Math  /  Data & Statistics

QuestionAn NHANES report gives data for 641 men aged 202920-29 years. The BMI of these 641 men was xˉ=25.6\bar{x}=25.6. On the basis of this sample, we want to estimate the BMI μ\mu in the population of all 23.2 million American men in this age group. Treat these data as an SRS from a Normally distributed population with standard deviation σ=7.2\sigma=7.2.
Source adapted from: Adapted from Fryar C. D., et al., Anthropometric reference data for children and adults: United States, 2011-2014. National Center for Health Statistics. Vital Health Statistics 3(2016), at htips:/www. ode.gov/nchs/data/ series/mr_04sr03_039.pAf.
Compare the three margins of error. How does increasing the sample size change the margin of error of a confidence interval when the confidence level and population standard deviation remain the same? Margin of error remains the same as nn increases. Margin of error is unaffected by the sample size nn. Margin of error increases as nn increases. Margin of error decreases as nn increases.

Studdy Solution

STEP 1

What is this asking? We're looking at how the margin of error of a confidence interval changes when we increase the sample size, while keeping the confidence level and population standard deviation the same. Watch out! Don't mix up sample size with population size!
We're interested in how *sample* size affects the margin of error.

STEP 2

1. Margin of Error Formula
2. Effect of Sample Size

STEP 3

Alright, let's kick things off with the formula for the margin of error!
The margin of error for a z-confidence interval is given by m=zσnm = z^* \cdot \frac{\sigma}{\sqrt{n}}, where zz^* is the **critical value** based on the desired confidence level, σ\sigma is the **population standard deviation**, and nn is the **sample size**.

STEP 4

Now, let's focus on how the sample size (nn) affects the margin of error.
Notice that nn is in the *denominator* of the fraction and it's inside a square root.

STEP 5

Let's imagine what happens when nn gets bigger.
As nn increases, n\sqrt{n} *also* increases.
Since n\sqrt{n} is in the denominator, a *larger* denominator means the *entire* fraction σn\frac{\sigma}{\sqrt{n}} gets *smaller*.

STEP 6

So, as nn gets bigger, σn\frac{\sigma}{\sqrt{n}} gets smaller.
Since the margin of error, mm, is just zz^* *times* this fraction, a *smaller* fraction means a *smaller* margin of error.

STEP 7

In a nutshell, as the sample size (nn) increases, the margin of error (mm) *decreases*!
This makes sense, right?
A bigger sample gives us more information, so we can be more confident in our estimate, leading to a smaller margin of error.

STEP 8

Margin of error decreases as nn increases.

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