# A poll in 2017 reported that 699 out of 1027 adults in a certain country believe that marijuana should be legalized. When this poll about the subject

Globokim8 2021-02-16 Answered
A poll in 2017 reported that 699 out of 1027 adults in a certain country believe that marijuana should be legalized. When this poll about the subject was first conducted in 1969, only 12% of rhe country supported legalizztion. Assume the conditions for using the CLT are met.
a) Find and interpet a 99% confidence interval for the proportion of adults in the country 2017 that believe marijuana should be legalized is (0.643, 0.718)
b) Find and interpret a 90%confidence interval for this population parameter. The 90% confidence interval for the proportion of adults in the country 2017 that believe marijuana should be legalized is (0.657, 0.705)
c)Find the margin of error for each of the confidence intervals found The margin of error of the 99% confidence interval is 0.039 and the margin of error of the 90% confidence interval is 0.025
d) Without computing it, how would the margin of error of an 80% confidence interval compare with the margin of error for 90% and 99% intervals? Construct the 80% confidence interval to see if your production was correct
How would a 80% interval compare with the others in the margin of error?
You can still ask an expert for help

Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it

## Expert Answer

Willie
Answered 2021-02-17 Author has 95 answers
Given Information:
Sample size $\left(n\right)=1027$
$X=699$
Sample proportion is given by the formula:
$\left(\stackrel{^}{p}\right)=\frac{X}{n}$
$=\frac{699}{1027}$
$\approx 0.6806232$
(a) To construct $99\mathrm{%}$ confidence interval:
Confidence level $=0.99$
Significance level $\alpha =1-0.99=0.01$
Confidence interval for proportion is obtained using the formula:
$C.I=\stackrel{^}{p}{z}_{a/2}\frac{\sqrt{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}}{n}$
Using a standard normal table, Z-critical value at $\frac{0.01}{2}=0.005$ significance level is 2.576
Substitute the values in the formula:
$C.I=0.6806232±2.576\frac{\sqrt{0.6806232\left(1-0.6806232\right)}}{1027}$
$=0.6806232±2.576×0.01455$
$=\left(0.6806232-0.037481,0.6806232+0.037481\right)$
$=\left(0.643,0.718\right)$
Thus, we are 99% confident that the population proportion of adults in the country in 2017 who believe marijuana should be legalized is between (0.643, 0.718)
(b) To construct 90% confidence interval:
Confidence level $=0.90$
Significance level $\alpha =1-0.90=0.10$
Confidence interval for proportion is obtained using the formula:
$C.I=\stackrel{^}{p}{z}_{a/2}\frac{\sqrt{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}}{n}$
Using a standard normal table, Z-critical value at $\frac{0.10}{2}=0.05$ significance level is 1.645
Substitute the values in the formula:
$C.I=0.6806232±1.645\frac{\sqrt{0.6806232\left(1-0.6806232\right)}}{1027}$
$=0.6806232±1.645×0.01455$
$=\left(0.6806232-0.023935,0.6806232+0.023935\right)$
$=\left(0.657,0.705\right)$
Thus, we are 90% confident that the population proportion of adults in the country in 2017 who believe marijuana should be legalized is between (0.657, 0.705)
(c) To find the margin of error for each of the confidence Intervals:
Margin of error is given by the formula:
$M.E={z}_{a/2}\frac{\sqrt{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}}{n}$
Margin of error of 99% confidence interval is:
$M.E=2.576\frac{\sqrt{0.6806231743\left(1-0.6806231743\right)}}{1027}$
$=2.576×0.01455$
$=0.037481$
$\approx 0.037$
Thus, margin of error for 99% confidence interval is 0.037
Margin of error of 90% confidence interval is:
$M.E=1.645\frac{\sqrt{0.6806231743\left(1-0.6806231743\right)}}{1027}$
$=1.645×0.01455$
$=0.023935\approx 0.024$
Thus, margin of error for 99% confidence interval is 0.024
(d) The margin of error of 80% confidence interval will be less than the margin of error of 90% and 99% confidence intervals because intervals get wider with an increasing confidence level.
To construct 80% confidence interval:
Confidence level $=0.80$
Significance level $\alpha =1-0.80=0.20$
Confidence interval for proportion is obtained using the formula:
$C.I=\stackrel{^}{p}{z}_{a/2}\frac{\sqrt{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}}{n}$
Using a standard normal table, Z-critical value at $\frac{0.20}{2}=0.10$ significance level is 1.2816
Substitute the values in the formula:
$C.I=0.6806232±1.2816\frac{\sqrt{0.6806232\left(1-0.6806232\right)}}{1027}$
$=0.6806232±1.2816×0.01455$
$=\left(0.6806232-0.0186473,0.6806232+0.0186473\right)$
$=\left(0.662,0.699\right)$ Thus, we are 80% confident that the population proportion of adults in the country in 2017 who believe marijuana should be legalized is between (0.662, 0.699)
The width of 80% confidence interval is narrower when compared to the confidence intervals of 99% and 90% levels.
###### Not exactly what you’re looking for?

Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it