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I want to conduct hypothesis testing to prove that there are no difference between two group's mean value.

Null hyp: μ_group1-μ_group2=0
Alt hyp : μ_group1-μ_group2 != 0

my first question is since I know all information about each group's population such as standard deviation, mean, etc... can I use hypothesis testing on whole population?

Second, does size of population(if 1st question's answer is "yes")/sample have to be same? so if I have population size of 300 for group1 and 100 for group2 I would need to sample same number from each group and do hypothesis testing?

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    $\begingroup$ If you _know+ values of $\mu_1$ and $\mu_2,$ then there is nothing to test. Just look to see if the two population means are equal. (Another discussion might center on whether any difference is large enough to be of practical importance.) // If the population parameters are unknown and the populations are known to be normal, then you can use a 2-sample Welch t test, which can accommodate different sizes. $\endgroup$ – BruceET Nov 15 '20 at 0:54
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    $\begingroup$ Thanks. So if I know mean μ1 > μ2 I can simply conclude that group1 has greater mean ? $\endgroup$ – Ambleu Nov 15 '20 at 1:10
  • $\begingroup$ It would be best to do a one-sided Welch t test if you suspect $\mu_1 > \mu_2.$ See Addendum. // Again here, if you know the population values, no test is needed. $\endgroup$ – BruceET Nov 15 '20 at 1:26
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Illustrating comment, using R:

set.seed(2020)
x1 = rnorm(500, 100, 15)
x2 = rnorm(100, 105, 17)

summary(x1); length(x1); sd(x1)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  53.32   89.29   98.93   99.18  109.45  148.02 
[1] 500          # size sample 1
[1] 15.96929     # sample SD sample 1
summary(x2); length(x2); sd(x2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  59.74   94.62  104.05  104.77  114.88  146.67 
[1] 100
[1] 17.11946

The two sample means $\bar X_1 = 99.18$ and $\bar X_2 = 104.77$ differ. The question is whether, in view of the variability of the data, this difference is large enough to be 'statistically significant' at the 5% level.

In the boxplots below, boxes are of different widths, as a reminder that sample sizes are quite different. The fact that the 'notches' in the sides of the boxplots do not overlap, is a preliminary clue that sample means may be significantly different.

 boxplot(x1, x2, varwidth=T, col="skyblue2", pch=20, notch=T)

enter image description here

A Welch t test (used because population variances are unequal), the small P-value $0.003 < 0.05$ indicates significant difference at the 5% level. This is not "proof" that the population means differ. However, we are unlikely to get such different sample means if the population means are the same.

t.test(x1, x2)

    Welch Two Sample t-test

data:  x1 and x2
t = -3.0129, df = 135.64, p-value = 0.003089
alternative hypothesis: 
  true difference in means is not equal to 0
95 percent confidence interval:
 -9.257022 -1.920342
sample estimates:
mean of x mean of y 
 99.18129 104.76998 

Addendum per comment. Here is a one-sided test. If $\bar X_1 > \bar X_2,$ then the test of $H_0: \mu_1 = \mu_2$ against $H_0: \mu_1 < \mu_2$ will have a P-value half the size of the two-sided test.

t.test(x1, x2, alt="less")

        Welch Two Sample t-test

data:  x1 and x2
t = -3.0129, df = 135.64, p-value = 0.001544
alternative hypothesis: 
  true difference in means is less than 0
95 percent confidence interval:
      -Inf -2.516599
sample estimates:
mean of x mean of y 
 99.18129 104.76998 
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