A whole lotta points here! :) Help me out, thanks!

2 answers:

8 0

Ok so basically I will define how to find it then maybe get a definition

ok so what you do is take the average of all the numbers
then you find the absolute value of the difference or deviation of each number from the average,
then you average all those new numbers and you get standard deviation

example
1,2,3
average is 2
2-1=1
2-2=0
3-2=1
add
1+0+1=2
2/3=0.66
standard deviation is 0.666

basically is is how much the numbers in the set deviate from the average

umka2103 [35]3 years ago

7 0

An amount of something calculated to specify the extent divergence of a whole cluster. <span />

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26. Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics

hjlf

The test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

<h3>What are null hypotheses and alternative hypotheses?</h3>

In null hypotheses, there is no relationship between the two phenomenons under the assumption or it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.

Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics course.

Then they are given a post-test once the professor has concluded lecturing on the material.

Pre-test and post-test scores for 4 students in an elementary statistics class are given below.

Then we have

$\mu _d = \mu _{post} - \mu _{pre}$

Then the null hypotheses and alternative hypotheses will be

H₀: $\mu _d = 0$

Hₐ: $\mu _d > 0$

Then the test statistic will be

$\rm \overline{x} _d = \dfrac{\Sigma x_d}{n} = \dfrac{15+12+10+1}{4}\\\\\overline{x} _d = 9.5$

Then

$\rm S_d = 6.02$

The test statistic value is given by

$\rm t = \dfrac{\overline{x} _d }{\dfrac{S_d}{\sqrtn}} \\\\t = \dfrac{9.5}{\dfrac{6.02}{\sqrt4}}\\\\t = 3.16$

Since this is a right-tailed test, so the critical value is given by

$\rm t_{n-1}(\alpha ) = t_3 (0.05) = 2.353$

Since the test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

Hence, we can conclude that $\mu _d > 0$ that is test scores have improved.

More about the null hypotheses and alternative hypotheses link is given below.

brainly.com/question/9504281

#SPJ1

7 0

2 years ago

PLEASE ANSER URGENT !!!!!

Lynna [10]

The answer is 6/13 since there are 12 distinct letters in the phrase and 12/26 is written as 6/13

3 0

3 years ago

FIST ONE TO AWNSER GETS BRAINLYIST AND 50 points IF CORRECT HURRY!!!!!!!!