# 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

Question
Confidence intervals
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?

2021-02-17
Given Information:
Sample size $$(n) = 1027$$
$$X = 699$$
Sample proportion is given by the formula:
$$(\widehat{p}) = \frac{X}{n}$$
$$= \frac{699}{1027}$$
$$\approx 0.6806232$$
(a) To construct $$99\%$$ confidence interval:
Confidence level $$= 0.99$$
Significance level $$\alpha=1−0.99=0.01$$
Confidence interval for proportion is obtained using the formula:
$$C.I = \widehat{p} z_{a/2}\frac{\sqrt{\widehat{p}(1-\widehat{p})}}{n}$$
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:
$$C.I=0.6806232 \pm 2.576 \frac{\sqrt{0.6806232(1−0.6806232)}}{1027}$$
$$= 0.6806232 \pm 2.576 \times 0.01455$$
$$= (0.6806232 − 0.037481, 0.6806232 + 0.037481)$$
$$= (0.643, 0.718)$$
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:
$$C.I=\widehat{p}z_{a/2}\frac{\sqrt{\widehat{p}(1-\widehat{p})}}{n}$$
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:
$$C.I = 0.6806232 \pm 1.645 \frac{\sqrt{0.6806232(1−0.6806232)}}{1027}$$
$$= 0.6806232 \pm 1.645 \times 0.01455$$
$$= (0.6806232 − 0.023935, 0.6806232 + 0.023935)$$
$$= (0.657, 0.705)$$
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:
$$M.E = z_{a/2}\frac{\sqrt{\widehat{p}(1-\widehat{p})}}{n}$$
Margin of error of 99% confidence interval is:
$$M.E=2.576\frac{\sqrt{0.6806231743(1−0.6806231743)}}{1027}$$
$$= 2.576 \times 0.01455$$
$$= 0.037481$$
$$\approx 0.037$$
Thus, margin of error for 99% confidence interval is 0.037
Margin of error of 90% confidence interval is:
$$M.E=1.645\frac{\sqrt{0.6806231743(1−0.6806231743)}}{1027}$$
$$= 1.645 \times 0.01455$$
$$= 0.023935 \approx 0.024$$
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:
$$C.I = \widehat{p} z_{a/2}\frac{\sqrt{\widehat{p}(1-\widehat{p})}}{n}$$
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:
$$C.I = 0.6806232 \pm 1.2816 \frac{\sqrt{0.6806232(1−0.6806232)}}{1027}$$
$$= 0.6806232 \pm 1.2816 \times 0.01455$$
$$= (0.6806232 − 0.0186473, 0.6806232 + 0.0186473)$$
$$= (0.662, 0.699)$$ 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.

### Relevant Questions

1. A researcher is interested in finding a 98% confidence interval for the mean number of times per day that college students text. The study included 144 students who averaged 44.7 texts per day. The standard deviation was 16.5 texts. a. To compute the confidence interval use a ? z t distribution. b. With 98% confidence the population mean number of texts per day is between and texts. c. If many groups of 144 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population number of texts per day and about percent will not contain the true population mean number of texts per day. 2. You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you believe the population standard deviation is approximately $$\displaystyle\sigma={40.4}$$ dollars. You would like to be 90% confident that your estimate is within 1.5 dollar(s) of average spending on the birthday parties. How many parents do you have to sample? n = 3. You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately $$\displaystyle\sigma={57.5}$$. You would like to be 95% confident that your estimate is within 0.1 of the true population mean. How large of a sample size is required?
The table below shows the number of people for three different race groups who were shot by police that were either armed or unarmed. These values are very close to the exact numbers. They have been changed slightly for each student to get a unique problem.
Suspect was Armed:
Black - 543
White - 1176
Hispanic - 378
Total - 2097
Suspect was unarmed:
Black - 60
White - 67
Hispanic - 38
Total - 165
Total:
Black - 603
White - 1243
Hispanic - 416
Total - 2262
Give your answer as a decimal to at least three decimal places.
a) What percent are Black?
b) What percent are Unarmed?
c) In order for two variables to be Independent of each other, the P $$(A and B) = P(A) \cdot P(B) P(A and B) = P(A) \cdot P(B).$$
This just means that the percentage of times that both things happen equals the individual percentages multiplied together (Only if they are Independent of each other).
