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elena-s [515]
4 years ago
9

u suspect that the true proportion of teens who smoke is different than 15%. Conduct a hypothesis test to test your theory. Use

a level of significance of LaTeX: \alpha = 0.05α = 0.05. Calculate the point estimator LaTeX: \hat{p}p ^, the test statistic, and the P-value for this test.
Mathematics
1 answer:
pentagon [3]4 years ago
8 0

Answer:

There is enough statistical evidence that the proportion of teenagers that smoke is different from 15%.

Step-by-step explanation:

The question is incomplete:

<em>"Gallup survey interviewed a nationally representative sample of 785 teenagers aged 13 to 17. Seventy-one teenagers in the survey acknowledged having smoked at least once in the past week."</em>

We have this null and alternative hypothesis:

H_0:p=0.15\\\\H_a: p\neq 0.15

The significance level is

\alpha=0.05

The sample proportion is:

\hat{p}=X/n=71/785=0.09

The standard deviation is

\sigma=\sqrt{\frac{p(1-p)}{n}} =\sqrt{\frac{0.15*0.85}{785}}=0.013

The z-statistic can be calculated as:

z=\frac{\hat{p}-p+0.5/n}{\sigma} =\frac{0.09-0.15+0.00}{0.013}=\frac{-0.06}{0.013}= -4.61

The P-value of z=-4.61 for this test is P=0

2P(|z|>-4.61)=0

The P-value is smaller than the significance level, so the effect is significant.

Then, the null hypothesis is rejected.

There is enough statistical evidence that the proportion of teenagers that smoke is different from 15%.

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You have decided to purchase a car for $19,000. The credit union requires a 10% down payment and will finance the balance with a
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<h2>○=> <u>Correct option</u> :</h2>

\color{plum} \bold{ \tt \:  \bold{C.  \$20,652.00}}

<h3>○=> <u>Steps to derive correct option</u> :</h3>

Selling price of a car = $19,000

Percentage of sales tax in the city = 8.3%

Sales tax :

= \tt  \frac{8.3}{10 \times 100}  \times 1900

= \tt  \frac{8.3 \times 19000}{1000}

= \tt  \frac{1577000}{1000}

\color{plum} = \tt  \$1577

Thus, the sales tax on the car = $1577

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\color{plum} = \tt  \$20652

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Assume ​Y=1​+X+u​, where X​, Y​, and ​u=v+X are random​ variables, v is independent of X​; ​E(v​)=0, ​Var(v​)=1​, ​E(X​)=1, and
kobusy [5.1K]

Answer:

a) E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1+

b) E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2

c) E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2

d) E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6

e) E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1

f) E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3

g) E(u) = E(v) +E(X) = 0+1=1

h) E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

Step-by-step explanation:

For this case we know this:

Y = 1+X +u

u = v+X

with both Y and u random variables, we also know that:

[tex] E(v) = 0, Var(v) =1, E(X) = 1, Var(X)=2

And we want to calculate this:

Part a

E(u|X=1)= E(v+X|X=1)

Using properties for the conditional expected value we have this:

E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1

Because we assume that v and X are independent

Part b

E(Y| X=1) = E(1+X+u|X=1)

If we distribute the expected value we got:

E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2

Part c

E(u|X=2)= E(v+X|X=2)

Using properties for the conditional expected value we have this:

E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2

Because we assume that v and X are independent

Part d

E(Y| X=2) = E(1+X+u|X=2)

If we distribute the expected value we got:

E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6

Part e

E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1

Part f

E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3

Part g

E(u) = E(v) +E(X) = 0+1=1

Part h

E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

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