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ACF and PACF plot has very small confidence level. How to interpret?

I'm rather new at programming at general so do forgive me if the question is rather basic.
I'm trying to determine my p, d, q values for an ARIMA model and I've already conducted an adfuller test that determined that my time series is stationary. However, when I plot out my ACF and PACF plots, I get the following:
ACF plot
PACF plot
From what I've read about the p values, I'm supposed to pick the value where the line first crosses the confidence interval except I'm not sure why my confidence intervals for both are that small? Does this mean that my MA value should be 2 according to the PACF plot? Any help in interpreting the graphs would be appreciated!
My code:
from statsmodels.graphics.tsaplots import plot_acf, plot_pacf
fig = plt.figure(figsize=(20, 12))
fig = plot_acf(train_set.dropna(), lags=10)
fig = plot_pacf(train_set.dropna(), lags=10)

The d component is used to make the data stationary by differencing, if the adf test (and kpss test) shows that the data is stationary, you can probably set it to 0. However, keep in mind that you cannot trust these tests by 100 %.
The confidence interval indicates whether the correlation is statistically significant, meaning that the correlation is very likely not to be random. All bars that cross the confidence interval are “real” correlations that you can use for modeling.
There are thousands of thumb rules to interpret these plots. I recommend the following:
If the ACF trails off, use an MA model with the significant and strong correlations from the PACF.
If the PACF trails off, use an AR model with the significant and strong correlations from the ACF.
You can also have a look here:
https://towardsdatascience.com/identifying-ar-and-ma-terms-using-acf-and-pacf-plots-in-time-series-forecasting-ccb9fd073db8
I guess you created the plots with statsmodels, in that case you shoud keep in mind that lag 0 (the first in the plots) is the correlation of the time series with itself, therefore it will always be +1 and significiant, you can ignore this lag.In your case, the ACF is trailing off, and the PACF has only one statistically significant and strong correlation with the first lag, perhaps you can also use 2, 3 and 4 but they are very weak. Best is of course if you just try it out. Or you can use pmdarima’s auto_arima() function:
https://alkaline-ml.com/pmdarima/tips_and_tricks.html
https://alkaline-ml.com/pmdarima/modules/generated/pmdarima.arima.auto_arima.html

How to reformat the Highest Density Interval (HDI) result on the posterior plot from percentage to float

I conducted a Bayesian mean difference test and obtained the posterior plot of the parameter estimation. I adjusted the HDI probability to 0.995; however, the plot_posterior function of Arviz rounds the probability value to 100% when displaying on the plot, as seen in the following figure. I need that plot to display 99.5%, which is the exact value of credible interval. Although the "round_to" argument allows controlling the formatting of floats, it didn't adjust the HDI percentage.

Negative confidence interval in linear regression despite all positive values

I am getting a negative confidence interval for a linear regression plot even though all data points are positive. Why is this happening? I believe this negative confidence interval will also affect my R^2 score?
Code used is:
sns.regplot(x = 'Consumer Confidence Index_1', y = 'Sales (ALV
sources)', data = df_mx2)
plt.show()
See graph pic here

One of the foundational assumptions for a linear regression is that the data is normally distributed about the line. In your case you have data on the right side and the left side with a big gap in the middle. As such, you should double check that a linear regression is appropriate for your analysis.
That being said, rest easy, the negative confidence interval will NOT effect your R² value.
The reason for the negative confidence interval has to do with the sparsity of data with x<42. If the three points on the right side were removed, the regression would have a positive slope intersecting the x axis around x=42. If that line were extended to x=30 or so it would be very negative. As such the data suggests that to hit the confidence threshold you have set, the confidence interval must be very large to include data that potentially lines up with the steeper regression line.
This can be interpreted as the data provides very little in the way of predictive ability below x=42.

How to use norm.ppf()?

