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defon
3 years ago
14

Which number is equal to 10 Superscript negative 3

Mathematics
2 answers:
jeka57 [31]3 years ago
5 0
Your Answer 0.001 (c)
timama [110]3 years ago
3 0

Answer:

1/1000 or 0.001.

Step-by-step explanation:

10 ^-3

= 1/10^3

= 1/1000.

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Here are 15 cards, numbered 1 through 15. if you pick a card, what is the probability that you choose an odd number or a two?
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9/15. There are 15 cards and 8 of them are odd and 1 is a 2. 
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Some scientists believe alcoholism is linked to social isolation. One measure of social isolation is marital status. A study of
alexandr402 [8]

Answer:

1) H0: There is significant evidence of independence between marital status and diagnostic

H1: There is significant evidence of an association between marital status and diagnostic

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

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

3) \chi^2 = \frac{(21-33.143)^2}{31.143}+\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) p_v = P(\chi^2_{2} >19.72)=0.00005222

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

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

Since the p values is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance, and we can say that we have associated 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:

                    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 significant evidence of independence between marital status and diagnostic

H1: There is significant evidence of an association between marital status and diagnostic

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

Part 2

The statistic to check the hypothesis is given by:

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

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:

                    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

Part 3

And now we can calculate the statistic:

\chi^2 = \frac{(21-33.143)^2}{31.143}+\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)=(2-1)(3-1)=2

And we can calculate the p value given by:

p_v = P(\chi^2_{2} >19.72)=0.00005222

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

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

Since the p values is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance, and we can say that we have associated between the two variables analyzed.

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