PLZZZZZ help me with this problem

A poll in 2017 reported that 705 out of 1023 adults in a certain country believe that marijuana should be legalized. When this p

just olya [345]

Answer:

1. d. (0.652, 0.726)

2. b. (0.661, 0.718)

a. The margin of error of a 90% confidence interval will be less than the margin of error for the 95% and 99% confidence intervals because intervals get wider with increasing confidence level.

Step-by-step explanation:

Data given and notation

n=1023 represent the random sample taken in 2017

X=705 represent the people who thinks that believe that marijuana should be legalized.

$\hat p =\frac{705}{1023}=0.689$ estimated proportion of people who thinks that believe that marijuana should be legalized.

z would represent the statistic in order to find the confidence interval

p= population proportion of people who thinks that believe that marijuana should be legalized.

Part 1

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 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.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

0.689 -2.58 sqrt((0.689(1-0.689))/(1023))=0.652

0.689 + 2.58sqrt((0.56(1-0.689))/{1023))=0.726

And the 99% confidence interval would be given (0.652;0.726).

We are 99% confident that about 65.2% to 72.6% of people believe that marijuana should be legalized

d. (0.652, 0.726)

Part 2

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.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

0.689 - 1.96((0.689(1-0.689))/(1023))=0.661

0.689 + 1.96 ((frac{0.56(1-0.689))/(1023))=0.718

And the 95% confidence interval would be given (0.661;0.718).

We are 95% confident that about 66.1% to 71.8% of people believe that marijuana should be legalized

b. (0.661, 0.718)

Part 3

Would be lower since the quantile $z_{\alpha/2}$ for a lower confidence is lower than a quantile for a higher confidence level.

The margin of error of a 90% confidence interval will be less than the margin of error for the 95% and 99% confidence intervals

If we calculate the 90% interval we got:

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution. The quantile for this case would be 1.64.

And replacing into the confidence interval formula we got:

0.689 - 1.64 ((0.689(1-0.689))/{1023))=0.665

0.689 + 1.64 ((0.56(1-0.689))/(1023))=0.713

And the 90% confidence interval would be given (0.665;0.713).

We are 90% confident that about 66.5% to 71.3% of people believe that marijuana should be legalized.

The intervals get wider with increasing confidence level.

So the correct answer is:

a. The margin of error of a 90% confidence interval will be less than the margin of error for the 95% and 99% confidence intervals because intervals get wider with increasing confidence level.

6 0

3 years ago

PLEASE HELP ASAP <br>are f(x)=4/x-2-1 and g(x)=3/x+2-2 inverses of each other?

GREYUIT [131]

Answer:

To find the inverse of:

$f (x)=\dfrac{4}{x-2}-1$