# A hybrid course is one that contains both online and classroom instruction. In a study performed at Middle Georgia State University, a software package was used as the main souce of instruction in a hybrid college algebra course. The software tracked the number of hours it took for each student to meet the objectives of the course. In a sample of 45 students, the mean number of hours was 80.5, with a standard deviation of 51.2. a. Construct a 95% confidence interval for the mean number of hours it takes for a student to meet the course objectives.

Question
Upper level algebra
A hybrid course is one that contains both online and classroom instruction. In a study performed at Middle Georgia State University, a software package was used as the main souce of instruction in a hybrid college algebra course. The software tracked the number of hours it took for each student to meet the objectives of the course. In a sample of 45 students, the mean number of hours was 80.5, with a standard deviation of 51.2.
a. Construct a 95% confidence interval for the mean number of hours it takes for a student to meet the course objectives.

2020-11-09
(a)Step 1
A.95% confidence interval for the mean number of hours it takes for the student to makes course of objectives is, First, compute f-critical value then find confidence interval.
The t critical value for the 95% confidence interval is,
$$\alpha=0.05$$
The sample size is small and two-tailed test. Look in the column headed and the row headed in the fdistribution table by using degree of freedom is,
d.f.=n-1
=45-1=44
The tcritical value for the 95% confidence interval is 2.015.
Step 2
$$95\% C.I.=\overline{x}\pm t_{c}\times s / \sqrt{n}$$
$$=80.50\pm2.015\times51.20/\sqrt{45}$$
$$=80.50\pm 15.38$$
= 65.12 to 95.88
65.12 to 95.884 95% confidence interval for the mean number of hours it takes for the student to makes course of objectives is.
Step 3
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### 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.
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You may need to use the appropriate appendix table or technology to answer this question.
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a.
To compute the confidence interval use a ? distribution.
b.
With $$90\%$$ confidence the population mean minutes of concentration is between ____ and ____ minutes.
c.
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a) Find a $$\displaystyle{95}\%$$ confidence interval for the improvement in traffic flow due to the new system.
b) Find a $$\displaystyle{98}\%$$ confidence interval for the improvement in traffic flow due to the new system.
c) A traffic engineer states that the mean improvement is between 581.6 and 726.6 vehicles per hour. With what level of confidence can this statement be made?
d) Approximately what sample size is needed so that a $$\displaystyle{95}\%$$
confidence interval will specify the mean to within $$\displaystyle\pm\ {50}$$ vehicles per hour?
e) Approximately what sample size is needed so that a $$\displaystyle{98}\%$$ confidence
interval will specify the mean to within $$\displaystyle\pm\ {50}$$ vehicles per hour?
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