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baherus [9]
3 years ago
5

⦁ In a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football. Construct a 95% confiden

ce interval for the proportion of adults in the United States whose favorite sport to watch is football.
Mathematics
1 answer:
fomenos3 years ago
5 0

Answer:

95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

Step-by-step explanation:

We are given that in a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football.

Firstly, the pivotal quantity for 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is given by;

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

where, \hat p = proportion of adults in the United States whose favorite sport to watch is football in a sample of 1219 adults = \frac{354}{1219}

           n = sample of US adults  = 1291

           p = population proportion of adults

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population proportion, p is ;

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

                                                    significance level 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

<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} } ]

                         = [ \frac{354}{1219}-1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} } , \frac{354}{1219}+1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} } ]

                         = [0.265 , 0.316]

Therefore, 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

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Step-by-step explanation:

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Read 2 more answers
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Answer:

(a)x_1=-2,x_2=1

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Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix A and the matrix B_j for each variable. The matrix A is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get A more easily.

2x_1+5x_2=1\\x_1+4x_2=2

\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]

To get B_1, replace in the matrix A the 1st column with the results of the equations:

B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]

To get B_2, replace in the matrix A the 2nd column with the results of the equations:

B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]

Apply the rule to solve x_1:

x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2

In the case of B2,  the determinant is going to be zero. Instead of using the rule, substitute the values ​​of the variable x_1 in one of the equations and solve for x_2:

2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1

(b) In this system, follow the same steps,ust remember B_3 is formed by replacing the 3rd column of A with the results of the equations:

2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0

\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]

B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]

B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]

B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]

x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}

x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}

x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}

6 0
3 years ago
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