Find 90% confidence interval

**madgab** · June 11th 2009, 06:32 AM

Random sample of math scores assumed to be $N(\mu_{x},\sigma^{2})$ : X1=644, X2=493, X3=532, X4=462, X5=565
Random sample of verbal scores assumed to be $N(\mu_{y},\sigma^{2})$ : Y1=623, Y2=472, Y3=492, Y4=661, Y5=540, Y6=502, Y7=549, Y8=518
How do I find a 90% confidence level for $\mu_{x}-\mu_{y}$ ?

**drichie** · June 11th 2009, 09:01 AM

hint:

If $X \sim N(\mu_X, \sigma^2_X)$ and $Y \sim N(\mu_Y, \sigma^2_Y)$ are independent normal random variables, then:
$V = X - Y \sim N(\mu_X - \mu_Y, \sigma^2_X + \sigma^2_Y).$

**matheagle** · June 11th 2009, 02:12 PM

This one is finally well worded.
You want to pool the sample variances because we are assuming that the two population variances are equal.

It will be a t with $n_1+n_2-2$ degrees of freedom.

$\bar X_{n_1}-\bar X_{n_2}\pm t_{n_1+n_2-2, .05}s_p\sqrt{{1\over n_1}+{1\over n_2}}$

see.. Unequal sample sizes, equal variance (Ignore uequal sample size comment)
at http://en.wikipedia.org/wiki/Student's_t-test

**madgab** · June 12th 2009, 07:25 AM

Thank you! That site was extremely helpful. I am now working on plugging all the correct values in to get a "normal" sounding answer. Thank you for helping me.

**matheagle** · June 12th 2009, 04:27 PM

Originally Posted by drichie

hint:

If $X \sim N(\mu_X, \sigma^2_X)$ and $Y \sim N(\mu_Y, \sigma^2_Y)$ are independent normal random variables, then:
$V = X - Y \sim N(\mu_X - \mu_Y, \sigma^2_X + \sigma^2_Y).$

NOTE that the two populations variances are equal $\sigma^2_X= \sigma^2_Y$ .
In this case you are supposed to pool the sample variances.

**madgab** · June 14th 2009, 08:27 AM

$\mu_{x}=539.2$

$\mu_{y}=544.625<br />$

$\sigma^{2}_{x}=3958.96$ (seems high)

$\sigma^{2}_{y}=3786.984372$

$t=\frac{\mu_{x}-\mu_{y}}{S_{xy}*\sqrt{1/5+1/8}}$

$=\frac{539.2-544.625}{62.04450793*0.5700877125}$

$=\frac{-5.425}{35.3708116}=-.1533750501$

This answer doesn't seem quite right.

Then how do I do the confidence interval using the 90% CI, 1.645?

I was going to just do $-5.425+/-1.645(\frac{\sqrt{3958.96+3786.984372}}{\sqrt{13}} )=-5.425+/-40.15424024$
but neither of my answers seemed to make any sense.

**madgab** · June 16th 2009, 06:44 AM

Thank you for all the help on this problem. I do not understand why the population variances are equal, or how you figure that. I have been going through different books and thought I could just use the equation that uses just the sample means and variances with the t value from the table. I'm sure I'm wrong, I just can't figure out how to tell that the pop variances are equal. Thanks

**matheagle** · June 16th 2009, 06:56 AM

Who ever gave you this problem stated that the variance of X and the variance of Y was $\sigma^{2}$ ,
otherwise they would have said

$X\sim N(\mu_{X},\sigma_X^{2})$ and $Y\sim N(\mu_{Y},\sigma_Y^{2})$ .

IN this case you need to pool the variances.
And it's a t percentile and not a z, as I stated a week ago.

Originally Posted by madgab

Random sample of math scores assumed to be $N(\mu_{x},\sigma^{2})$ : X1=644, X2=493, X3=532, X4=462, X5=565
Random sample of verbal scores assumed to be $N(\mu_{y},\sigma^{2})$ : Y1=623, Y2=472, Y3=492, Y4=661, Y5=540, Y6=502, Y7=549, Y8=518
How do I find a 90% confidence level for $\mu_{x}-\mu_{y}$ ?

**madgab** · June 16th 2009, 07:04 AM

I got it! Thank you! I'm sorry for bothering you again. It all makes sense now.

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