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Inessa [10]
3 years ago
8

The claim is that the proportion of adults who smoked a cigarette in the past week is less than 0.30​, and the sample statistics

include nequals1491 subjects with 462 saying that they smoked a cigarette in the past week. Find the value of the test statistic.
Mathematics
1 answer:
IrinaK [193]3 years ago
6 0

Answer: 0.84

Step-by-step explanation:

let p be the population proportion of adults who smoked a cigarette in the past week.

As per given , we have

H_a: p

Sample size : n= 1491

The sample proportion of adults smoked a cigarette= \hat{p}=\dfrac{x}{n}=\dfrac{462}{1491}=0.30985915493\approx0.31

The test statistic for proportion is given by :-

z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}

Substitute all the values , we get

z=\dfrac{0.31-0.30}{\sqrt{\dfrac{0.30(1-0.30)}{1491}}}\\\\=\dfrac{0.01}{\sqrt{0.000141}}\\\\=\dfrac{0.01}{0.011874342087}=0.84261497731\approx0.84

Hence, the value of the test statistic = 0.84

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If you were to use the substitution method to solve the following system, choose the new equation after the expression equivalen
adell [148]

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option D

2x − 7(−3x − 17) = 4

Step-by-step explanation:

Given in the question two equations

<h3>Equation 1</h3>

2x - 7y = 4

<h3>Equation 2</h3>

3x + y = -17

Rearranging equation 2 in terms of y

3x + y = -17

y = -17 - 3x

Put this value of y in Equation 1

2x - 7y = 4

2x - 7(-17 - 3x) = 4

So,

If you use the substitution method to solve the following system, new equation will be

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2 years ago
Arrange the entries of matrix A in increasing order of their cofactors values
givi [52]

To find the cofactor of

A=\left[\begin{array}{ccc}7&5&3\\-7&4&-1\\-8&2&1\end{array}\right]

We cross out the Row and columns of the respective entries and find the determinant of the remaining 2\times 2 matrix with the alternating signs.


Ac_{11}=\left|\begin{array}{ccc}4&-1\\2&1\end{array}\right|


Ac_{11}=4\times 1- -1\times 2


Ac_{11}=4+ 2

Ac_{11}=6




Ac_{12}=-\left|\begin{array}{ccc}-7&-1\\-8&1\end{array}\right|


Ac_{12}=-(-7\times 1- -1\times -8)


Ac_{12}=-(-7- 8)

Ac_{12}=15




Ac_{21}=-\left|\begin{array}{ccc}5&3\\2&1\end{array}\right|


Ac_{21}=-(5\times 1- 3\times 2)


Ac_{21}=-(5-6)


Ac_{21}=1







A_c{23}=-\left|\begin{array}{ccc}7&5\\-8&2\end{array}\right|


Ac_{23}=-(7\times 2 -8\times 5)


Ac_{23}=-(14-40)


Ac_{23}=26




A_c{31}=\left|\begin{array}{ccc}5&3\\4&-1\end{array}\right|


Ac_{31}=5\times -1 -4\times 3


Ac_{31}=-5-12


Ac_{31}=-17


A_c{33}=\left|\begin{array}{ccc}7&5\\-7&4\end{array}\right|


Ac_{33}=7\times 4- -7\times 5


Ac_{33}=28+35


Ac_{33}=63


Therefore in increasing order, we have;

Ac_{31}=-17,Ac_{21}=1,Ac_{11}=6,Ac_{23}=26,Ac_{12}=15, Ac_{33}=63



7 0
3 years ago
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