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jeyben [28]
3 years ago
10

Helppp pleasee!! If you can answer ASAP I would really appreciate it thank youuu.

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
7 0

Answer:

1. 4/7 = 16/28

Check:

(4/7)(4/4) = 16/28 = 16/28

2. 40/15 = 8/3

Check:

(40/15)/(5/5) = 8/3

3. 15/6 = 5/2

Check:

(15/6)/(3/3) = 5/2

4. 8/11 = 56/77

Check:

(8/11)(7/7) = 56/77

5. 9/2 = 63/14

Check:

(9/2)(7/7) = 63/14

~

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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
Y+5<br> ——=0 solve for y<br> X-1
Ira Lisetskai [31]

Answer:

y = -5

Step-by-step explanation:

see the pic for the steps . .

5 0
4 years ago
A tree casts a shadow that is 8 feet long, a 3 foot mailbox cast a shadow that is 5 feet long. how tall is the tree?
madam [21]
Set up two fractions:
8/x and 3/5
Set them equal to each other and cross multiply to get:
8*5 = 3x
40 = 3x
13.3 = x or 13 and 1/3 feet tall. 
7 0
3 years ago
A number cube was rolled as part of an experiment. The results are in the table below. The fraction StartFraction 1 over x EndFr
Svetlanka [38]

Answer:

6

Step-by-step explanation:

3 0
3 years ago
Jalen made $533.75 while working at his job. He makes $15.25 an hour, how many hours did he work to make that amount of money?
mixer [17]

Answer:

35 hours

Step-by-step explanation:

Typically, people get their paycheck after 2 weeks.

Let say he worked 8 hours a day, a typical amount of hours for the average human who has a occupation.

Multiply the amount of hours by the amount payed every hour.

So 15.25 x 8 is 122

( so in 8 hours jalen made $122)

Add 8 hours, which would now be 16 hours worked.

16 x 15.25 = 244

(so in 16 hours jalen made $244)

Add another 8 hours of work time which would now be 24 hours (1 full day)

24x5.25=366

Add 10 hours on now which would be 34 hours worked (2 days = 48 hours meaning she worked 1 day and 10 hours )

34x15.25=518.50

(getting closer to the amount she earned)

So if we add another hour

which would be 35 hours worked (1 day 11 hours)

35 x 15.25 = 533.75

Therefore, Jalen worked 35 hours.

5 0
3 years ago
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