Question

Calculating the degrees of freedom, the sample variance, and the estimated standard error for evaluations using the t statistic You are planning to evaluate the mean of a single continuous variable from a study with a sample of n = 10 using the t statistic. What are the degrees of freedom for the sample? a.10 b.8 c.11 d.9 With another study, where you also plan on evaluating a mean using the t statistic, you have a sample of n = 41 that has an SS of 600. What is the variance for the sample? a.15 b.3.87 c.24.49 d.360,000 For a sample of n = 25 that has a sample variance of 400, what is the estimated standard error for the sample? a.4 b.17 c.16 d.4.08

Answer 1

Answer:

a) For the first part we have a sample of n =10 and we want to find the degrees of freedom, and we can use the following formula:

$df = n-1= 10-1=9$

d.9

b) $s^2 = \frac{SS}{n-1}= \frac{600}{41-1}= 15$

a.15

c) For this case we have the sample size n = 25 and the sample variance is $s^2 =400$ , the standard error can founded with this formula:

$SE = \frac{s^2}{\sqrt{n}}= \frac{400}{\sqrt{25}}= 80$

Step-by-step explanation:

Part a

For the first part we have a sample of n =10 and we want to find the degrees of freedom, and we can use the following formula:

$df = n-1= 10-1=9$

d.9

Part b

From a sample we know that n=41 and SS= 600, where SS represent the sum of quares given by:

$SS = \sum_{i=1}^n (X_i -\bar X)^2$

And the sample variance for this case can be calculated from this formula:

$s^2 = \frac{SS}{n-1}= \frac{600}{41-1}= 15$

a.15

Part c

For this case we have the sample size n = 25 and the sample variance is $s^2 =400$ , the standard error can founded with this formula:

$SE = \frac{s^2}{\sqrt{n}}= \frac{400}{\sqrt{25}}= 80$