Question

A poll in 2017 reported that 705 out of 1023 adults in a certain country believe that marijuana should be legalized. When this poll about the same subject was first conducted in​ 1969, only​ 12% of the adults of the country supported legalization. Assume the conditions for using the CLT are met. Complete question 1 through​3 below.
1) Find and interpret a 99​% confidence interval for the proportion of adults in the country in 2017 that believe marijuana should be legalized.

Select one:

a. (0.661, 0.717)
b. (0.096, 0.143)
c. (0.094, 0.146)
d. (0.652, 0.726)

2)Find and interpret a 95​% confidence interval for this population parameter.

Select one:

a. (0.096, 0.143)
b. (0.661, 0.718)
c. (0.1, 0.139)
d. (0.652, 0.726)

3) Without computing​ it, how would the margin of error of a 90​% confidence interval compare with the margin of error for the 95​% and 99​% ​intervals? Construct the 90​% confidence interval to see if your prediction was correct. How would a 90​% interval compare with the others in the margin of​ error?

Select one:

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.
b. 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 narrower with increasing confidence level.
c. The margin of error of a 90​% confidence interval will be more than the margin of error for the 95​% and 99​% confidence intervals because intervals get wider with increasing confidence level.
d. The margin of error of a 90​% confidence interval will be greater than the margin of error for the 95​% confidence interval and less than the margin of error for the 99​% confidence interval because intervals get wider with increasing confidence level.

Answer 1

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.