MHB Confidence interval of difference

[mathmari]

Hey! :giggle:

From two populations of small and medium enterprises, we collect two random and independent samples of size $n_1 = 64$ and $n_2 = 36$ enterprises respectively. If we calculate for the companies of the first sample average sales of $9$ thousand euros and a variation of $4$, while for the companies of the second sample average sales of $7$ thousand euros and a variation of $9$ and in addition we assume unequal population variances, then

a) estimate a confidence interval with a probability of $95\%$ of the population sales difference

b) what sample size should we consider when the second population to construct a $95\%$ confidence interval for estimating the population average sales amount, with a sampling error of $0.6$ thousand euros, if we know that the random sales variables $X\sim N (8,4 )$.I have done the following :

a) We have that $$\overline{x}-\overline{y}=9-7=2$$ and $$s_{\bar{x}-\bar{y}}\approx \sqrt{\frac{s_x^2}{n_x}+\frac{s_y^2}{n_y}}=\sqrt{\frac{4}{64}+\frac{9}{36}}=\sqrt{\frac{5}{16}}=\frac{\sqrt{5}}{4}$$
Since $C=95\%$ then $z^{\star}=1.96$ and then the confidence intervalis $$\left ( 2-1.96\cdot \frac{\sqrt{5}}{4}, 2+1.96\cdot \frac{\sqrt{5}}{4}\right )$$ Is that correct?

b) Could you give me a hint?
:unsure:

 

[I like Serena]

Hey mathmari!

It says 9 thousand euros with a variation of 4.
That is a bit confusing and ambiguous.
Is it a variance? Or is it a standard deviation?
And what its unit? Is it thousands of euros?

If we assume a standard deviation of 4 thousand euros is intended, then we have $s_1=4$ instead of $s_1^2=4$. :unsure:

For b we presumably have $\mu=8$ and $\sigma=4$.
If I'm not mistaken, they are asking for an n such that the the margin of error of the confidence interval is 0.6.
That error is 1.96 SE, where SE is the so called standard error.
It is $SE=\frac \sigma{\sqrt n}$.