You mention at least two fundamentally different uses of the chi-squared distribution.
Confidence interval of normal variance. Let $X_1, \dots, X_{10}$
be a sample of size $n = 10$ from $Norm(\mu = 100, and \sigma = 20).$
Suppose $\mu$ and $\sigma$ are unknown to us. Then a 95% confidence
interval (CI) for $\sigma^2$ is based on the fact that
$Q = (n-1)S^2/\sigma^2 \sim Chisq(df = 9),$ were $S^2$ is the sample variance.
In particular, we use $q_L$ and $q_U$ such that $P(q_L < Q < q_U) = .95.$
A probability-symetric CI would use quantiles .025 and .975 of $Chisq(9)$
for the $q$'s: $q_L = 2.70$ and $q_U = 19.02.$
Then a 95% CI for $\sigma^2$ is $(9S^2/q_U, 9S^2/q_L).$ In particular, for the data generated below,
where $S = 20.023,$ a 95% CI for $\sigma^2$ is $(189.7, 1336.2)$. Upon taking
square roots, a 95% CI for $\sigma$ is $(13.8, 36.6).$
x = rnorm(10, 100, 20); s = sd(x); s # simulated dataset and its SD
## 20.02325
q = qchisq(c(.975,.025), 9); q # quantiles for CI
## 19.022768 2.700389
ci = 9*s^2/q; ci # 95% CI for pop var
## 189.6871 1336.2423
sqrt(ci) # 95% CI for pop SD
## 13.77270 36.55465
Now for the units: Both $\sigma$ and $S$ have the units of the data (perhaps meters); both $\sigma^2$ and $S^2$ have squared units ($m^2$), the quantiles $q_L$ and $q_U$ have no units; the CI for $\sigma^2$ has squared units, and the CI for $\sigma$ has the same units as the data. If you convert to feet
instead of meters, of course $S$ and the CI for $\sigma$ will change accordingly.
But the relevant chi-squared distribution and its quantiles will not change.
Goodness-of-fit tests. Suppose I roll a die $n = 60$ times in an attempt to
see whether it is fair. I observe counts $X = (10, 14, 11, 12, 3)$ for
die faces 1 through 6, respectively. Of course, for a fair die the expected
counts are $E = 10$ for each face. A chi-squared goodness-of-fit (GOF) test
has test statistic $Q = \sum_{i=1}^6 (X_i - E)^2/E = 7.0.$ Because expected
counts exceed 5, it is safe to assume that $Q \stackrel{aprx}{\sim} Chisq(df=5).$
If $Q$ exceeds the 95th quantile $q = 16.92$ of $Chisq(5)$, we would judge the
die to be unfair, but $Q = 7.0$ does not provide evidence of unfairness.
Now if I were to roll the die $n=600$ times and observe counts $X = (100, 140, 110, 120, 30)$ with expected counts $E = 100,$ then $Q = 70,$ and I would conclude the die to be unfair. Notice that the degrees of freedom parameter is still 5, because there are still 6 faces (categories).
The sample size $n$ is not directly
reflected in the chi-squared distribution used for the test. However, in a real experiment,
the value of $Q$ may be influenced by $n$. My "counts" $X$ for the 600-roll "experiment" are fake. I merely
multiplied the truthful counts for the 60-roll experiment by 10. But that
produces totally unrealistic counts for a 600-roll experiment. (By analogy,
we are not surprised to get 6 heads in 10 tosses of a fair coin, but we would
be astonished to get 600 heads in 1000 tosses.)
One must never judge goodness of fit just by looking at a bar chart of various
categories, because bar charts do not give visual information on how much
data there is. Bar charts for my real 60-roll experiment and my fake 600-roll
experiment would look the same.