I couldn't understand how to properly use this function, could someone please explain it to me?
Let's say I have:
a mean of 172.7815
a standard deviation of 4.1532
N = 50 (50 samples)
When I'm asked to calculate the (95%) margin of error using norm.ppf() will the code look like below?
norm.ppf(0.95, loc=172.78, scale=4.15)
or will it look like this?
norm.ppf(0.95, loc=0, scale=1)
Because I know it's calculating the area of the curve to the right of the confidence interval (95%, 97.5% etc...see image below), but when I have a mean and a standard deviation, I get really confused as to how to use the function.

The method norm.ppf() takes a percentage and returns a standard deviation multiplier for what value that percentage occurs at.
It is equivalent to a, 'One-tail test' on the density plot.
From scipy.stats.norm:
ppf(q, loc=0, scale=1) Percent point function (inverse of cdf — percentiles).
Standard Normal Distribution
The code:
norm.ppf(0.95, loc=0, scale=1)
Returns a 95% significance interval for a one-tail test on a standard normal distribution (i.e. a special case of the normal distribution where the mean is 0 and the standard deviation is 1).
Our Example
To calculate the value for OP-provided example at which our 95% significance interval lies (For a one-tail test) we would use:
norm.ppf(0.95, loc=172.7815, scale=4.1532)
This will return a value (that functions as a 'standard-deviation multiplier') marking where 95% of data points would be contained if our data is a normal distribution.
To get the exact number, we take the norm.ppf() output and multiply it by our standard deviation for the distribution in question.
A Two-Tailed Test
If we need to calculate a 'Two-tail test' (i.e. We're concerned with values both greater and less than our mean) then we need to split the significance (i.e. our alpha value) because we're still using a calculation method for one-tail. The split in half symbolizes the significance level being appropriated to both tails. A 95% significance level has a 5% alpha; splitting the 5% alpha across both tails returns 2.5%. Taking 2.5% from 100% returns 97.5% as an input for the significance level.
Therefore, if we were concerned with values on both sides of our mean, our code would input .975 to represent a 95% significance level across two-tails:
norm.ppf(0.975, loc=172.7815, scale=4.1532)
Margin of Error
Margin of error is a significance level used when estimating a population parameter with a sample statistic. We want to generate our 95% confidence interval using the two-tailed input to norm.ppf() since we're concerned with values both greater and less than our mean:
ppf = norm.ppf(0.975, loc=172.7815, scale=4.1532)
Next, we'd take the ppf and multiply it by our standard deviation to return the interval value:
interval_value = std * ppf
Finally, we'd mark the confidence intervals by adding & subtracting the interval value from the mean:
lower_95 = mean - interval_value
upper_95 = mean + interval_value
Plot with a vertical line:
_ = plt.axvline(lower_95, color='r', linestyle=':')
_ = plt.axvline(upper_95, color='r', linestyle=':')

James' statement that norm.ppf returns a "standard deviation multiplier" is wrong. This feels pertinent as his post is the top google result when one searches for norm.ppf.
'norm.ppf' is the inverse of 'norm.cdf'. In the example, it simply returns the value at the 95% percentile. There is no "standard deviation multiplier" involved.
A better answer exists here:
How to calculate the inverse of the normal cumulative distribution function in python?

You can figure out the confidence interval with norm.ppf directly, without calculating margin of error
upper_of_interval = norm.ppf(0.975, loc=172.7815, scale=4.1532/np.sqrt(50))
lower_of_interval = norm.ppf(0.025, loc=172.7815, scale=4.1532/np.sqrt(50))
4.1532 is sample standard deviation, not the standard deviation of the sampling distribution of the sample mean. So, scale in norm.ppf will be specified as scale = 4.1532 / np.sqrt(50), which is the estimator of standard deviation of the sampling distribution.
(The value of standard deviation of the sampling distribution is equal to population standard deviation / np.sqrt(sample size). Here, we did not know the population standard deviation and the sample size is more than 30, so sample standard deviation / np.sqrt(sample size) can be used as a good estimator).
Margin of error can be calculated with (upper_of_interval - lower_of_interval) / 2.

calculate the amount for the 95% percentile and draw a vertical line and an annotation with the amount
mean=172.7815
std=4.1532
N = 50
results=norm.rvs(mean,std, size=N)
pct_5 = norm.ppf(.95,mean,std)
plt.hist(results,bins=10)
plt.axvline(pct_5)
plt.annotate(pct_5,xy=(pct_5,6))
plt.show()

As other answers pointed out, norm.ppf(1-alpha) returns the value on the (1-alpha)x100-th percentile of a normal distribution specified by the parameters passed to the it. For example in the OP, it returns the 95th percentile of a normal distribution with mean 172.78 and standard deviation 4.15.
If you're looking for a function that returns the same value (N-th percentile on the normal distribution) as a function of alpha instead, there's the inverse survival function, norm.isf(alpha), which tells you the number at which (1-alpha) is above it.
from scipy.stats import norm
alpha = 0.05
v1 = norm.isf(alpha)
v2 = norm.ppf(1-alpha)
np.isclose(v1, v2) # True

How to interpret the upper/lower bound of a datapoint with confidence intervals?

Given a list of values:
>>> from scipy import stats
>>> import numpy as np
>>> x = list(range(100))
Using student t-test, I can find the confidence interval of the distribution at the mean with an alpha of 0.1 (i.e. at 90% confidence) with:
def confidence_interval(alist, v, itv):
return stats.t.interval(itv, df=len(alist)-1, loc=v, scale=stats.sem(alist))
x = list(range(100))
confidence_interval(x, np.mean(x), 0.1)
[out]:
(49.134501289005009, 49.865498710994991)
But if I were to find the confidence interval at every datapoint, e.g. for the value 10:
>>> confidence_interval(x, 10, 0.1)
(9.6345012890050086, 10.365498710994991)
How should the interval of the values be interpreted? Is it statistically/mathematical sound to interpret that at all?
Does it goes something like:
At 90% confidence, we know that the data point 10 falls in the interval (9.6345012890050086, 10.365498710994991),
aka.
At 90% confidence, we can say that the data point falls at 10 +- 0.365...
So can we interpret the interval as some sort of a box plot of the datapoint?

