Additional steps when fail to reject null hypothesis My understanding of the hypothesis test below is:
$H_0: \theta_0$ = 0 ; $H_A: \theta_0 \neq$ 0
given $\alpha$ = .05  if my p-value is < .05  I reject the null in favor of the alternative then:

*

*Look at 95% CI to get an idea of the precision

*Determine if the effect size is meaningful

if p-value >.05 fail to reject $H_0$
My question is now what does one do specifically
Do you still look at the 95% CI?
I would assume a narrow interval that barely includes zero might suggest maybe the setup lacked power to detect an effect. However what about the opposite case where the interval is exceptionally wide?
What does the wide CI suggest and does it invalidate your failure to reject the null?
 A: Suppose you have $n = 10$ observations in vector x in R from a normal distribution
and wish to test $H_0: \mu = 0$ against $H_a: \mu \ne 0.$ at the 5% level.
x
[1]  7.04  1.94 -2.42  3.85  3.58 -5.70 -4.86 -3.14  4.50  4.04

summary(x);  sd(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 -5.700  -2.960   2.760   0.883   3.993   7.040 
[1] 4.491528

stripchart(x)


The sample mean is a little above $0;$ the stripchart shows six observations
above $0$ and four below. But in view of the relatively large spread of the
of the data, there seems to be inadequate evidence that $\mu \ne 0.$
A t test in R, gives P-value 0.5496, which is not below 5%. That is pretty
much the end of the story. Your ten observations don't contain information
for you to believe $\mu$ differs from $0.$

If you're P-hacking, you might try a Wilcoxon
signed rank test, a median test, or some kind of permutation test. But the truth is that your ten observations
are not enough to detect a population mean $\mu$ different from $0,$ a t test was
an appropriate one, and you should accept its verdict.

t.test(x)

        One Sample t-test

data:  x
t = 0.62168, df = 9, p-value = 0.5496
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 -2.330046  4.096046
sample estimates:
mean of x 
    0.883 

The 95% confidence interval $(-2.33, 4.10)$ includes $0.$
A confidence interval might be considered an interval of potential
values of $\mu$ that could not be rejected in view of the data
at hand. So it will do nothing to change the verdict that $H_0$
cannot be rejected.
If you have reason--beyond the data at hand--to believe that $\mu \ne 0,$
and sufficient budget to continue investigation, then a reasonable "additional step"
might be to perform an experiment with more observations.
If you have a specific
amount $\Delta$ in mind by which you believe $\mu$ differs from $0,$ you might
use a power and sample size computation to find a sample size $n$ that would
give a good chance of rejecting $H_0.$ if such a difference is really present. Part of that computation
requires you to guess at the population standard deviation $\sigma.$ The sample
standard deviation $S = 4.49$ of your current data is an estimate of $\sigma.$
However, if $\mu = 0$ is really true, then more investigation is not going to be useful.
