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Ipatiy [6.2K]
2 years ago
14

Researchers ask a random sample of 1,001 adults nationwide whether they favor or oppose the legalization of marijuana. Fifty-fiv

e of respondents say they oppose it. If the researchers increase the sample size of the poll by a factor of 4 (to n=4,004), which of the following effects on the length of the 95% confidence interval for the proportion is most likely to be observed?
a. It will decrease by a factor of 4
b. It will decrease by a factor of 2
c. It will increase by a factor of 2
d. It will increase by a factor of 4
Mathematics
1 answer:
natima [27]2 years ago
4 0

Answer:

b. It will decrease by a factor of 2

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence interval 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The lower end of the interval is given by:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}}

The upper end of the interval is given by:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}}

The length of the interval is the subtraction of the upper end by the lower end, so it is:

L = 2z\sqrt{\frac{\pi(1-\pi)}{n}}

This means that the length is inverse proportional to the square root of the size of the sample.

So, if the sample size is multiplied by 4, the length of the interval is going to decrease by a factor of 2.

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use the polygon tool to draw an image of the given polygon under a dilation with a scale factor of 1/3 and center of dilation at
Sveta_85 [38]

Answer:

x-y and AxR

Step-by-step explanation:

5 0
2 years ago
the probability that Jenya receives spam email is 4%.if she receives 520 email in a week about how much will be spam
wariber [46]
520*.04=20.8, or closer to 21 spam emails.
6 0
2 years ago
2.14x - 42.9 for x = 22.4 ?<br> A. -18.36<br> B. 5.026<br> C. 47.939<br> D. 90.839
Fynjy0 [20]

Answer:

Step-by-step explanation:

Your answer would be B., because you would multiply

2.14 x  22.4   =

47.939  

then you will subtract that by 42.9

47.939 - 42.9 =  

5.026

7 0
2 years ago
Which describes money that is received from sales of goods or services?
Lana71 [14]

the answer would be revenue

7 0
2 years ago
A construction company is considering submitting bids for contracts of three different projects. The company estimates that it h
julsineya [31]

Answer:

a.P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\

b. E(x) = 0.3

c. S(x)=0.5196

d. E=5,000

Step-by-step explanation:

The probability that the company won x bids follows a binomial distribution because we have n identical and independent experiments with a probability p of success and (1-p) of fail.

So, the PMF of X is equal to:

P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\

Where p is 0.1 and it is the chance of winning. Additionally, n is 3 and it is the number of bids. So the PMF of X is:

P(x)=\frac{3!}{x!(3-x)!}*0.1^{x}*(1-0.1)^{n-x}\\

For binomial distribution:

E(x)=np\\S(x)=\sqrt{np(1-p)}

Therefore, the company can expect to win 0.3 bids and it is calculated as:

E(x) = np = 3*0.1 = 0.3

Additionally, the standard deviation of the number of bids won is:

S(x)=\sqrt{np(1-p)}=\sqrt{3(0.1)(1-0.1)}=0.5196

Finally, the probability to won 1, 2 or 3 bids is equal to:

P(1)=\frac{3!}{1!(3-1)!}*0.1^{1}*(1-0.1)^{3-1}=0.243\\P(2)=\frac{3!}{2!(3-2)!}*0.1^{2}*(1-0.1)^{3-2}=0.027\\P(3)=\frac{3!}{3!(3-3)!}*0.1^{3}*(1-0.1)^{3-3}=0.001

So, the expected profit for the company is equal to:

E=-10,000+50,000(0.243)+100,000(0.027)+150,000(0.001)\\E=5,000

Because there is a probability of 0.243 to win one bid and it will produce 50,000 of income, there is a probability of 0.027 to win 2 bids and it will produce 100,000 of income and there is a probability of 0.001 to win 3 bids and it will produce 150,000 of income.

5 0
2 years ago
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