In short
Your call gives the interval of confidence for the mean parameter of a normal law of unknown parameters of which you observed 100 observations with an average of 10 and a stdv of 29. It is furthermore not sound to interpret it, since your distribution is clearly not normal, and because 10 is not the observed mean.
TL;DR
There are a lot misconceptions floating around confidence intervals, most of which seemingly stems from a misunderstanding of what we are confident about. Since there is some confusion in your understanding of confidence interval maybe a broader explanation will give a deeper understanding of the concepts you are handling, and hopefully definitely rule out any source of error.
Clearing out misconceptions
Very briefly to set things up. We are in a situation where we want to estimate a parameter, or rather, we want to test a hypothesis for the value of a parameter parameterizing the distribution of a random variable. e.g: Let's say I have a normally distributed variable X with mean m and standard deviation sigma, and I want to test the hypothesis m=0.
What is a parametric test
This a process for testing a hypothesis on a parameter for a random variable. Since we only have access to observations which are concrete realizations of the random variable, it generally procedes by computing a statistic of these realizations. A statistic is roughly a function of the realizations of a random variable. Let's call this function S, we can compute S on x_1,...,x_n which are as many realizations of X.
Therefore you understand that S(X) is a random variable as well with distribution, parameters and so on! The idea is that for standard tests, S(X) follows a very well known distribution for which values are tabulated. e.g: http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
What is a confidence interval?
Given what we've just said, a definition for a confidence interval would be: the range of values for the tested parameter, such that if the observations were to have been generated from a distribution parametrized by a value in that range, it would not have probabilistically improbable.
In other words, a confidence interval gives an answer to the question: given the following observations x_1,...,x_n n realizations of X, can we confidently say that X's distribution is parametrized by such value. 90%, 95%, etc... asserts the level of confidence. Usually, external constraints fix this level (industrial norms for quality assessment, scientific norms e.g: for the discovery of new particles).
I think it is now intuitive to you that:
The higher the confidence level, the larger the confidence interval. e.g. for a confidence of 100% the confidence interval would range across all the possible values as soon as there is some uncertainty
For most tests, under conditions I won't describe, the more observations we have, the more we can restrain the confidence interval.
At 90% confidence, we know that the data point 10 falls in the interval (9.6345012890050086, 10.365498710994991)
It is wrong to say that and it is the most common source of mistakes. A 90% confidence interval never means that the estimated parameter has 90% percent chance of falling into that interval. When the interval is computed, it covers the parameter or it does not, it is not a matter of probability anymore. 90% is an assessment of the reliability of the estimation procedure.
What is a student test?
Now let's come to your example and look at it under the lights of what we've just said. You to apply a Student test to your list of observations.
First: a Student test aims at testing a hypothesis of equality between the mean m of a normally distributed random variable with unknown standard deviation, and a certain value m_0.
The statistic associated with this test is t = (np.mean(x) - m_0)/(s/sqrt(n)) where x is your vector of observations, n the number of observations and s the empirical standard deviation. With no surprise, this follows a Student distribution.
Hence, what you want to do is:
compute this statistic for your sample, compute the confidence interval associated with a Student distribution with this many degrees of liberty, this theoretical mean, and confidence level
see if your computed t falls into that interval, which tells you if you can rule out the equality hypothesis with such level of confidence.
I wanted to give you an exercise but I think I've been lengthy enough.
To conclude on the use of scipy.stats.t.interval. You can use it one of two ways. Either computing yourself the t statistic with the formula shown above and check if t fits in the interval returned by interval(alpha, df) where df is the length of your sampling. Or you can directly call interval(alpha, df, loc=m, scale=s) where m is your empirical mean, and s the empirical standard deviatation (divided by sqrt(n)). In such case, the returned interval will directly be the confidence interval for the mean.
So in your case your call gives the interval of confidence for the mean parameter of a normal law of unknown parameters of which you observed 100 observations with an average of 10 and a stdv of 29. It is furthermore not sound to interpret it, beside the error of interpretation I've already pointed out, since your distribution is clearly not normal, and because 10 is not the observed mean.
Resources
You can check out the following resources to go further.
wikipedia links to have quick references and an elborated overview
https://en.wikipedia.org/wiki/Confidence_interval
https://en.wikipedia.org/wiki/Student%27s_t-test
https://en.wikipedia.org/wiki/Student%27s_t-distribution
To go further
http://osp.mans.edu.eg/tmahdy/papers_of_month/0706_statistical.pdf
I haven't read it but the one below seems quite good.
https://web.williams.edu/Mathematics/sjmiller/public_html/BrownClasses/162/Handouts/StatsTests04.pdf
You should also check out p-values, you will find a lot of similarities and hopefully you understand them better after reading this post.
https://en.wikipedia.org/wiki/P-value#Definition_and_interpretation

Confidence intervals are hopelessly counter-intuitive. Especially for programmers, I dare say as a programmer.
Wikipedida uses a 90% confidence to illustrate a possible interpretation:
Were this procedure to be repeated on numerous samples, the fraction of calculated confidence intervals (which would differ for each sample) that encompass the true population parameter would tend toward 90%.
In other words
The confidence interval provides information about a statistical parameter (such as the mean) of a sample.
The interpretation of e.g. a 90% confidence interval would be: If you repeat the experiment an infinite number of times 90% of the resulting confidence intervals will contain the true parameter.
Assuming the code to compute the interval is correct (which I have not checked) you can use it to calculate the confidence interval of the mean (because of the t-distribution, which models the sample mean of a normally distributed population with unknown standard deviation).
For practical purposes it makes sense to pass in the sample mean. Otherwise you are saying "if I pretended my data had a sample mean of e.g. 10, the confidence interval of the mean would be [9.6, 10.3]".
The particular data passed into the confidence interval does not make sense either. Numbers increasing in a range from 0 to 99 are very unlikely to be drawn from a normal distribution.