The next Collection we will begin working on will be Collection 5, "Decisions That Matter." This Collection begins on page 251 i
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Answer:
For this case the statistic calculated is
So the p value is a very high value and using the significance level given we see that
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and there is enough evidence to conclude that we have a difference between the two means at 10% of significance.
Step-by-step explanation:
Data given and notation
represent the mean for the sample for A
represent the mean for the sample B
represent the sample standard deviation for the sample A
represent the sample standard deviation for the sample B
sample size for the group poisoned
sample size for the group unpoisoned
t would represent the statistic (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to check if the two means are equal or not , the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
For this case is better apply a t test to compare means, and the statistic is given by:
(1)
t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
Calculate the statistic
For this case the statistic calculated is
Statistical decision
The significance level is 0.1 , but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:
The p value on this case is given
So the p value is a very high value and using the significance level given we see that
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and there is enough evidence to conclude that we have a difference between the two means at 10% of significance.
Given
We can rewrite it in matrix form as:
![\left[\begin{array}{cccc}1&-1&2&4\\5&2&-1&3\\0&0&0&0\\0&0&0&0\end{array}\right] \left[\begin{array}{c}w\\x\\y\\z\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right] \\ \\ \left[\begin{array}{cccc}1&-1&2&4\\5&2&-1&3\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \ -5R_1+R_2\rightarrow R_2](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-1%262%264%5C%5C5%262%26-1%263%5C%5C0%260%260%260%5C%5C0%260%260%260%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dw%5C%5Cx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C0%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%20%20%5C%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-1%262%264%5C%5C5%262%26-1%263%5C%5C0%260%260%260%5C%5C0%260%260%260%5Cend%7Barray%7D%5Cright%7C%5Cleft.%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C0%5Cend%7Barray%7D%5Cright%5D%5C%20%5C%20%5C%20%5C%20%5C%20-5R_1%2BR_2%5Crightarrow%20R_2)
![\left[\begin{array}{cccc}1&-1&2&4\\0&7&-11&-17\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \ \frac{1}{7} R_2\rightarrow R_2 \\ \\ \left[\begin{array}{cccc}1&-1&2&4\\0&1&-\frac{11}{7}&-\frac{17}{7}\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \ R_1+R_2\rightarrow R_1](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-1%262%264%5C%5C0%267%26-11%26-17%5C%5C0%260%260%260%5C%5C0%260%260%260%5Cend%7Barray%7D%5Cright%7C%5Cleft.%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C0%5Cend%7Barray%7D%5Cright%5D%5C%20%5C%20%5C%20%5C%20%5C%20%20%5Cfrac%7B1%7D%7B7%7D%20R_2%5Crightarrow%20R_2%20%5C%5C%20%20%5C%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-1%262%264%5C%5C0%261%26-%5Cfrac%7B11%7D%7B7%7D%26-%5Cfrac%7B17%7D%7B7%7D%5C%5C0%260%260%260%5C%5C0%260%260%260%5Cend%7Barray%7D%5Cright%7C%5Cleft.%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C0%5Cend%7Barray%7D%5Cright%5D%5C%20%5C%20%5C%20%5C%20%5C%20R_1%2BR_2%5Crightarrow%20R_1)
![\\ \\ \left[\begin{array}{cccc}1&0&\frac{3}{7}&\frac{11}{7}\\0&1&-\frac{11}{7}&-\frac{17}{7}\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right] \\ \\ \Rightarrow w= -\frac{3}{7} y- \frac{11}{7} z \\ x=\frac{11}{7} y+ \frac{17}{7} z \\ y=free \\ z=free \\ \\ =y\left\ \textless \ -\frac{3}{7},\frac{11}{7}1,0\right\ \textgreater \ +z\left\ \textless \ - \frac{11}{7},\frac{17}{7},0,1\right\ \textgreater \](https://tex.z-dn.net/?f=%20%5C%5C%20%20%5C%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26%5Cfrac%7B3%7D%7B7%7D%26%5Cfrac%7B11%7D%7B7%7D%5C%5C0%261%26-%5Cfrac%7B11%7D%7B7%7D%26-%5Cfrac%7B17%7D%7B7%7D%5C%5C0%260%260%260%5C%5C0%260%260%260%5Cend%7Barray%7D%5Cright%7C%5Cleft.%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C0%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20w%3D%20-%5Cfrac%7B3%7D%7B7%7D%20y-%20%5Cfrac%7B11%7D%7B7%7D%20z%20%5C%5C%20x%3D%5Cfrac%7B11%7D%7B7%7D%20y%2B%20%5Cfrac%7B17%7D%7B7%7D%20z%20%5C%5C%20y%3Dfree%20%5C%5C%20z%3Dfree%20%5C%5C%20%20%5C%5C%20%3Dy%5Cleft%5C%20%5Ctextless%20%5C%20-%5Cfrac%7B3%7D%7B7%7D%2C%5Cfrac%7B11%7D%7B7%7D1%2C0%5Cright%5C%20%5Ctextgreater%20%5C%20%2Bz%5Cleft%5C%20%5Ctextless%20%5C%20-%20%5Cfrac%7B11%7D%7B7%7D%2C%5Cfrac%7B17%7D%7B7%7D%2C0%2C1%5Cright%5C%20%5Ctextgreater%20%5C%20)
Thus, the solution set is a span of
We can rewrite it in matrix form as:
Thus, the solution set is a span of