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Answer:
not factorable
Step-by-step explanation:
The easiest way to factor is to find two numbers of the "a" value (the coefficient before ) and the factors of the c value (10) that ADD up to give you -9, in this case it would be:
1 -10 --> 1(-10) gives you -10
1 -1 --> 1(-1) gives you -1
adding these together -10 + (-1) = -11 which does not equal -9.
You could stop here and conclude that this is not factorable by inspection.
Another method: using the quadratic formula to find its roots.
Roots: and
Answer:
Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed. So we can say that the 3 remedies ar approximately equally effective.
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:
NyQuil Robitussin Triaminic Total
No relief 11 13 9 33
Some relief 32 28 27 87
Total relief 7 9 14 30
Total 50 50 50 150
We need to conduct a chi square test in order to check the following hypothesis:
H0: There is no difference in the three remedies
H1: There is a difference in the three remedies
The level os significance assumed for this case is
The statistic to check the hypothesis is given by:
The table given represent the observed values, we just need to calculate the expected values with the following formula
And the calculations are given by:
And the expected values are given by:
NyQuil Robitussin Triaminic Total
No relief 11 11 11 33
Some relief 29 29 29 87
Total relief 10 10 10 30
Total 50 50 50 150
And now we can calculate the statistic:
Now we can calculate the degrees of freedom for the statistic given by:
And we can calculate the p value given by:
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(3.81,4,TRUE)"
Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed.