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Vanyuwa [196]
3 years ago
9

Could a set of three vectors in ℝ^4 span all ℝ^4? Explain. What about n vectors in ℝ^m when n is less than m?

Mathematics
1 answer:
Elena-2011 [213]3 years ago
8 0

Answer:

Check the ecplanation

Step-by-step explanation:

A set of three vectors in R^{4} represents a matrix of 3 column vectors, and each vector containing 4 entries (that is, a matrix of 4 rows, and 3 columns).

Let A be that 4x 3 matrix. The columns of A span R^{m}. if and only if A has a pivot position in each row. So, there are at most 3 pivot positions in the matrix A, but the number of rows is 4, therefore, there exist at least one row not having a pivot position. If A does not have a pivot position in at least one row, then the columns of A do not span R^{4}. It implies that the set of 3 vectors of A does not span all of R^{4}.

In general, the set of n vectors in R^{m} represents a matrix of in rows, and n columns (an in x matrix). So, there are at most n pivot positions in the matrix A, but n is less than the number of rows. In therefore, there exist at least one row that does not contain a pivot position.

And, hence the set of n vectors of A does not span all of R^{m}. for n < m

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Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 ​students, she finds 2 w
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Answer:

A 95​% confidence interval for the proportion of students who eat cauliflower on​ Jane's campus is [0.012, 0.270].

Step-by-step explanation:

We are given that Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 ​students, she finds 2 who eat cauliflower.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

                              P.Q.  =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of students who eat cauliflower

           n = sample of students

           p = population proportion of students who eat cauliflower

<em>Here for constructing a 95% confidence interval we have used a One-sample z-test for proportions.</em>

<u>So, 95% confidence interval for the population proportion, p is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                   of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

Now, in Agresti and​ Coull's method; the sample size and the sample proportion is calculated as;

n = n + Z^{2}__(\frac{_\alpha}{2})

n = 24 + 1.96^{2} = 27.842

\hat p = \frac{x+\frac{Z^{2}__(\frac{\alpha}{2}_)  }{2} }{n} = \hat p = \frac{2+\frac{1.96^{2}   }{2} }{27.842} = 0.141

<u>95% confidence interval for p</u> = [ \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ]

 = [ 0.141 -1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } } , 0.141 +1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } } ]

 = [0.012, 0.270]

Therefore, a 95​% confidence interval for the proportion of students who eat cauliflower on​ Jane's campus [0.012, 0.270].

The interpretation of the above confidence interval is that we are 95​% confident that the proportion of students who eat cauliflower on​ Jane's campus is between 0.012 and 0.270.

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Step-by-step explanation:

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