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.
Side note to help you understand the interpretation: The upper limit of the confidence interval is referred to as the upper confidence limit. For example, if the confidence interval is between -2 lbs and -1 lb, -1 is the upper confidence limit. This is the most conservative estimate of the mean. That is, the statistician can claim it has been statistically shown that the average weight loss is at least a pound (with a certain level of confidence) because it is the upper confidence limit.
A.
We have given,
Sample Size = 400
Mean = -2.5 pounds
Standard Deviation = 10 pounds
Confidence interval = 95%
z value at 95% confidence interval (z0.025) = 1.96
Following formula can be used to calculate the confidence interval:
Where,
X is the sample mean
Z is the z score
σ is standard deviation
n is the sample size
The confidence intervals are ( -2.5 + 0.98 , -2.5 - 0.98) ( -1.52 , -3.48)
The Upper Confidence interval is -1.52
B. Following formula can be used to calculate the sample size:
Given Data:
The Margin of Error is already calculated in Part A (E) = 0.98
Z = 1.96
After calculating, we get n > = 400.
The sample size should be greater than 400.
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