The point (0,0) is a solution to which of these inequalities?

A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben

juin [17]

Answer:

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

5 0

3 years ago

Solve the system using the substituiton method<br> branliest offered<br> 3x-7y=10<br> x-4y=5

fiasKO [112]

Hi there!

To solve this problem using substitution, we need to set x equal to a value and substitute it into the other equation.

Since
x - 4y = 5,
x = 5 + 4y

3x - 7y = 10
3(5 + 4y) - 7y = 10
15 + 12y - 7y = 10
15 + 5y = 10
5y = -5
y = -1

Now that we know y is -1, we can substitute it back into the equation to find x:

y = -1
x - 4y = 5
x - 4(-1) = 5
x - (-4) = 5
x + 4 = 5
x = 1

So, your answers are x = 1 and y = -1.

Hope this helps!

6 0

2 years ago

Recently, Frank took a one-hundred question aptitude test where each correct answer scored 5 points, each incorrect answer score

Lyrx [107]

Im guessing he answered 30 correct, let me know if im correct

3 0

3 years ago

Read 2 more answers

Rename 4,180 as hundreds and tens

My name is Ann [436]

Four thousand one hundred eighty

6 0