Question 1181382: Refer to the technology output given to the right that results from measured hemoglobin levels (g/dL) in 100 randomly selected adult females.
The confidence level of 95% was used.
a. Express the confidence interval in the format that uses the "less than" symbol. Assume that the original listed data use two decimal places,
and round the confidence interval limits accordingly.
b. Identify the best point estimate of μ and the margin of error.
c. In constructing the confidence interval estimate of μ,
why is it not necessary to confirm that the sample data appear to be from a population with a normal distribution?
TInterval
(12.800,13.318)
x=13.059
Sx=1.307
n=100
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's the breakdown of how to interpret the technology output and answer the questions:
**(a) Confidence Interval Format:**
The technology output gives the 95% confidence interval as (12.800, 13.318). In the "less than" symbol format, this is expressed as:
12.80 g/dL < μ < 13.32 g/dL
**(b) Point Estimate and Margin of Error:**
* **Best Point Estimate of μ:** The best point estimate of the population mean (μ) is the sample mean (x̄), which is given as x̄ = 13.059 g/dL.
* **Margin of Error:** The margin of error is half the width of the confidence interval. You can calculate it as follows:
Margin of Error = (Upper Limit - Lower Limit) / 2
Margin of Error = (13.318 - 12.800) / 2
Margin of Error = 0.518 / 2
Margin of Error = 0.259 g/dL
**(c) Normality Assumption:**
Because the sample size is large (n = 100), the Central Limit Theorem applies. The Central Limit Theorem states that the distribution of sample means will be approximately normal, *regardless* of the shape of the population distribution, as long as the sample size is sufficiently large (generally considered to be n ≥ 30). Therefore, it is *not* necessary to confirm that the sample data appear to be from a normally distributed population when constructing this confidence interval.
|
|
|
