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Katyanochek1 [597]
3 years ago
11

Put the following in order for the most area in the tails of the distribution. ​(a) Standard Normal Distribution ​(b) Student's​

t-Distribution with 1515 degrees of freedom. ​(c) Student's​ t-Distribution with 2020 degrees of freedom.
Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
5 0

Answer:

(b) (c) (a)

Step-by-step explanation:

Standard Normal distribution has a higher peak in the center, with more area in this región, hence it has less area in its tails.

Student's​ t-Distribution has a shape similar to the Standard Normal Distribution, with the difference that the shape depends on the degree of freedom. When the degree of freedom is smaller the distribution becomes flatter, so it has more area in its tails.

Student's​ t-Distributionwith 1515 degrees of freedom has mores area in the tails than the Student's​ t-Distribution with 2020 degrees of freedom and the latter has more area than Standard Normal Distribution

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In the past decades there have been intensive antismoking campaigns sponsored by both federal and private agencies. In one study
astra-53 [7]

Answer:

C. z = 2.05

Step-by-step explanation:

We have to calculate the test statistic for a test for the diference between proportions.

The sample 1 (year 1995), of size n1=4276 has a proportion of p1=0.384.

p_1=X_1/n_1=1642/4276=0.384

The sample 2 (year 2010), of size n2=3908 has a proportion of p2=0.3621.

p_2=X_2/n_2=1415/3908=0.3621

The difference between proportions is (p1-p2)=0.0219.

p_d=p_1-p_2=0.384-0.3621=0.0219

The pooled proportion, needed to calculate the standard error, is:

p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{1642+1415}{4276+3908}=\dfrac{3057}{8184}=0.3735

The estimated standard error of the difference between means is computed using the formula:

s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.3735*0.6265}{4276}+\dfrac{0.3735*0.6265}{3908}}\\\\\\s_{p1-p2}=\sqrt{0.00005+0.00006}=\sqrt{0.00011}=0.0107

Then, we can calculate the z-statistic as:

z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.0219-0}{0.0107}=\dfrac{0.0219}{0.0107}=2.05

z=2.05

4 0
3 years ago
1. Consider drawing a single card from a well shuffled 52-card deck. Let E1 be an event of getting a heart card and E2 be an eve
Paraphin [41]
<span>1) Find P(E1UE2)
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<span>There are 26 red cards in a 52 card deck, so the probability of choosing a red card is = 26/52 = 1/2
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2)</span></span><span>the probability of drawing a blue ball on the first draw: 4 /10
</span>the probability of drawing a white ball on the second drawn: 6/9 (because there is less one ball from the previous draw).
the probability of the cases together is 4/15 ( 4 /10 x 6/9) <span>since they are independent cases.</span>
6 0
3 years ago
What is a counterexample for the conjecture?
Dmitriy789 [7]

Answer:

A counterexample for a conjecture is the statement that disproves a conjecture.

Step-by-step explanation:

To find : What is a counterexample for the conjecture?

Solution :

A conjecture is an educated guess but not yet proven. It is possible that next example shown the conjecture wrong.

A counterexample is an example that disproves or disagree a conjecture.

For example : Prime numbers - 3,7,11,23

Conjecture  - All prime numbers are odd

Counterexample - 2

→ 2 is a prime number but not odd, it is an even number.

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