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3 years ago

Bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb

Mathematics

1 answer:

Sophie [7]3 years ago

3 0

Answer:

Ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc

Step-by-step explanation:

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A book claims that more hockey players are born in January through March than in October through December. The following data sh

astra-53 [7]

Answer:

$\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295$

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

$df=(categories-1)=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >18.295)=0.00038$

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year

Step-by-step explanation:

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

H0: There is no difference of birthdates distributed throughout the year

H1: There is a difference between birthdates distributed throughout the year

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

The statistic to check the hypothesis is given by:

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

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total}{4}$

And replacing we got:

$E_{1} =\frac{67+56+30+37}{4}=47.5$

And now we can calculate the statistic:

$\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295$

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

$df=(categories-1)=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >18.295)=0.00038$

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year

3 0

4 years ago

If a denotes some event, what does upper a overbara denote? if p(a)equals=0.9930.993, what is the value of p(upper a overba

masya89 [10]

P(A) = 0.993 is the probability of event A
_
P(A) is the probability of not A (note that I am trying to place a bar over A. I hope you can see it, else it is because the editor did not show it as I intended)

Then, use the fact that the probability of A plus the probability not A is equal to 1:
                  _
=> P(A) + P(A) = 1
       _
=> P(A) = 1 - P(A) = 1 - 0.993 = 0.007

So, the answers are:
      _
1) P(A) = 0.07
             _
2) Yes, A is unlikely because its probability is very low.

5 0

3 years ago

Write a system of equations for (4,-3)

qaws [65]

y = 4x - 3 would be it !!!!

7 0

3 years ago

Someone help me I need to understand

Firlakuza [10]

Answer:

Easy you describe the sum of each arithmetic series. GOOD LUCK

Step-by-step explanation:

3 0

4 years ago

The cost of equal-priced DVD and the number purchased

Vilka [71]

Answer:

Linear relationship

Step-by-step explanation:

ok <3

7 0

3 years ago