Does this data set contradict the assumption of the physical model? In a problem physical model suggests that increase in temperature should not exceed 5 C. So, 8 independent measurements are recorded.
$6.4 ,4.3 ,5.7, 4.9, 6.5, 5.9, 6.4, 5.1$ (sample) and $\alpha = 0.05$.
Does this data set contradict the assumption of the physical model?
I did the following:
Since we need Hypothesis testing then $H_0: \mu\le 5$ (Claim) and $H_a:\mu\ge 5$ then mean $= 5.65$ with $s = 0.81.$
Is it the correct way of solving it? Should I continue with finding critical value?
 A: One-sample t test of $H_0: \mu \le 5$ against $H_a: \mu > 5.$
[Notice that alternative hypothesis should have $>,$ not $\ge.]$
Your data are as follows (entered into R statistical software):
x = c(6.4, 4.3, 5.7, 4.9, 6.5, 5.9, 6.4, 5.1)
mean(x);  sd(x)
[1] 5.65
[1] 0.8106435

Here is a stripchart (dotplot) of your data. The vertical line is
located at $\bar X.$

So, $\bar X = 5.65$ and $S = 0.8106$ are the sample mean and
standard deviation. You should verify this by using the formulas.
Because the sample mean $\bar X > 5$ you may wonder if the
population mean $\mu$ is also greater than $5.$ The question is
whether $5.65$ is enough greater than $5$ to be called 'significantly'
greater at the 5% level, and thus reject $H_0.$
Here are results from the required t test computed using R:
t.test(x, mu=5, alte="greater")

        One Sample t-test

data:  x
t = 2.2679, df = 7, p-value = 0.02883
alternative hypothesis: true mean is greater than 5
95 percent confidence interval:
 5.107003      Inf
sample estimates:
mean of x 
     5.65 

The first thing you should do is to compute the t statistic
$T = \frac{\bar X - 5}{S/\sqrt{n}}$ to see if you get $T = 2.2679$
as in the printout.
I don't know for sure whether (according to your textbook or class notes) you are supposed to look at
the P-value or the critical value to decide whether to reject
the null hypothesis. [I see you mention 'critical value' in your Question.]

*

*The P-value is based on the assumption that the null hypothesis
is true. In that case $T$ is distributed according to Student's
t distribution with degrees of freedom DF $= n - 1 = 7.$ The
probability $P(T \ge 2.2679)$ is the probability of getting
a more extreme result (in a positive direction) than the observed
value 2.2679 is called the P-value. The output tells you
that the P-value is 0.02883. If the P-value is smaller than 0.05 = 5%,
then you reject the null hypothesis at the 5% level. (Generally speaking you can't determine the exact P-value by looking at a printed table of t distributions. P-values are 'creatures' of the computer age.)


*You can find the critical value $c = 1.895$ (to some number of decimal places) by looking a printed t table. Look in the row for DF = 7 and
in the column for values that cut probability 0.05 from the upper tail
of the distribution. If the t statistic is greater than $c = 1.895,$ then you reject the null hypothesis. Because you have $T = 2.2679,$ you do reject.
The figure below shows the density function of Student's t distribution with DF = 7.

*

*The heavy black vertical line is at $T = 2.2679.$ The area under
the curve to the right of this line represents the P-value.


*The dotted red vertical line is at $c =1.895.$ You reject the null hypothesis because the black line is to the right of the red line. This is a 5% critical value; the area under the curve to the right of the dotted red line is 5%.
.