Therefore, if a person's race is independent of whether they were killed being unarmed then the percentage of black people that are killed while being unarmed should equal the percentage of blacks times the percentage of Unarmed. Let's check this. Multiply your answer to part a (percentage of blacks) by your answer to part b (percentage of unarmed).
Remember, the previous answer is only correct if the variables are Independent.
d) Now let's get the real percent that are Black and Unarmed by using the table?
If answer c is "significantly different" than answer d, then that means that there could be a different percentage of unarmed people being shot based on race. We will check this out later in the course.
Let's compare the percentage of unarmed shot for each race.
e) What percent are White and Unarmed?
f) What percent are Hispanic and Unarmed?
If you compare answers d, e and f it shows the highest percentage of unarmed people being shot is most likely white.
Why is that?
This is because there are more white people in the United States than any other race and therefore there are likely to be more white people in the table. Since there are more white people in the table, there most likely would be more white and unarmed people shot by police than any other race. This pulls the percentage of white and unarmed up. In addition, there most likely would be more white and armed shot by police. All the percentages for white people would be higher, because there are more white people. For example, the table contains very few Hispanic people, and the percentage of people in the table that were Hispanic and unarmed is the lowest percentage.
Think of it this way. If you went to a college that was 90% female and 10% male, then females would most likely have the highest percentage of A grades. They would also most likely have the highest percentage of B, C, D and F grades
The correct way to compare is "conditional probability". Conditional probability is getting the probability of something happening, given we are dealing with just the people in a particular group.
g) What percent of blacks shot and killed by police were unarmed?
h) What percent of whites shot and killed by police were unarmed?
i) What percent of Hispanics shot and killed by police were unarmed?
You can see by the answers to part g and h, that the percentage of blacks that were unarmed and killed by police is approximately twice that of whites that were unarmed and killed by police.
j) Why do you believe this is happening?
Do a search on the internet for reasons why blacks are more likely to be killed by police. Read a few articles on the topic. Write your response using the articles as references. Give the websites used in your response. Your answer should be several sentences long with at least one website listed. This part of this problem will be graded after the due date.
The bulk density of soil is defined as the mass of dry solidsper unit bulk volume. A high bulk density implies a compact soilwith few pores. Bulk density is an important factor in influencing root development, seedling emergence, and aeration. Let X denotethe bulk density of Pima clay loam. Studies show that X is normally distributed with $$\displaystyle\mu={1.5}$$ and $$\displaystyle\sigma={0.2}\frac{{g}}{{c}}{m}^{{3}}$$.
(a) What is thedensity for X? Sketch a graph of the density function. Indicate onthis graph the probability that X lies between 1.1 and 1.9. Findthis probability.
(b) Find the probability that arandomly selected sample of Pima clay loam will have bulk densityless than $$\displaystyle{0.9}\frac{{g}}{{c}}{m}^{{3}}$$.
(c) Would you be surprised if a randomly selected sample of this type of soil has a bulkdensity in excess of $$\displaystyle{2.0}\frac{{g}}{{c}}{m}^{{3}}$$? Explain, based on theprobability of this occurring.
(d) What point has the property that only 10% of the soil samples have bulk density this high orhigher?
(e) What is the moment generating function for X?
A psychologist is interested in constructing a $$99\%$$ confidence interval for the proportion of people who accept the theory that a person's spirit is no more than the complicated network of neurons in the brain. 68 of the 702 randomly selected people who were surveyed agreed with this theory. Round answers to 4 decimal places where possible.
a)
With $$99\%$$ confidence the proportion of all people who accept the theory that a person's spirit is no more than the complicated network of neurons in the brain is between ____ and ____.
b)
If many groups of 702 randomly selected people are surveyed, then a different confidence interval would be produced from each group. About ____ percent of these confidence intervals will contain the true population proportion of all people who accept the theory that a person’s spirit is no more than the complicated network of neurons in the brain and about ____ percent will not contain the true population proportion.
You may need to use the appropriate appendix table or technology to answer this question.
