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aleksandrvk [35]
3 years ago
12

Suppose we’re conducting a t-test for a population mean with the following hypotheses. LaTeX: H_0 H 0 : LaTeX: \mu=50 μ = 50 LaT

eX: H_a H a : LaTeX: \mu\ne50 μ ≠ 50 A sample of size 30 has a t-statistic of LaTeX: -2.43 − 2.43 . Find the corresponding P-value. Do not round.
Mathematics
1 answer:
blsea [12.9K]3 years ago
5 0

Answer: 0.02055867

Step-by-step explanation:

Given : H_0: \mu=50\\\\ H_a: \mu\neq 50, since the alternative hypothesis is two-tailed , so the test is a two tailed test.

Sample size : n= 30

Degree of freedom for t-distribution = df=n-1=30-1=29

Test statistic= t= -2.43

By using the normal t-distribution table ( or calculator ) , we have

The corresponding P-value for two tailed test = 0.02055867

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3x^2+6x-9=0 complete the square
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You could simplify this work by factoring "3" out of all four terms, as follows:

3(x^2 + 2x - 3) =3(0) = 0

Hold the 3 for later re-insertion.  Focus on "completing the square" of x^2 + 2x - 3.

1.  Take the coefficient (2) of x and halve it:  2 divided by 2 is 1
2.   Square this result:  1^2 = 1
3.   Add this result (1) to x^2 + 2x, holding the "-3" for later:
                    x^2 +2x 
4    Subtract (1) from x^2 + 2x + 1:     x^2 + 2x + 1               -3 -1  =    0, 
       or      x^2 + 2x + 1 - 4 = 0
5.   Simplify, remembering that x^2 + 2x + 1 is a perfect square:

                        (x+1)^2 - 4 = 0

We have "completed the square."  We can stop here.  or, we could solve for x:  one way would be to factor the left side:

            [(x+1)-2][(x+1)+2]=0     The solutions would then be:

             x+1-2=0=> x-1=0, or x=1, and
             x+1 +2 = 0 => x+3=0, or x=-3.  (you were not asked to do this).


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2 years ago
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Answer:

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3 years ago
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Step-by-step explanation:

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