)

yes, answer is different from a)