Two $t$-curves have degrees of freedom $12$ and $20$ , respectively. Which one more closely resembles the standard normal curve? Explain your answer.

The degrees of freedom for the two -curves are $12$and $20$respectively.

The formula used:$t=\frac{\overline{x}-\mu }{s/\sqrt{n}}$,$Z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}$

$t-$Curve with degrees of freedom $20$resembles the typical normal curve more closely than With degrees of freedom, $t$curve $12$

We know that the $t-$distribution with $(n-1)$degrees of freedom is followed by the studentized form of $\overline{x},t=\frac{\overline{x}-\mu }{s/\sqrt{n}}$, where $n$is the sample size. When a result, The degrees of freedom increase as the sample size grows. As a result, the sample size for a $t-$curve with degrees of freedom $20$(which equals $21$) is larger than the sample size for a $t-$curve with degrees of freedom $12$(which is equal to $13$).

Both the conventional normal curve and the $t-$curve are now symmetric and bell-shaped at about $0$degrees. The only difference is that the $t-$curve is wider than the standard normal curve. This is because the studentized variable $t$is obtained by replacing the unknown population s.d., $\sigma$(which is constant) in the standardized version $\overline{x},Z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}$with the random variable sample s.d., $s$(estimate of $\sigma$from the sample). If the sample size is expanded, samples will now carry more information about the population. For a fixed sample size, all potential sample standard deviations cluster closer to the actual value of the unknown population$\sigma$as the sample size increases. i.e. When the sample size is fixed, the variability in all possible sample $S.D.$ values decreases.

As a result, for a fixed sample size, the variation in all potential values of the studentized variable $t$reduces as the sample size increases. As the sample size grows, i.e. as the degrees of freedom grows, the spread of the $t-$curve reduces, and the shape of the $t$curve approaches that of the standard normal curve. As a result, the degrees of freedom $t-$curve the$t-$curve with degrees of freedom resembles the conventional normal curve more closely than $20$