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svetlana [45]
3 years ago
15

Suppose you are a television network executive interested in how a new pilot show, Statistics, will be received by the general U

.S. television-viewing population. You hire several market research firms; each recruits a random sample of television viewers and asks them to watch the pilot of Statistics. Each firm then gathers 9 respondents and asks each of them, "On a scale of 1 to 10, where 1 means you’ll never watch Statistics again, and 10 means you expect to never miss an episode of Statistics, how do you rate Statistics?" Assume that if you asked every member of the general U.S. television-viewing population about the pilot, the mean response would be 6.5 with a standard deviation of 1.5. Each market research firm reports the mean rating of its random sample. Assume each market research firm recruits a different sample, and that you hired exactly enough market research firms such that all possible samples of 9 U.S. television viewers were sampled and their mean ratings reported. Which of the following are true about the sampling distribution of the mean ratings reported to you by the market research firms?
1. The shape of the distribution of mean ratings depends on the shape of the ratings of the general U.S. television-viewing population.
2. The mean of the distribution of mean ratings is 0.20.
3. The shape of the distribution of mean ratings is approximately normal.
4. The standard deviation of the distribution of mean ratings is 0.30.


If you were really a television network executive, you would not hire multiple market research firms to each recruit a different sample of respondents. Instead, you would likely hire only one market research firm. If this is the case, why should you care about the theoretical distribution of mean ratings of multiple samples if you are only going to hire one firm and get one sample and one mean from that sample?

1. You need to know the theoretical distribution of possible values of the sample mean ratings to know whether the market research firm recruited the right sample.
2. To draw conclusions about the population mean using a sample mean, you need to know the theoretical distribution of the possible values the sample mean ratings could have.
3. You should not care about the possible distribution of mean ratings. You should only care about the mean and perhaps the highest and lowest ratings.
Mathematics
1 answer:
a_sh-v [17]3 years ago
8 0

Answer:

Step-by-step explanation:

Hello!

If X₁, X₂, ..., Xₙ be the n random variables that constitute a sample, then any function of type θ = î (X₁, X₂, ..., Xₙ) that depends solely on the n variables and does not contain any parameters known, it is called the estimator of the parameter.

When the function i (.) It is applied to the set of the n numerical values ​​of the respective random variables, a numerical value is generated, called parameter estimate θ.

This follows the concepts:

1) The function i (.) It is a function of random variables, so it is also a random variable, that is to say, that every estimator is a random variable.

2) From the above, it follows that Î has its probability distribution and therefore mathematical hope, E (î), and variance, V (î).

In this example, the variable of interest is X: response of a U.S. television viewer surveyed over the Statistics show.

The mean response was μ=6.5 and the standard deviation σ=1.5

If many samples of U.S. television viewers with size n=9 were taken and the sample mean of the user's ratings was calculated. The resulting sample mean will be a random variable and its distribution will have the same shape as the population it was originally calculated from.

The correct option is:

1. The shape of the distribution of mean ratings depends on the shape of the ratings of the general U.S. television-viewing population.

Since is impossible to study whole populations, due to economic, materials, and time reasons, especially when the populations are very big. Then since you cannot study the population mean directly, it is best to do it through the sample mean. For this, you need to know its theoretical distribution.

Correct option:

2. To conclude the population mean using a sample mean, you need to know the theoretical distribution of the possible values the sample mean ratings could have.

I hope you have a SUPER day!

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Answer:

The matrix form of the system of equations is \left[\begin{array}{ccccc}1&1&1&1&-3\\1&-1&-2&1&2\\2&0&1&-1&1\end{array}\right] \left[\begin{array}{c}x&y&w&z&u\end{array}\right] =\left[\begin{array}{c}5&4&3\end{array}\right]

The reduced row echelon form is \left[\begin{array}{ccccc|c}1&0&0&1/4&0&3\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]

The vector form of the general solution for this system is \left[\begin{array}{c}x&y&w&z&u\end{array}\right]=u\left[\begin{array}{c}-\frac{1}{6}&\frac{5}{2}&0&\frac{2}{3}&1\end{array}\right]+w\left[\begin{array}{c}-\frac{1}{6}&-\frac{3}{2}&1&\frac{2}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]

Step-by-step explanation:

  • <em>Convert the given system of equations to matrix form</em>

We have the following system of linear equations:

x+y+w+z-3u=5\\x-y-2w+z+2u=4\\2x+w-z+u=3

To arrange this system in matrix form (Ax = b), we need the coefficient matrix (A), the variable matrix (x), and the constant matrix (b).

so

A= \left[\begin{array}{ccccc}1&1&1&1&-3\\1&-1&-2&1&2\\2&0&1&-1&1\end{array}\right]

x=\left[\begin{array}{c}x&y&w&z&u\end{array}\right]

b=\left[\begin{array}{c}5&4&3\end{array}\right]

  • <em>Use row operations to put the augmented matrix in echelon form.</em>

An augmented matrix for a system of equations is the matrix obtained by appending the columns of b to the right of those of A.

