 2022-09-23

Confidence Intervals (A level)
Firstly let me apologise for asking this question on here - I see this as getting a sledgehammer to crack a nut, but I have nobody else that I can ask for advice on this topic.
I'm an A level Mathematics student so please forgive me for lack of/poor notation that you may normally come to expect/be familiar with.
I am looking at the specification for my exam that is coming up and these are three sections:
A) Construct symmetric confidence intervals for the mean of a normal distribution with known variance
B) Construct symmetric confidence intervals from large samples, for the mean of a normal distribution with unknown variance.
C) Construct symmetric confidence intervals from small samples, for the mean of a normal distribution with unknown variance using the t -distribution. Marvin Hughes

Step 1
The difference between B and C is in the choice of critical value. A typical symmetric confidence interval for a location parameter has the form
$\text{point estimate}±\text{critical value}×\text{standard error},$
where in turn
$\text{standard error}=\frac{\text{standard deviation}}{\sqrt{\text{sample size}}},$
and
$\text{critical value}×\text{standard error}=\text{margin of error}.$
The choice of critical value is informed by the nature of the sampling distribution. When a normally distributed population has unknown variance, the sampling distribution of the mean is Student's t-distributed; however, when the sample size is large, the difference in critical values is negligible; i.e.,
$\underset{\nu \to \mathrm{\infty }}{lim}{t}_{\nu ,\alpha }^{\ast }={z}_{\alpha }^{\ast },$
where ${t}_{\nu ,\alpha }^{\ast }$ is the upper $\alpha$ quantile of the Student's t distribution with $\nu$ degrees of freedom, and ${z}_{\alpha }^{\ast }$ is the upper $\alpha$ quantile of the standard normal distribution. So for scenario B, you will use ${z}_{\alpha /2}^{\ast }$ for a two-sided CI with large sample size as an approximation; for scenario C, you will use ${t}_{\nu ,\alpha /2}^{\ast }$ for the critical value.
Step 2
The difference between A and B lies in the standard deviation. In scenario A, it is presumed to be known, thus the standard error will use the population standard deviation $\sigma$ in the numerator. In scenario B, it is unknown, thus you will use the unbiased estimator of the standard deviation
$s=\sqrt{\frac{1}{n-1}\sum _{i=1}^{n}\left({x}_{i}-\overline{x}{\right)}^{2}}.$
In practice, the factor of $n-1$ rather than n makes little difference when n is large as in this case. But in scenario C, as $\sigma$ is also unknown and must be estimated from the sample, you must use $n-1$, otherwise your CI will not have the required coverage probability even if you correctly use the t-distribution quantile for the critical value.

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