Comparing 2 sample means of 2 different samples

[cse63146]

Homework Statement

sample 1: size -10, mean -124, variance -1681
sample 2: size -5, mean -68, variance - 481

Assuming equal variance, independent and normality holds

a)Is there any evidence that that the mean in population 2 is is less than the mean in population 1? Use alpha = 0.01. What is the p-value?

b)Test the 2 variances. Alpha = 0.05

Homework Equations

The Attempt at a Solution

for a), would I use the following test statistic$$t = \frac{\overline{x} - c}{s/\sqrt{n}}$$~$$t_{n-1}$$. Just not sure how to incorporate the 2 different sample variances.

 

[statdad]

For a), don't you want to test $$H_0 \colon \mu_1 - \mu_2 > 0$$? What test statistic is appropriate for that?

 

[cse63146]

For 2 independent Normal samples, with common unknown variance $$\sigma^2$$

$$\frac{(\overline{y}_1 - \overline{y}_2) - (\mu_1 - \mu_2)}{S_p \sqrt{n^{-1}_1 -+n^{-1}_2}}$$ is a t distribution with (n₁ + n₂ - 2) degrees of freedom.

Where S²_p = $$\frac{(n_1 - 1)S^2_1 + (n_2 -1)S^2_2}{n_1 + n_2 - 2} = 1311.8$$

Not sure how to use the formula though since I don't know mu1 and mu2

 

[statdad]

Your complete hypotheses would be (I typed $$H_0$$ instead of $$H_a$$ and > rather than $$\le$$ at first, sorry)

$$\begin{align*} H_0 \colon & \mu_1 - \mu_2 \le 0 \\ H_a \colon & \mu_1 - \mu_2 > 0 \end{align*}$$

You don't need to know the actual value of $$\mu_1 - \mu_2$$ to use the test statistic: you use the boundary between the two possibilities, which is 0.

 

[cse63146]

$$H_0 \colon & \mu_1 - \mu_2 = 0$$

is that another acceptable way to write H₀?

 

[Mark44]

$$H_0 \colon & \mu_1 - \mu_2 = 0$$

is that another acceptable way to write H₀?

No, you want the inequality, since that's what matches your problem statement.

Is there any evidence that that the mean in population 2 is is less than the mean in population 1?

From the quote just above (which represents the alternate hypothesis), the null hypothesis is that

$$\mu_1 - \mu_2 \leq 0$$

As you wrote it, the probability of the null hypothesis would be zero.

 

[cse63146]

For some reason I thought that it was asking if there was a difference between the 2 means. My mistake. Thanks.

 

[statdad]

"the probability of the null hypothesis would be zero."

Not sure what you mean here (that would be the only way to write the null in the case of the alternative being two-sided), but many authors would write $$H_0 \colon \mu_1 - \mu_2$$ as $$H_0 \colon \mu_1 - \mu_2 = 0$$. It's an odd convention, but it does happen.

 

[Mark44]

My mistake. I was thinking in terms of variables rather than statistics, if that makes any sense. IOW, along these lines: Pr(z = k) = 0.

 

[cse63146]

So I computed the p-value using a t-test, and got the value of 0.0079. Since alpha = 0.01, I can say that the p-value is statistically significant at the 1% level, and I reject the null hypothesis and accept the alternate hypothesis?

 

[statdad]

My mistake. I was thinking in terms of variables rather than statistics, if that makes any sense. IOW, along these lines: Pr(z = k) = 0.

That's rather what I thought - but I have learned to be hesitant making assumptions about the posts of others.

So I computed the p-value using a t-test, and got the value of 0.0079. Since alpha = 0.01, I can say that the p-value is statistically significant at the 1% level, and I reject the null hypothesis and accept the alternate hypothesis?

I haven't checked your calculations, but if the numbers are correct, so is your final statement. However, "accept the alternative hypothesis" is rarely the form a faculty member or researcher wants as the final answer: if you know that the alternative is to be accepted, you should clearly state what this means about the two population means.

 

[cse63146]

Does this sound better:

At the 1% level, the data provides evidence against the null hypothesis in favour of the alternate hypothesis, which implies that the mean of population 2 is less than the mean if population 1.