IM ALMOST DONE. I NEED HELP PLEASE!!

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 husband and wife have normal vision, although both of the couple's fathers are red–green color-blind, an inherited X-linked re

Mekhanik [1.2K]

Answer:

1/2

Step-by-step explanation:

Since the recessive gene is carried in the x chromosome and a father can only pass it to a daughter, the father doesn't have the disease nor carries it.

The mother carries the recessive gene because her father passed her his conditioned gene (only having one x chromosome), while her mother gave her a normal one so she can have normal vision.

The chance of having a boy or a girl is 1/2 and the chance of this baby of having both recessive gene is null because in order to have a girl that is color blind both her x chromosomes must have the recessive gene, but the father can't pass it since he doesn't carries it. Therefore the combined probability is 1/2*(1-0) = 1/2.

7 0

3 years ago

Which function in vertex form is equivalent to f(x) = x2 + 6x + 3?

Georgia [21]

F(x)=x^2+6x+3
seperate x terms
f(x)=(x^2+6x)+3
take 1/2 of 6 and square it
6/2=3
3^2=9
add negative and positive inside parenthasees
f(x)=(x^2+6x+9-9)+3
factor perfect square
f(x)=((x+3)^2-9)+3
get rid of parenthasees
f(x)=(x+3)^2-9+3
f(x)=(x+3)^2-6

last one is answer

7 0

4 years ago

Read 2 more answers

Let A= { 15, 25, 35, 45, 55, 65 } and B= { 25, 45, 65} What is A∪B?

Karolina [17]

Hello:
<span>A= { 15, 25, 35, 45, 55, 65 } and B= { 25, 45, 65}
</span> A∪B = <span>{15, 25, 35, 45, 55, 65 } ....(answer : d )</span>

4 0

4 years ago

Read 2 more answers

Please answer this question

const2013 [10]

Answer: