To be explicit: Suppose you have the (fictitious) data sampled using R below:
set.seed(2021)
x = rnorm(20, 50, 7)
summary(x); length(x); sd(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
36.54 49.86 52.48 52.76 60.42 62.11
[1] 20 # sample size
[1] 7.621391 # sample standard deviation
A boxplot and a stripchart (dotplot) of the data are shown below:
par(mfrow = c(2,1))
boxplot(x, col="skyblue2", horizontal=T)
stripchart(x, pch=20)
abline(v=mean(x), col="red")
par(mfrow = c(1,1))

t.test(x, mu = 55)
One Sample t-test
data: x
t = -1.3151, df = 19, p-value = 0.2041
alternative hypothesis: true mean is not equal to 55
95 percent confidence interval:
49.19194 56.32578
sample estimates:
mean of x
52.75886
Then (because this is simulated data) we know population mean is $\mu = 50,$
sample mean is $52.76,$ which is not significantly different from hypothetical mean $\mu_0 = 55.$ A 95% CI for $\mu$ is $(49.19, 56.33).$ which is centered at $\bar X = 52.76$ and contains $\mu_0 = 55.$
In a real application you would never know that $\mu =50,$ exactly. The best point estimate is $\hat \mu = \bar X = 52.76$ and we can be 95% confident
that $\mu$ is in interval $(49.19, 56.33).$
A narrower 90% confidence interval is $(49.81, 55.71).$
t.test(x, conf.lev=.90)$conf.int
[1] 49.81208 55.70564
attr(,"conf.level")
[1] 0.9
And a wider 99% CI is $(47.88, 57,63).$
t.test(x, conf.lev=.99)$conf.int
[1] 47.88327 57.63445
attr(,"conf.level")
[1] 0.99
To consolidate these relationships, you should look at the formula in
your text or class notes for a one-sample t.test and use the summary
information printed above for $n, \bar X, S_x$ to make all three confidence intervals, 90%, 95%, and 00%.
Then check your hand computations with the results above from R.