1. ## df..stats

n1=28
mean1=72.75
SD1=5.37225

n1=24
mean1=72.625
SD1=7.69987

Im finding a 90% confidence interval of the difference.

What is the df?
What is the t*?
I dont get it.

2. Originally Posted by Morgan82
n1=28
mean1=72.75
SD1=5.37225

n1=24
mean1=72.625
SD1=7.69987

Im finding a 90% confidence interval of the difference.

What is the df?
What is the t*?
I dont get it.
Are the samples independent? I'll assume that they are.

The samples are small size so the t-distribution should be used. Since $s_1^2$ and $s_2^2$ do NOT differ by an order of magnitude, it can be assumed that df = $n_1 + n_2 - 2 = 50.$

$\hat{\delta} = \mu_1 - \mu_2 = 0.125$.

$se(\hat{\delta}) = \sqrt{\frac{5.37225^2}{28} + \frac{7.69987^2}{24}} = ......$

Two tailed therefore require $t_{0.05} = 1.676$ (refer table given at Student's t-distribution - Wikipedia, the free encyclopedia or use the calculator found at http://www.anu.edu.au/nceph/surfstat...e/tables/t.php).

The confidence interval is $\hat{\delta} \pm t_{\alpha} \, se(\hat{\delta}) = .....$

Since the degree of freedom is so large you could use the z-distribution without too much error.

3. Originally Posted by mr fantastic
Are the samples independent? I'll assume that they are.

The samples are small size so the t-distribution should be used. Since $s_1^2$ and $s_2^2$ do NOT differ by an order of magnitude, it can be assumed that df = $n_1 + n_2 - 2 = 50.$

$\hat{\delta} = \mu_1 - \mu_2 = 0.125$.

$se(\hat{\delta}) = \sqrt{\frac{5.37225^2}{28} + \frac{7.69987^2}{24}} = ......$

Two tailed therefore require $t_{0.05} = 1.676$ (refer table given at Student's t-distribution - Wikipedia, the free encyclopedia).

The confidence interval is $\hat{\delta} \pm t_{\alpha} \, se(\hat{\delta}) = .....$

Since the degree of freedom is so large you could use the z-distribution without too much error.

Thanks alot! I really appreciate your help! I think i got it!