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4 years ago

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Suppose we are testing the null hypothesis H 0 mu=20 and the alternative H iu =20 , normal population with sigma=6 A random samp

le of nine observations are drawn from the population, and we find the sample mean of these observations x = 17 The value of the test statistic z is approximatel

1 answer:

cupoosta [38]4 years ago

8 0

Answer:

The value of the test statistic is $z = -1.5$

Step-by-step explanation:

The formula for the test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the statistic, $\mu$ is the mean, $\sigma$ is the standard deviation and n is the number of observations.

In this problem, we have that:

$\mu = 20, X = 17, \sigma = 6, n = 9$

So

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{17 - 20}{\frac{6}{\sqrt{3}}}$

$z = -1.5$

The value of the test statistic is $z = -1.5$

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Answer:

x = 3 && y = - 1

Step-by-step explanation:

Just solving the two equations

the first equation

y = 4/3 x +3

y - 4/3 x = 3 ..............................................(1)

the second equation

y = -2/3x -3

y + 2/3 x = -3 ..............................................(2)

by substracting 2 from 1

-4/3x - 2/3x = -6

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4/3x +2/3x = 6

6/3x = 6

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Answer:

Step-by-step explanation:

Product of slope of perpendicular lines = -1

6x + 5y = 30

Write this equation in y = mx + b form

5y = -6x + 30

y = $\frac{-6}{5}x+\frac{30}{5}$

$y=\frac{-6}{5}x + 6$

Slope of this line m₁ = -6/5

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m₂ = -1÷m₁ = -1 * $\frac{5}{-6}$

$m_{2}=\frac{5}{6}$ & (-6 , -7)

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$y - (-7) = \frac{5}{6}(x - [-6])\\\\y + 7 = \frac{5}{6}x + 6 *\frac{5}{6}\\\\y = \frac{5}{6}x +5-7\\\\y=\frac{5}{6}x-2$

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The probability that a randomly selected teenager watched a rented video at least once during a week was 0.75. What is the proba

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Answer:

There is a 75.65% probability that at least 5 teenagers in a group of 7 watched a rented movie at least once last week.

Step-by-step explanation:

For each teenager, there are only two possible outcomes. Either they watched a rented video at least once during a week, or they did not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

The probability that a randomly selected teenager watched a rented video at least once during a week was 0.75. This means that $p = 0.75$ .

What is the probability that at least 5 teenagers in a group of 7 watched a rented movie at least once last week?

Group of 7, so $n = 7$ .

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7)$ .

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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$P(X = 6) = C_{7,6}.(0.75)^{6}.(0.25)^{1} = 0.3115$

$P(X = 7) = C_{7,7}.(0.75)^{7}.(0.25)^{0} = 0.1335$

So

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) = 0.3115 + 0.3115 + 0.1335 = 0.7565$ .

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