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Assoli18 [71]
3 years ago
6

4 it is claimed that less than half of households in the United States have carbon monoxide detectors. A survey of 1005 househol

ds resulted in 462 who have carbon monoxide detectors. Find the value of the test statistic Z
Mathematics
1 answer:
Crank3 years ago
3 0

Answer:

The value of the test statistic is -2.54.

Step-by-step explanation:

A hypothesis test is conducted to determine whether the proportion of households in the United States having carbon monoxide detectors is less than 50% or not.

The hypothesis is:

H_{o}: The proportion of households in the United States having carbon monoxide detectors is less than 50%, i.e. <em>p</em> < 0.50.

H_{a}: The proportion of households in the United States having carbon monoxide detectors is not less than 50%, i.e. <em>p</em> ≥ 0.50.

The test statistic is:

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

<u>Given:</u>

Sample size, <em>n</em> = 1005

Number of households that have carbon monoxide detectors is, <em>X</em> = 462.

The sample proportion is: \hat p=\frac{X}{n}= \frac{462}{1005}= 0.4597\approx0.46

Compute the value of the test statistic as follows:

z=\frac{\hat p -p}{\sqrt{\frac{p(1-p)}{n} }}\\=\frac{0.46-0.50}{\sqrt{\frac{0.50(1-0.50)}{1005}}} \\=-2.54

Thus, the value of the test statistic is -2.54.

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91%

Step-by-step explanation:

The first step is to calculate the increase

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Please help!!!!
valentina_108 [34]
ANSWER

\frac{ {t}^{2}  + 4t - 12}{ {t}^{2} - 4 }  =  \frac{t  + 6}{ t + 2}
where,
t \ne - 2



EXPLANATION

We want to simplify the rational expression

\frac{ {t}^{2}  + 4t - 12}{ {t}^{2} - 4 }


We can observe that the numerator of the given rational expression is a quadratic trinomial and the denominator is a difference of two squares



We need to split the middle term in the numerator and rewrite the denominator as a difference of two squares to obtain,




\frac{ {t}^{2}  + 4t - 12}{ {t}^{2} - 4 }  =  \frac{{t}^{2}  + 6t - 2t - 12}{ {t}^{2}  -  {2}^{2} }



We now factor to obtain,

\frac{ {t}^{2}  + 4t - 12}{ {t}^{2} - 4 }  =  \frac{t (t+ 6) - 2(t  + 6)}{ (t  -  2)(t + 2)}


This implies that,

\frac{ {t}^{2}  + 4t - 12}{ {t}^{2} - 4 }  =  \frac{(t - 2) (t  + 6)}{ (t  -  2)(t + 2)}


We now cancel out common factors to obtain,

\frac{ {t}^{2}  + 4t - 12}{ {t}^{2} - 4 }  =  \frac{(t  + 6)}{ (t + 2)}


The restriction is that, the denominator cannot be zero.

Thus

t + 2 \ne0


This implies that,

t  \ne - 2
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Step-by-step explanation:

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