Money reports that the average annual cost of the first year of owning and caring for a large dog in 2017 is $1,448. The Irish Red and White Setter Association of America has requested a study to estimate the annual first-year cost for owners of this breed. A sample of 50 will be used. Based on past studies, the population standard deviation is assumed known with $$\displaystyle\sigma=\{230}.$$ $$\begin{matrix} 1,902 & 2,042 & 1,936 & 1,817 & 1,504 & 1,572 & 1,532 & 1,907 & 1,882 & 2,153 \\ 1,945 & 1,335 & 2,006 & 1,516 & 1,839 & 1,739 & 1,456 & 1,958 & 1,934 & 2,094 \\ 1,739 & 1,434 & 1,667 & 1,679 & 1,736 & 1,670 & 1,770 & 2,052 & 1,379 & 1,939\\ 1,854 & 1,913 & 2,163 & 1,737 & 1,888 & 1,737 & 2,230 & 2,131 & 1,813 & 2,118\\ 1,978 & 2,166 & 1,482 & 1,700 & 1,679 & 2,060 & 1,683 & 1,850 & 2,232 & 2,294 \end{matrix}$$ (a) What is the margin of error for a $$95\%$$ confidence interval of the mean cost in dollars of the first year of owning and caring for this breed? (Round your answer to nearest cent.) (b) The DATAfile Setters contains data collected from fifty owners of Irish Setters on the cost of the first year of owning and caring for their dogs. Use this data set to compute the sample mean. Using this sample, what is the $$95\%$$ confidence interval for the mean cost in dollars of the first year of owning and caring for an Irish Red and White Setter? (Round your answers to nearest cent.)$_______ to \$________
You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 58 ​dates, the mean record high daily temperature in a certain city has a mean of $$\displaystyle{83.43}^{{\circ}}{F}$$. Assume the population standard deviation is $$\displaystyle{14.02}^{{\circ}}{F}$$. $$\displaystyle{90}\%=$$ $$\displaystyle{95}\%=$$ Which interval is wider?
You are given the sample mean and standard deviation of the population. Use this information to construct the​ $$\displaystyle{90}\%{\quad\text{and}\quad}​{95}\%$$ confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.
From a random sample of 66 ​dates, the mean record high daily temperature in a certain city has a mean of $$\displaystyle{85.69}^{\circ}{F}$$. Assume the population standard deviation is $$\displaystyle{13.60}^{\circ}{F}.$$
The​ $$90\%$$ confidence interval is
The​ $$95\%$$ confidence interval is
Which interval is​ wider?
Interpret the results.
The owner of a large equipment rental company wants to make a rather quick estimate of the average number of days a piece of ditch-digging equipment is rented out per person per time. The company has records of all rentals, but the amount of time required to conduct an audit of all accounts would be prohibitive. The owner decides to take a random sample of rental invoices. Fourteen different rentals of ditch-diggers are selected randomly from the files, yielding the following data. She wants to use these data to construct a $$99\%$$ confidence interval to estimate the average number of days that a ditch-digger is rented and assumes that the number of days per rental is normally distributed in the population. Use only the appropriate formula and/or statistical table in your textbook to answer this question. Report your answer to 2 decimal places, using conventional rounding rules.
The presidential election is coming. Five survey companies (A, B, C, D, and E) are doing survey to forecast whether or not the Republican candidate will win the election. Each company randomly selects a sample size between 1000 and 1500 people. All of these five companies interview people over the phone during Tuesday and Wednesday. The interviewee will be asked if he or she is 18 years old or above and U.S. citizen who are registered to vote. If yes, the interviewee will be further asked: will you vote for the Republican candidate? On Thursday morning, these five companies announce their survey sample and results at the same time on the newspapers. The results show that a% (from A), b% (from B), c% (from C), d% (from D), and e% (from E) will support the Republican candidate. The margin of error is plus/minus 3% for all results. Suppose that $$\displaystyle{c}{>}{a}{>}{d}{>}{e}{>}{b}$$. When you see these results from the newspapers, can you exactly identify which result(s) is (are) not reliable and not accurate? That is, can you identify which estimation interval(s) does (do) not include the true population proportion? If you can, explain why you can, if no, explain why you cannot and what information you need to identify. Discuss and explain your reasons. You must provide your statistical analysis and reasons.
We wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled, 192 had kids. Based on this, plot a $$99\%$$ confidence interval for the proportion of adult residents who are parents in a given county.
Express your answer in the form of three inequalities. Give your answers in decimal fractions up to three places $$\displaystyle<{p}<$$ Express the same answer using a point estimate and a margin of error. Give your answers as decimals, to three places.
$$\displaystyle{p}=\pm\pm$$