Noah and Clare are studying the noise level outside of their school in the morning. Noah's
sample contains 12 time periods, and Clare's sample contains 25 time periods.
How do you think Noah's reported margin of error compares to Clare's? Explain your
reasoning.
Noah's margin of error will be higher
his
than Clare's, because
sample is larger.



Answer :

To compare the margin of error between Noah's and Clare's samples, let's first understand the relation between sample size and margin of error. The margin of error for a sample is inversely proportional to the square root of the sample size. This means that as the sample size increases, the margin of error decreases.

Given:
- Noah's sample size: 12 time periods
- Clare's sample size: 25 time periods

Let's denote the margin of error as [tex]\( ME \)[/tex]. The margin of error can be represented approximately as:

[tex]\[ ME \propto \frac{1}{\sqrt{n}} \][/tex]

where [tex]\( n \)[/tex] is the sample size.

For Noah's sample size ([tex]\( n = 12 \)[/tex]):

[tex]\[ ME_{\text{Noah}} \propto \frac{1}{\sqrt{12}} \approx \frac{1}{3.464} \approx 0.2886751345948129 \][/tex]

For Clare's sample size ([tex]\( n = 25 \)[/tex]):

[tex]\[ ME_{\text{Clare}} \propto \frac{1}{\sqrt{25}} = \frac{1}{5} = 0.2 \][/tex]

When comparing these values:
- [tex]\( ME_{\text{Noah}} \approx 0.2886751345948129 \)[/tex]
- [tex]\( ME_{\text{Clare}} = 0.2 \)[/tex]

Observing these results, we see that Noah's margin of error is approximately [tex]\( 0.2886751345948129 \)[/tex], which is higher than Clare's margin of error of [tex]\( 0.2 \)[/tex].

Therefore, Noah's reported margin of error will be higher than Clare's because his sample size is smaller, and the margin of error is inversely proportional to the square root of the sample size. This makes Noah's margin of error larger due to his smaller sample size compared to Clare's larger sample size.