How do you make 6.83 as a mixed number?

Crash testing is a highly expensive procedure to evaluate the ability of an automobile to withstand a serious accident. A simple

polet [3.4K]

Answer:

95% confidence interval for the difference in the proportion is [-0.017 , 0.697].

Step-by-step explanation:

We are given that a simple random sample of 12 small cars were subjected to a head-on collision at 40 miles per hour. Of them 8 were "totaled," meaning that the cost of repairs is greater than the value of the car.

Another sample of 15 large cars were subjected to the same test, and 5 of them were totaled.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of small cars that were totaled = $\frac{8}{12}$ = 0.67

$\hat p_2$ = sample proportion of large cars that were totaled = $\frac{5}{15}$ = 0.33

$n_1$ = sample of small cars = 12

$n_2$ = sample of large cars = 15

$p_1$ = population proportion of small cars that are totaled

$p_2$ = population proportion of large cars that were totaled

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

So, 95% confidence interval for the difference between population population, ( $p_1-p_2$ ) is ;

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 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < $p_1-p_2$ < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ) = 0.95

<u>95% confidence interval for</u> $p_1-p_2$ = [ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ]

= [ $(0.67-0.33)-1.96 \times {\sqrt{\frac{0.67(1-0.67)}{12}+\frac{0.33(1-0.33)}{15} } }$ , $(0.67-0.33)+1.96 \times {\sqrt{\frac{0.67(1-0.67)}{12}+\frac{0.33(1-0.33)}{15} } }$ ]

= [-0.017 , 0.697]

Therefore, 95% confidence interval for the difference between proportions l and 2 is [-0.017 , 0.697].

6 0

4 years ago

By accident, four burned-out bulbs have been mixed in with 20 good ones. Ken is replacing old bulbs in his house. If he selects

ollegr [7]

We will be using the fact that $P(A \, and \,B) = P(A) \cdot P(B)$ , where $P(A)$ is the probability that the first bulb works and $P(B)$ is the probability that the second bulb works.

The probability that the first bulb works is $\frac{20}{24} = \frac{5}{6}$ . However, when we take one out (given that the first bulb works) we now have 19 working bulbs and 4 bad ones. This means that the probability that the second bulb works is $\frac{19}{23}$ .

Now, we can find our final probability:

$\dfrac{5}{6} \cdot \dfrac{19}{23} = \boxed{\dfrac{95}{138}}$

7 0

4 years ago

Can someone please help me? 20 Pts!!!

Romashka-Z-Leto [24]

I think it’s c but im not sure

5 0

3 years ago

Read 2 more answers

Find the slope of the line through points (5,2) and (-6,3).

fenix001 [56]

Answer:

-1/11

Step-by-step explanation:

The slope of a line when given two points is found by

m = (y2-y1)/(x2-x1)

= (3-2)/(-6-5)

=1/-11

= -1/11

3 0

3 years ago

Read 2 more answers

Don Alfredo quiete construir un corral cuadrado para sus gallinas, utizando metros de malla lineal para redearlo. Si el Zootecni

konstantin123 [22]

Answer:

Don Alfredo debe pagar $272,200 para encerrar por completo el corral.

Step-by-step explanation:

Dado que Don Alfredo quiere construir un corral cuadrado para sus gallinas, utilizando metros de malla lineal para rodearlo, si el zootecnista le recomendó construir uno que tenga 121 metros cuadrados, para determinar cuánto dinero debe pagar don Alfredo en la ferretería para encerrar por completo el corral si el metro lineal de malla tiene un valor de $ 6.200 el metro se deben realizar los siguientes cálculos:

121 metros cuadrados = √121 = 11 (x 11) = 11 x 4 = 44 metros lineales

44 x 6,200 = X

272,800 = X

Por lo tanto, Don Alfredo debe pagar $272,200 para encerrar por completo el corral.

8 0

3 years ago