Which of the following best describes an Extraneous Solution?
2 answers:
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Using a t-distribution calculator and finding the p-value, the correct option regarding the conclusion is given by:
a) the p-value is 0.02. We reject h0 at the 5% significance level because the p-value 0.02 is less than 0.05.
<h3>What is the relation between the p-value and the conclusion?</h3>It also involves the significance level, as follows.
- If the p-value is less than the significance level, we reject the null hypothesis
.
- If it is more, we do not reject.
In this problem, a t-distribution calculator for a right-tailed with <em>t = 2.15 and 25 - 1 = 24 df</em> is used to find a p-value of 0.02.
It is less than 0.05, hence option a is correct.
More can be learned about p-values at brainly.com/question/26454209
We know that
(ad)/(bd)=d/d time a/b=a/b since d's cancel
also
if a/b=c/d in simplest form, then a=c and b=d
we have
p/(x^2-5x+6)=(x+4)/(x-2)
therefor
p/(x^2-5x+6)=d/d times (x+4)/(x-2)
p/(x^2-5x+6)=d(x+4)/d(x-2)
therefor
p=d(x+4) and
x^2-5x+6=d(x-2)
we can solve last one
factor
(x-6)(x+1)=d(x-2)
divide both sides by (x-2)
[(x-6)(x+1)]/(x-2)=d
sub
p=d(x+4)
p=([(x-6)(x+1)]/(x-2))(x+4)
(ad)/(bd)=d/d time a/b=a/b since d's cancel
also
if a/b=c/d in simplest form, then a=c and b=d
we have
p/(x^2-5x+6)=(x+4)/(x-2)
therefor
p/(x^2-5x+6)=d/d times (x+4)/(x-2)
p/(x^2-5x+6)=d(x+4)/d(x-2)
therefor
p=d(x+4) and
x^2-5x+6=d(x-2)
we can solve last one
factor
(x-6)(x+1)=d(x-2)
divide both sides by (x-2)
[(x-6)(x+1)]/(x-2)=d
sub
p=d(x+4)
p=([(x-6)(x+1)]/(x-2))(x+4)