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Pie
2 years ago
7

Can anyone answer this for me?

Mathematics
1 answer:
3241004551 [841]2 years ago
6 0
The chapter test and retest medians are almost the same

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Four students decide to flip a coin a number of times and record their results. Dan flips his coin 5 times, Kelly flips her coin
Tcecarenko [31]

Answer:

Judah

Step-by-step explanation:

Because the more times you do the same thing over and over again then average it the more accurate it becomes

7 0
3 years ago
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
3 years ago
Which values of a, b, and c represent the answer in simplest form?<br><br> 7/9 * 4/9 = a b/c
myrzilka [38]

Answer:

Option 2: A=1, B=3, C=4

Step-by-step explanation:

  1. 7/9÷4/9  = 7/9*9/4
  2. cancel 9 on each side
  3. =7/4
  4. = 1 3/4

4 0
2 years ago
Solve linear equation 1/4 x + 18 = X
Anton [14]

Answer:

24

Step-by-step explanation:

1 / 4 x + 18 = x

x / 4 + 18 = x

x = x / 4 + 18

x - x / 4 = 18

( 4x / 4 ) - ( x/4 ) = 18

( 4x - x ) / 4 = 18

3x / 4 = 18

3x = 18 * 4

x = ( 18 * 4 ) / 3

  = 6 * 4

x = 24

3 0
2 years ago
I will give branly to the best and fast answer :D
Anton [14]

Answer:

its fifteen fiths

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
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