Help??? I Don’t Understand

HELP THE TEACHER IS ASKING FOR MY ANSWERS CUZ I WAS SUPPOSED TO DO IT AND I NEVER WAS PREPARED PLSSS HELP IM NEXT PLS DRAW IT FO

tatuchka [14]

Step-by-step explanation:

1 is the least and 3 is the greatest

3 0

2 years ago

Ira described the data on a dot plot as having a cluster from 0 to 3, a gap at 4, a peak at 1, and being skewed right. Which dot

rjkz [21]

Answer:

it would be a

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

*PLEASE HELP WILL GIVE BRAINLIEST*

Likurg_2 [28]

The second one i think

7 0

3 years ago

Read 2 more answers

If gas costs $2.19 per gallon, how many whole gallons of gas could you buy with $17.00? (Make sure to answer with a whole number

ankoles [38]

Answer:

2.19*17.00

=37.23

=$37

Step-by-step explanation:

4 0

2 years ago

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

3 years ago