Ask question

# 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 th

Question
Confidence intervals
asked 2021-02-25
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.

## Answers (1)

2021-02-26
Step 1
Solution:
Let X be the number of people agreed with the theory and n be the sample number of people.
From the given information, $$\displaystyle{X}={68}{\quad\text{and}\quad}{n}={702}.$$
Step 2
The sample proportion is
$$\displaystyle\hat{{p}}=\frac{X}{{n}}$$
$$\displaystyle=\frac{68}{{702}}$$
$$\displaystyle={0.0969}$$
Step 3
The confidence level is 0.99.
$$\displaystyle{1}-\alpha={0.99}$$
$$\displaystyle\alpha={1}-{0.99}$$
$$\displaystyle\alpha={0.01}$$
Then, the $$99\%$$ confidence interval for 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
$$\displaystyle\hat{{p}}\pm{Z}_{{\frac{\alpha}{{2}}}}{\left(\sqrt{{\frac{{\hat{{p}}{\left({1}-\hat{{p}}\right)}}}{{n}}}}\right)}={0.0969}\pm{Z}_{{\frac{0.01}{{2}}}}{\left(\frac{\sqrt{{{0.0969}{\left({1}-{0.0969}\right)}}}}{{702}}\right)}$$
$$\displaystyle{0.0969}\pm{Z}_{{{0.005}}}{\left(\frac{\sqrt{{{0.0969}{\left({1}-{0.0969}\right)}}}}{{702}}\right)}$$
$$\displaystyle{0.0969}\pm{\left({2.58}\right)}{\left(\frac{\sqrt{{{0.0969}{\left({1}-{0.0969}\right)}}}}{{702}}\right)}$$ [using the excel function = NORM.INV(0.005, 0, 1)]
$$\displaystyle={0.0969}\pm{0.0288}$$
$$\displaystyle={\left({0.0969}-{0.0288},{0.0969}+{0.0288}\right)}$$
$$\displaystyle={\left({0.0681},{0.1257}\right)}$$
Thus, 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 0.0681 and 0.1257.
Step 4
b)
If many groups of 702 randomly selected people are surveyed, then a different confidence interval would be produced from each group. About 99 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 $$\displaystyle{100}-{99}={1}$$ percent will not contain the true population proportion.

### Relevant Questions

asked 2020-11-16
A newsgroup is interested in constructing a $$90\%$$ confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 555 randomly selected Americans surveyed, 437 were in favor of the initiative. Round answers to 4 decimal places where possible.
a)
With $$90\%$$ confidence the proportion of all Americans who favor the new Green initiative is between ____ and ____.
b)
If many groups of 555 randomly selected Americans were 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 Americans who favor the Green initiative and about ____ percent will not contain the true population proportion.
asked 2020-11-09
A researcher is interested in finding a $$90\%$$ confidence interval for the mean number minutes students are concentrating on their professor during a one hour statistics lecture. The study included 117 students who averaged 40.9 minutes concentrating on their professor during the hour lecture. The standard deviation was 11.8 minutes. Round answers to 3 decimal places where possible.
a.
To compute the confidence interval use a ? distribution.
b.
With $$90\%$$ confidence the population mean minutes of concentration is between ____ and ____ minutes.
c.
If many groups of 117 randomly selected students are studied, then a different confidence interval would be produced from each group. About ____ percent of these confidence intervals will contain the true population mean minutes of concentration and about ____ percent will not contain the true population mean number of minutes of concentration.
asked 2020-12-30
You are interested in finding a 95% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 14 randomly selected physical therapy patients. Round answers to 3 decimal places where possible.
$$9 6 10 15 19 6 23 26 19 16 11 25 16 11$$
a. To compute the confidence interval use a t or z distribution.
b. With 95% confidence the population mean number of visits per physical therapy patient is between ___ and ___ visits.
c. If many groups of 14 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About ___ percent of these confidence intervals will contain the true population mean number of visits per patient and about ___ percent will not contain the true population mean number of visits per patient.
asked 2021-02-23
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?
asked 2020-10-23
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.
asked 2020-12-29
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.
asked 2020-11-01
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.
DATA: 3 1 3 2 5 1 2 1 4 2 1 3 1 1
asked 2021-02-25
Give a full and correct answer Why is it important that a sample be random and representative when conducting hypothesis testing? Representative Sample vs. Random Sample: An Overview Economists and researchers seek to reduce sampling bias to near negligible levels when employing statistical analysis. Three basic characteristics in a sample reduce the chances of sampling bias and allow economists to make more confident inferences about a general population from the results obtained from the sample analysis or study: * Such samples must be representative of the chosen population studied. * They must be randomly chosen, meaning that each member of the larger population has an equal chance of being chosen. * They must be large enough so as not to skew the results. The optimal size of the sample group depends on the precise degree of confidence required for making an inference. Representative sampling and random sampling are two techniques used to help ensure data is free of bias. These sampling techniques are not mutually exclusive and, in fact, they are often used in tandem to reduce the degree of sampling error in an analysis and allow for greater confidence in making statistical inferences from the sample in regard to the larger group. Representative Sample A representative sample is a group or set chosen from a larger statistical population or group of factors or instances that adequately replicates the larger group according to whatever characteristic or quality is under study. A representative sample parallels key variables and characteristics of the large society under examination. Some examples include sex, age, education level, socioeconomic status (SES), or marital status. A larger sample size reduced sampling error and increases the likelihood that the sample accurately reflects the target population. Random Sample A random sample is a group or set chosen from a larger population or group of factors of instances in a random manner that allows for each member of the larger group to have an equal chance of being chosen. A random sample is meant to be an unbiased representation of the larger population. It is considered a fair way to select a sample from a larger population since every member of the population has an equal chance of getting selected. Special Considerations: People collecting samples need to ensure that bias is minimized. Representative sampling is one of the key methods of achieving this because such samples replicate as closely as possible elements of the larger population under study. This alone, however, is not enough to make the sampling bias negligible. Combining the random sampling technique with the representative sampling method reduces bias further because no specific member of the representative population has a greater chance of selection into the sample than any other. Summarize this article in 250 words.
asked 2021-02-16
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?
asked 2020-11-27
A random sample of 5805 physicians in Colorado showed that 3332 provided at least some charity care (i.e., treated poor people at no cost).
a)
Let p represent the proportion of all Colorado physicians who provide some charity care. Find a point estimate for p. (Round your answer to four decimal places.)
b)
Find a $$99\%$$ confidence interval for p. (Round your answer to three decimal places.)
lower limit $$\displaystyle=?$$
upper limit $$\displaystyle=?$$
c)
Is the normal approximation to the binomial justified in this problem? Explain
...