student's $t$-distribution

Question

Random sample of $457$

Sample mean = $3.59$

Sample standard deviation $1.045$

Confidence interval from $3.49$ to $3.69$

What is the confidence level?

How can I get the answer when sample size is so large? I tried to find the number from the student's $t$-distribution table but the maximum was $100$.

Is $3.59 + \dfrac{1.045x}{\sqrt{457}} = 3.69$ right? $x$ is what I couldn't find from the table.

Please help.

Answer 1

Hint: when the number of degrees of freedom / data points is very large-scale, the $t$ distribution is very close to a normal one.

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Quoting link,

A Student's $t$ distribution with mean $\mu$, scale $\sigma^2$ and $n$ degrees of freedom converges in distribution (ie their density functions converge) to a normal distribution with mean $\mu$ and variance $\sigma^2$ when the number of degrees of freedom $n \to \infty $.

Or from Wikipedia,

Whereas a normal distribution describes a full population, t-distributions describe samples drawn from a full population; accordingly, the t-distribution for each sample size is different, and the larger the sample, the more the distribution resembles a normal distribution.

From another angle, recall that the $t$ statistic is the $z$ statistic where we replaced the true population standard deviation $\sigma$ by the sample's standard error $s_n$. As $n$ grows, $s_n \to \sigma$ and, intuitively, $t_n \to z \sim N(0,1)$.

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In your case, what you need to calculate is the cumulative probability (the area under the density curve) of a Student $t_{456}$ distribution between $-t' = \frac{3.49-3.59}{1.045/\sqrt{457}}$ and $t' = \frac{3.69-3.59}{1.045/\sqrt{457}}$.

If you want to use tables (which I find is a good idea for training), take a $z$ table and look for $t' \simeq z' = \frac{3.69-3.59}{1.045/\sqrt{457}} = 2.05$; this gives you a value of $P(z < z') = 0.9798$ (area under the curve excluding the right tail). Given that you want the area under the curve excluding both tails, the value you are looking for is $P(-z' < z < z') = 0.9798 - (1-0.9798) = 0.96$.

If you only have a $t$ table, since you know that $t_n \to z$, just use the bottommost row as an approximation. Using this one (which is particularly handy since it gives values for $df=100$ and $df=1000$ as well as $z^*$), your value of $t_n' = 2.05$ falls very close to the column corresponding to a right-tail area of $0.2$, or a confidence interval of $96\%$. QED