NEED HELP ASAP, WILL GIVE BRAINLIEST

It seems to you that fewer than half of people who are registered voters in the City of Madison do in fact vote when there is an

Ad libitum [116K]

Answer:

a) Going to public places like restaurants, parks, theaters, etc in Madison and asking voters.

b) The 80% CI to estimate the true proportion of registered voters in the City of Madison who vote in non-presidential elections is (0.5533, 0.6667).

c) We are 80% sure that our confidence interval contains the true proportion of registered voters in the City of Madison who vote in non-presidential elections.

d) The lower limit of the interval is higher than 0.5. This means that it does seem that MORE than half of registered voters in the City of Madison vote in non-presidential elections.

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 $1 - \frac{\alpha}{2}$ .

(a) How might a simple random sample have been gathered?

Going to public places like restaurants, parks, theaters, etc in Madison and asking voters.

(b) Construct an 80% CI to estimate the true proportion of registered voters in the City of Madison who vote in non-presidential elections.

You take an SRS of 200 registered voters in the City of Madison, and discover that 122 of them voted in the last non-presidential election. This means that $n = 200, \pi = \frac{122}{200} = 0.61$ .

We want to build an 80% CI, so $\alpha = 0.20$ , z is the value of Z that has a pvalue of $1 - \frac{0.20}{2} = 0.90[tex], so [tex]z = 1.645$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{200}} = 0.61 - 1.645\sqrt{\frac{0.61*0.39}{200}} = 0.5533$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{200}} = 0.61 + 1.645\sqrt{\frac{0.61*0.39}{200}} = 0.6667$

The 80% CI to estimate the true proportion of registered voters in the City of Madison who vote in non-presidential elections is (0.5533, 0.6667).

(c) Interpret the interval you created in part (b).

We are 80% sure that our confidence interval contains the true proportion of registered voters in the City of Madison who vote in non-presidential elections.

(d) Based on your CI, does it seem that fewer than half of registered voters in the City of Madison vote in non-presidential elections? Explain.

The lower limit of the interval is higher than 0.5. This means that it does seem that MORE than half of registered voters in the City of Madison vote in non-presidential elections.

4 0

2 years ago

Solve for x: 2 over 3 equals the quantity x minus 1 end quantity over 5

sweet-ann [11.9K]

Answer:

13 over 3

Step-by-step explanation:

Hi Jakeyriabryant! I hope you’re fine!

I hope I have understood the problem well.

If so, what the exercise raises is the following equality:

(x-1) / 5 = 2/3

From this equation you must clear the "x".

First, we pass the 5 that is dividing on the side of the x, to the other side and passes multiplying

(X – 1) / 5 = 2/3

(X – 1) = (2/3)*5

X – 1 = 10/3

Then we pass the one that is subtracting from the side of the x, to the other side and passes adding

X = 10/3 + 1

Remember that to add or subtract fractions they must have the same denominator or a common denominator (in this case we can write 1 as fraction 3/3). Then,

X = 10/3 + 3/3

X = 13/3

I hope I've been helpful!

Regards!

7 0

3 years ago

A certain town never has two sunny days in a row. Each day is classified as being either sunny, cloudy (but dry), or rainy. If i

11111nata11111 [884]

Answer:

the proportion of days that are Sunny is 0.2

Step-by-step explanation:

Given the data in the question;

Using markov chain;

3 states; Sunny(1), Cloudy(2) and Rainy(3)

Now, based on given conditions, the transition matrix can be obtained in the following way;

$\left[\begin{array}{ccc}0&0.5&0.5\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]$

so let the proportion of sunny, cloudy and rainy days be S, C and R respectively.

such that, from column 1

S = 0.25C + 0.25R -------------let this be equation 1

from column 2

0.5C = 0.5S + 0.25R

divided through by 0.5

C = S + 0.5R ---------------------- let this be equation 2

now putting equation 2 into equation;

S = 0.25(S + 0.5R) + 0.25R

S = 0.25S + 0.125R + 0.25R

S - 0.25S = 0.375R

0.75S = 0.375R

S = 0.375R / 0.75

S = 0.5R

Therefore,

from equation 2; C = S + 0.5R

input S = 0.5R

C = 0.5R + 0.5R

C = R

Now, we know that, the sum of the three proportion should be equal to one;

so

S + C + R = 1

since C = R and S = 0.5R

we substitute

0.5R + R + R = 1

2.5R = 1

R = 1/2.5

R = 0.4

Hence, the proportion of days that are Rainy is 0.4

C = R

C = 0.4

Hence, the proportion of days that are Cloudy is 0.4

S = 0.5R

S = 0.5(0.4)

S = 0.2

Hence, the proportion of days that are Sunny is 0.2

8 0

3 years ago

A gold, a silver, and a bronze medal are awarded in an Olympic event. In how many possible ways can the medals be awarded for a

RideAnS [48]

Answer:

720 possible ways

Step-by-step explanation:

The gold is awarded to the first position, the silver is awarded to the second position while the bronze is awarded to the third position.

The first position can be taken by any of the 10 runners

Now, the second position can be taken by remaining 9 runners

while the third position can be taken by the renaming 8 runners.

Thus, the number of ways in which these medals can be awarded = 10 * 9 * 8 = 720 ways

3 0

3 years ago

#14

Delicious77 [7]

Answer:

The answer is equivalent

5 0

2 years ago