So for our system the augmented matrix is:

\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\1&-1&-2&1&2&4\\2&0&1&-1&1&3\end{array}\right]

To transform the augmented matrix to reduced row echelon form we need to follow this row operations:

  • add -1 times the 1st row to the 2nd row

\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&-2&-3&0&5&-1\\2&0&1&-1&1&3\end{array}\right]

  • add -2 times the 1st row to the 3rd row

\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&-2&-3&0&5&-1\\0&-2&-1&-3&7&-7\end{array}\right]

  • multiply the 2nd row by -1/2

\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&3/2&0&-5/2&1/2\\0&-2&-1&-3&7&-7\end{array}\right]

  • add 2 times the 2nd row to the 3rd row

\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&3/2&0&-5/2&1/2\\0&0&2&-3&2&-6\end{array}\right]

  • multiply the 3rd row by 1/2

\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&3/2&0&-5/2&1/2\\0&0&1&-3/2&1&-3\end{array}\right]

  • add -3/2 times the 3rd row to the 2nd row

\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]

  • add -1 times the 3rd row to the 1st row

\left[\begin{array}{ccccc|c}1&1&0&5/2&-4&8\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]

  • add -1 times the 2nd row to the 1st row

\left[\begin{array}{ccccc|c}1&0&0&1/4&0&3\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]

  • <em>Find the solutions set and put in vector form.</em>

<u>Interpret the reduced row echelon form:</u>

The reduced row echelon form of the augmented matrix is

\left[\begin{array}{ccccc|c}1&0&0&1/4&0&3\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]

which corresponds to the system:

x+1/4\cdot z=3\\y+9/4\cdot z-4u=5\\w-3/2\cdot z+u=-3

We can solve for <em>z:</em>

<em>z=\frac{2}{3}(u+w+3)</em>

and replace this value into the other two equations

<em>x+1/4 \cdot (\frac{2}{3}(u+w+3))=3\\x=-\frac{u}{6} -\frac{w}{6}+\frac{5}{2}</em>

y+9/4 \cdot (\frac{2}{3}(u+w+3))-4u=5\\y=\frac{5u}{2}-\frac{3w}{2}+\frac{1}{2}

No equation of this system has a form zero = nonzero; Therefore, the system is consistent. The system has infinitely many solutions:

<em>x=-\frac{u}{6} -\frac{w}{6}+\frac{5}{2}\\y=\frac{5u}{2}-\frac{3w}{2}+\frac{1}{2}\\z=\frac{2u}{3}+\frac{2w}{3}+2</em>

where <em>u</em> and <em>w</em> are free variables.

We put all 5 variables into a column vector, in order, x,y,w,z,u

x=\left[\begin{array}{c}x&y&w&z&u\end{array}\right]=\left[\begin{array}{c}-\frac{u}{6} -\frac{w}{6}+\frac{5}{2}&\frac{5u}{2}-\frac{3w}{2}+\frac{1}{2}&w&\frac{2u}{3}+\frac{2w}{3}+2&u\end{array}\right]

Next we break it up into 3 vectors, the one with all u's, the one with all w's and the one with all constants:

\left[\begin{array}{c}-\frac{u}{6}&\frac{5u}{2}&0&\frac{2u}{3}&u\end{array}\right]+\left[\begin{array}{c}-\frac{w}{6}&-\frac{3w}{2}&w&\frac{2w}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]

Next we factor <em>u</em> out of the first vector and <em>w</em> out of the second:

u\left[\begin{array}{c}-\frac{1}{6}&\frac{5}{2}&0&\frac{2}{3}&1\end{array}\right]+w\left[\begin{array}{c}-\frac{1}{6}&-\frac{3}{2}&1&\frac{2}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]

The vector form of the general solution is

\left[\begin{array}{c}x&y&w&z&u\end{array}\right]=u\left[\begin{array}{c}-\frac{1}{6}&\frac{5}{2}&0&\frac{2}{3}&1\end{array}\right]+w\left[\begin{array}{c}-\frac{1}{6}&-\frac{3}{2}&1&\frac{2}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]

7 0
4 years ago
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krek1111 [17]

We want our exponential function to look like

y = ab^x.

Let a = the initial y-value.

Our initial value is the first number given for f(x). So, a = 3.

Let b = the number that is needed to go from 3 to 6 to 12 to 24 to 48.

We find b by division.

So, b = the next number divided by the previous.

So, b = 6/3 = 2.

We now plug in our values into the general formula above.

y = ab^x

Answer: y = (3)(2)^x

4 0
3 years ago
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