You wish to test the following claim ($H_a$) at a significance level of alpha=0.10.
$H_o: \mu=70.4$
$H_a: \mu<70.4$
You believe the population is normally distributed and you know the standard deviation is $\sigma=7.6$, sample mean $M=69.3$ for a sample size of $n=51$.
What is the critical value for this test?
I cannot understand how to find/calculate a critical value for this test. I believe it must be based on $\alpha$, but apparently this is beyond my scope of intuition as 0.1, 0.9, 1-$\alpha$/2, etc. has not worked. Any help would be greatly provided in verbose yet simple terms. This is an intro to stats course for non-majors.
You do not give a full description of the problem and your notation is a bit sketchy. There are two plausible possibilities.
(a) The population standard deviation $\sigma = 7.6$ is known and the sample mean $\bar X = 69.3.$ Then this is a z test and the test statistic is $Z = \frac{\bar X - 70.4}{\sigma/\sqrt{n}}.$ This is a left-tailed test at the 10% level, so you will reject $H_0: \mu = 70.4$ in favor of the alternative $H_a: \mu < 70.4,$ if $Z < - 1.2816.$
Here the 'critical value' $-1.2816$ cuts 10% of the area from the lower tail of the standard normal distribution. You can find the critical value using software or printed normal tables. If using printed tables, you may need to find that 1.2816 cuts 10% from the upper tail of the standard normal distribution, and then use the symmetry of the normal distribution.
(b) The sample standard deviation $S = 7.6$ is used to estimate the unknown population standard deviation and the sample mean $\bar X = 69.3.$ Then this is a t test and the test statistic is $T = \frac{\bar X - 70.4}{S/\sqrt{n}}.$ This is a left-tailed test at the 10% level, so you will reject $H_0: \mu = 70.4$ in favor of the alternative $H_a: \mu < 70.4,$ if $T < -1.2987.$
Here the 'critical value' $-1.2987$ cuts 10% of the area from the lower tail of Student's t distribution with $n = 51 - 1 = 50$ degrees of freedom. You can find the critical value using software or printed normal tables. If using printed tables, you need to look on line $df= 50$ of the table of t distributions to find that 1.2987 cuts 10% from the upper tail of the distribution $\mathsf{T}(df=50),$ and then use the symmetry of the t distribution.