Question

STATISTICS...help please! I need to comment upon "the effect of varying the sample size and the confidence level on the relationship between
(i) the sample size and the sampling error
(ii) the confidence level and the sampling error.

Answer 1

The wording is slightly confusing. I know that as n increases, the sample error decreases
but I'm not sure if "confidence level" refers to the width of the confidence level or not...

Answer 2

Consider the confidence interval for a population mean with \((1-\alpha)100\%\) confidence:
\[\left(\bar{x}-Z_{\alpha/2}\frac{\sigma}{\sqrt n},~\bar{x}+Z_{\alpha/2}\frac{\sigma}{\sqrt n}\right)\]
Increasing the sample size will make the \(Z_{\alpha/2}\dfrac{\sigma}{\sqrt n}\) approach 0, so the confidence interval would approach \((\bar{x},\bar{x})\), or simply \(\mu=\bar{x}\).

Increasing the confidence level decreases the value of \(\alpha\), which will increase the value of \(Z_{\alpha/2}\). This means the \(Z_{\alpha/2}\dfrac{\sigma}{\sqrt n}\) will get larger, and so the interval would get wider.

Answer 3

Thanks for your reply!!
Oooh I see... I have 2 questions 1) with notation, can I represent an interval like you have done, or do I need square brackets or..?
2) So would the relationship between increasing the confidence level and the sampling error be connected by the interval width?
i.e. As \[Z _{\alpha/2}\] increases, w also increases, but
\[w=2\times1.96\sqrt{\frac{ p(1-p) }{ n }}\]
Therefore 'w' will be proportional to the Sampling error
\[SE=\sqrt{\frac{ p(1-p) }{ n }}\]
\[SE∝\frac{ 1}{ \sqrt{n} }∝w\]
Therefore, the sampling error will increase?
(I'm dealing with proportions btw)

Answer 4

Fortunately, the CI for proportions has a similar structure, so the conclusions don't change.

(1) I'm not entirely sure about whether closed brackets would be appropriate. I've only ever seen texts provide the interval formula with open parentheses. There must be some mathematical reason as to why an open interval is preferred. This link might interest you:
http://stats.stackexchange.com/questions/15872/are-confidence-intervals-open-or-closed-intervals

(2) Yes, you're right. Because the sampling error increases, the interval width also increases. This means the sample mean is not as good as it could be. A higher confidence level is considered "good" if the interval is small, but generally the interval gets wider. It can get smaller if the sample size also increases, but I assumed \(n\) is kept constant in the second case.

Answer 5

Cool, thanks for the link!
And that makes a lot of sense! Thank-you so so much! I really appreciate your help :)
Have a great week!

Answer 6

You're welcome!