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dimaraw [331]
4 years ago
14

How can you tell the difference between additive comparison and multiplication comparison

Mathematics
2 answers:
Gala2k [10]4 years ago
7 0
Ok Multiplication comparison-Problem that ask "Times as many"

Additive comparison -Problem asking "How many more"
Damm [24]4 years ago
4 0


In an additive comparison, we find the relation between two amounts by asking or telling how much more is one compared to the other.

In multiplicative comparison, there two different sets being compared. The first set contains a certain numbers of items. The second set contains multiple copies of the first set. Any two factors and their product can be read as a comparison.

HOPE THIS HELPS!!!!!!!

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Some scientists believe alcoholism is linked to social isolation. One measure of social isolation is marital status. A study of
frez [133]

Answer:

1) H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

2) The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

3) \chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72

4) df=(rows-1)(cols-1)=(3-1)(2-1)=2

And we can calculate the p value given by:

p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

                    Diag. Alcoholic   Undiagnosed Alcoholic    Not alcoholic    Total

Married                     21                              37                            58                116

Not Married              59                             63                            42                164

Total                          80                             100                          100              280

Part 1

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

The level os significance assumed for this case is \alpha=0.05

Part 2

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

Part 3

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{total col * total row}{grand total}

And the calculations are given by:

E_{1} =\frac{80*116}{280}=33.143

E_{2} =\frac{100*116}{280}=41.429

E_{3} =\frac{100*116}{280}=41.429

E_{4} =\frac{80*164}{280}=46.857

E_{5} =\frac{100*164}{280}=58.571

E_{6} =\frac{100*164}{280}=58.571

And the expected values are given by:

                    Diag. Alcoholic   Undiagnosed Alcoholic    Not alcoholic    Total

Married             33.143                       41.429                        41.429                116

Not Married     46.857                      58.571                        58.571                164

Total                   80                              100                             100                 280

And now we can calculate the statistic:

\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72

Part 4

Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(3-1)(2-1)=2

And we can calculate the p value given by:

p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

7 0
4 years ago
A.) Find h(g(x)) if<br> h(x)= 2x+9<br> and<br> g(x) = 4*
goblinko [34]

Answer:

h(4)=17

Step-by-step explanation:

h(g(x)), h(x)= 2x+9; g(x)=4

2(4)+9= 8+9=17

3 0
2 years ago
Which point is located at (2, -3)?point Hpoint Epoint Gpoint K
Fynjy0 [20]

Answer:

Since i dont have a graph il just tell you this way. on the x axis, go right 2 times, then go down 3 times and that will be your answer.

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Someone help please! k(-1)=______<br>​
prohojiy [21]

Answer:

<h2>k(-1) = 7</h2>

Step-by-step explanation:

Look at the picture.

8 0
3 years ago
Pls help: √5 multiplied by √245
Luden [163]

I believe it is 35!

I hope this helps you, and don't forget to give this answer brainliest if it really helped!

5 0
2 years ago
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