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expeople1 [14]
3 years ago
6

n a Gallop poll of 1012 randomly selected adults, 9% said that cloning of humans should be allowed. We are going to use a .05 si

gnificance level to test the claim that less than 10% of all adults say that cloning of humans should be allowed. The test statistic for this test is -1.06. What is the p-value for this test statistic?
Mathematics
1 answer:
Vanyuwa [196]3 years ago
4 0

Answer:0.1446

Step-by-step explanation:

Let p be the population proportion of adults say that cloning of humans should be allowed.

As per given , the appropriate set of hypothesis would be :-

H_0: p\geq0.10\\\\ H_a: p

Since H_a is left-tailed and sample size is large(n=1012) , so we perform left-tailed z-test.

Given : The test statistic for this test is -1.06.

P-value for left tailed test,

P(z<-1.06)=1-P(z<1.06)        [∵P(Z<-z)=1-P(Z<z]

=1-0.8554=0.1446          (Using z-value table.)

Hence, the p-value for this test statistic= 0.1446

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3 years ago
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Three listening stations located at (3300, 0), (3300, 1100), and (-3300, 0) monitor an explosion. The last two stations detect t
erastovalidia [21]

Answer:

The coordinates of the explosion is (3300, -2750)

Step-by-step explanation:

I have plot the points and attached to thus answer for easy understanding.

Now, from the question, since station A is the first to hear the explosion, we'll make it the foci of the parabola in the graph I attached and it will be horizontal since the distance between station C and A is much more than that between station B and A. Thus, the reason why station C will have to be the other foci with the hyperbola centred at the origin.

Now, sound travels at a speed of 1100 ft/s and station B is located 1100 ft from station A. Thus, the explosion would likely have occurred at a point on the line x = 3300ft . Since station A is 3300ft from centre C = 3300,hence C² = 3300² = 10,890,000. Since it takes 4 seconds longer for the sound to reach station C than A, the sound has traveled 4(1100)= 4400 ft.

Thus, 4400 = d1 = d2 = 2a

So,2a = 4400 and so, a =2200

a² = 2200² = 4,840,000 where d1 is the distance from station C to the explosion and d2 is the distance from station A to the explosion. To find b², let's use the equation ;

c² = a² + b² and so; b² = c² - a² = 10,890,000 - 4,840,000 = 6,050,000

Equation of hyperbola is given as;

(x²/a²) - (y²/b²) = 1

Plugging in the values of a² and b², we obtain ;

(x²/4,840,000) - (y²/6,050,000) = 1

Since we have deduced that the explosion must occur on the line x= 3300, we'll put in 3300 for x to obtain ;

(3300²/4,840,000) - (y²/6,050,000) = 1

2.25 - 1 = (y²/6,050,000)

y² = (1.25 x 6,050,000)

y² = 7562500

y = √7562500

y = ± 2750

Due to the fact that the explosion will occur at a point further from station B than from station A, the explosion will take place in quadrant 4. Thus, we will take the negative value of y which is - 2750.

So explosion will occur at the coordinate (3300, -2750)

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