Likelyhood ratio test hypotheses and normal distribution

[Hummingbird25]

Homework Statement

Given the normal distribution

$$X_{ij} \sim N(\mu_i, \omega^2)$$ where i = 1,2 and j = 1,...,n

deduce that $$H_{0\mu}: \mu_1 = \mu _2$$

The Attempt at a Solution

Do I take in the Likelyhood function here?

and use it to analyse the case?

Sincerely Hummingbird

p.s. I have reading in Wiki that the Null hypo is rejected by the likehood ratio test, could be what I am expected to show here?

 

[EnumaElish]

I have reading in Wiki that the Null hypo is rejected by the likehood ratio test, could be what I am expected to show here?

You are taking 2 samples from 2 different normal distributions, where sample size is n for each sample. You are supposed to calculate the sample averages then test the null hyp. using a z-test, assuming their common variance $\omega^2$ is known (given).

If you don't know the variance you'll need to estimate it from the combined sample, then use a t-test for equality of means (assuming equal variances and equal sample sizes).

 

[Hummingbird25]

You are taking 2 samples from 2 different normal distributions, where sample size is n for each sample. You are supposed to calculate the sample averages then test the null hyp. using a z-test, assuming their common variance $\omega^2$ is known (given).

If you don't know the variance you'll need to estimate it from the combined sample, then use a t-test for equality of means (assuming equal variances and equal sample sizes).

The sample average of the two norm distributions is that

$$\overline{x} = \frac{\sum_{i=1}^{2}f_i}{n}$$??

Sincerely
Hummingbird

 

[EnumaElish]

No.

$$\overline{x_1} = \frac{\sum_{j=1}^{n}x_{1j}}{n}$$

Same for i = 2.

 

[Hummingbird25]

Hello again EnomaElish and thank you,

$$\overline{x_1} = \frac{\sum_{j=1}^{n}x_{1j}}{n}$$

$$\overline{x_2} = \frac{\sum_{j=1}^{n}x_{2j}}{n}$$

Then I say by the z-test then the null hypotheses is rejected if the variance isn't given since the samples aren't drawn from the same population.

But the null hypotheses is accepted if the means are equal which can be tested using the student t-test.

Is this it?

Sincerely Hummingbird

 

[EnumaElish]

Is the variance given, or assumed known?

Possible answer 1:
Yes, the variance is given (or the problem assumes it is known).
What you need to do: use the z test for equality of means to determine whether or not the two means are equal. (You are not supposed to use the t test in this case.)

Possible answer 2:
No, the variance is not given (nor does the problem assume the variance is known).
What you need to do: calculate the common variance from the combined sample. Then use the t test for equality of means to determine whether or not the two means are equal. (You are not supposed to use the z test in this case.)