3
$\begingroup$

The r.v. $X$ represents the time taken by a computer in company $1$ in order to perform a certain job, and $Y$ represents the same thing but for company $2$. A sample of $n_X = 12$ computers are taken from company $1$, and we obtain: $\bar x = 65$, $s_X ^2 = 279$. A sample of $n_Y = 8$ computers are taken from company $2$ and we get $\bar y = 48$, $s_Y ^2 = 224$.

I am required to find a $.95$ confidence interval for the difference between the means of the two populations.

What I did:

Because $\bar x > \bar y$ let's find the C.I for the difference $\mu_X - \mu_Y$. To do this, we note that:

The variances are unknown, and $n_X + n_Y - 2 = 18 \le 30$ is small. Then, we must consider:

$$T = \frac{(\bar X - \bar Y) - (\mu_X - \mu_Y)}{\hat \sigma \sqrt{\frac1{n_X} + \frac1{n_Y}}}$$

Where:

$$\hat \sigma^2 = \frac{n_X S_X ^2 + n_Y S_Y ^2}{n_X + n_Y -2}$$

$T$ has a t-student distribution with degrees of freedom $\nu = n_X + n_Y - 2 = 18$.

$$- t \le T \le t \iff - t \le \frac{(\bar X - \bar Y) - (\mu_X - \mu_Y)}{\hat \sigma \sqrt{\frac1{n_X} + \frac1{n_Y}}} \le t \iff ... \iff \\ (\bar X - \bar Y) - t \hat \sigma \sqrt{\frac1{n_X} + \frac1{n_Y}} \le \mu_X - \mu_Y \le (\bar X - \bar Y) + t \hat \sigma \sqrt{\frac1{n_X} + \frac1{n_Y}}$$

Now we find $t$ from the table, and replace all the known values to get:

C.I $= \left[ 0.457, 33.542 \right]$

I don't care about the part with calculations, but my question is:

Is my work correct?

The next part of the question is asking to find whether we can say the company $1$ has faster computers than company $2$ at a risk $.05$. I know how to do this by testing the hypothesis $\mu_X = \mu_Y$ against $\mu_X > \mu_Y$. But is there a way to do it that makes use of the first part?

$\endgroup$
  • $\begingroup$ Please add the [self-study] tag & read its wiki. $\endgroup$ – gung - Reinstate Monica Sep 3 '15 at 1:49
  • $\begingroup$ @gung: I added it. the specific problem that I encountered is the whether I interpreted the question and applied the formula correctly. $\endgroup$ – George Sep 3 '15 at 1:52
  • $\begingroup$ Crossposted on Math: math.stackexchange.com/q/1419054/23353 $\endgroup$ – apnorton Sep 3 '15 at 13:38
1
$\begingroup$

Your formulas are correct, but the calculations might not be very accurate.

n1<-12
n2<-8
x_bar<-65
y_bar<-48
sx_2<-279
sy_2<-224

sp<-sqrt(((n1-1)*sx_2+(n2-1)*sy_2)/(n1+n2-2))


t<-qt(0.975,n1+n2-2)#with 0.975 and 18 df


CL1<-(x_bar-y_bar)-t*sp*sqrt(1/n1+1/n2) #lower 95% CI
CL1

#1.608831

CL2<-(x_bar-y_bar)+t*sp*sqrt(1/n1+1/n2) #higher 95% CI
CL2

# 32.39117

95% CI: 1.61-32.4

Since the 95% CI does not include 0, I think you can say company 1 's computer is faster than company 2 at 0.05 type I error level.

$\endgroup$
  • $\begingroup$ thanks. I define the $s_X^2$ to be the sample variance and not the estimation of $\sigma_X ^2$. Sorry for the confusion. I guess with that my calculations are correct? $\endgroup$ – George Sep 3 '15 at 2:25
  • $\begingroup$ Yes, your formulas and procedures are correct, but may need to check the calculation. $\endgroup$ – Deep North Sep 3 '15 at 2:27
  • $\begingroup$ Pardon me, but I see that you used $\frac{(n_1 -1)s_X^2 + (n_2 - 1)s_Y ^2}{n_1 + n_2 - 2}$ instead of $\frac{n_1 s_X^2 + n_2 s_Y ^2}{n_1 + n_2 - 2}$. so it seems to me that you are thinking of $s_X^2$ as the point estimation of the population variance, while it's in fact the sample variance. So i want to know whether there's a misunderstanding or an error with my formula. thanks for your help. $\endgroup$ – George Sep 3 '15 at 2:32
  • $\begingroup$ My question is: do you use the symbol $s_X ^2$ to denote the unbiased estimator of the population variance? I define $s_X ^2$ to be the sample variance, and $\frac{n_X}{n_X - 1} \times s_X ^2$ to be the unbiased estimator of the population variance. Some people define $s_X ^2$ otherwise (i.e. it being the unbiased estimator). So I want to know if there's a misunderstanding. thanks again for your time. $\endgroup$ – George Sep 3 '15 at 2:45
  • $\begingroup$ Ok, I see the confusion, some people define sample variance as $s=\sum (x_i-\bar{x})/(n-1)$ some people define it as $s=\sum (x_i-\bar{x})/n$ $\endgroup$ – Deep North Sep 3 '15 at 3:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.