a) What is an exponential distribution (include an APA citation)? ___________________________

b) When would you use an exponential distribution? ________________________

c) What is a binomial distribution (include an APA citation)?_______________________

d) When would you use a binomial distribution? ___________________

3. Task

Run these commands in R, then use your own words to describe what the resulting numbers represent.  You can get some information about the functions by using the help commands in R (such as ?pbinom to get information about the pbinom() command in R):

a) pbinom(q=5, size=10, prob=1/6)

b)

n=10
p=.5
x=9
pbinom(x, n, p)

c) punif(5, min=1, max=10) - punif(4, min=1, max=10)

1.

a. A researcher wishes to estimate the proportion of adults who have​ high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.3 with 99% confidence?

b.

it is believed that people prefer rubies over other gems. In a recent simple random survey of 150 people, 63 said they would prefer a ruby over other gems. Use this sample data to complete a hypothesis test to determine if a majority of people would prefer a ruby. over other gems at the 0.01 significance level.

Be sure to include all the steps for a complete hypothesis test - start and end in context, test conditions, show formulas and numbers used, clearly state REJECT or FAIL TO reject.

c.

If 12 jurors are randomly selected from a population that is 45% Hispanic, what is the probability that 2 or fewer jurors will be Hispanic?

Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping interval. Based on a certain article, the mean of the x distribution is about $49 and the estimated standard deviation is about $8.

(a) Consider a random sample of n = 60 customers, each of whom has 10 minutes of unplanned shopping time in a supermarket. From the central limit theorem, what can you say about the probability distribution of x, the average amount spent by these customers due to impulse buying? What are the mean and standard deviation of the x distribution?

The sampling distribution of x is not normal.
The sampling distribution of x is approximately normal with mean μ_x = 49 and standard error σ_x = $8.    
The sampling distribution of x is approximately normal with mean μ_x = 49 and standard error σ_x = $0.13.
The sampling distribution of x is approximately normal with mean μ_x = 49 and standard error σ_x = $1.03.

Is it necessary to make any assumption about the x distribution? Explain your answer.

It is not necessary to make any assumption about the x distribution because μ is large.
It is necessary to assume that x has an approximately normal distribution.    
It is not necessary to make any assumption about the x distribution because n is large.
It is necessary to assume that x has a large distribution.

(b) What is the probability that x is between $47 and $51? (Round your answer to four decimal places.)


(c) Let us assume that x has a distribution that is approximately normal. What is the probability that x is between $47 and $51? (Round your answer to four decimal places.)


(d) In part (b), we used x, the average amount spent, computed for 60 customers. In part (c), we used x, the amount spent by only one customer. The answers to parts (b) and (c) are very different. Why would this happen?

The sample size is smaller for the x distribution than it is for the x distribution.
The standard deviation is smaller for the x distribution than it is for the x distribution.    
The x distribution is approximately normal while the x distribution is not normal.
The mean is larger for the x distribution than it is for the x distribution.
The standard deviation is larger for the x distribution than it is for the x distribution.

In this example, x is a much more predictable or reliable statistic than x. Consider that almost all marketing strategies and sales pitches are designed for the average customer and not the individual customer. How does the central limit theorem tell us that the average customer is much more predictable than the individual customer?

The central limit theorem tells us that small sample sizes have small standard deviations on average. Thus, the average customer is more predictable than the individual customer.
The central limit theorem tells us that the standard deviation of the sample mean is much smaller than the population standard deviation. Thus, the average customer is more predictable than the individual customer.    

The weights for newborn babies is approximately normally distributed with a mean of 5lbs and a standard deviation of 1.5lbs. Consider a group of 1,000 newborn babies:

How many would you expect to weigh between 4-7lbs?

How many would you expect to weigh less than 6lbs?

How many would you expect to weigh more than 5lbs?

How many would you expect to weigh between 5-10lbs?

In 2018 and 2019, Health Canada commissioned two national surveys on the use of cannabis among people aged 20 to 24. They found 175 users out of the 1000 surveyed individuals in 2018 and 230 out of the 1100 sample in 2019.

1. (a) Test the hypothesis that there has been a change in the proportion of cannabis users in the 20-24 age

   group population between 2018 and 2019. Use the critical value approach and a 0.05 level of

   significance.

2. (b) Find the p-value for your result in (a) above.

3. (c) Calculate a 95% two-sided confidence interval for the true difference using the data provided.

4. (d) Explain how the p-value and the confidence interval are or are not consistent with your result in

   part (a) above.

Use the accompanying paired data consisting of registered boats​ (tens of​ thousands) and manatee fatalities from boat encounters. Let x represent the number of registered boats and let y represent the corresponding number of manatee deaths. Use the given number of registered boats and the given confidence level to construct a prediction interval estimate of manatee deaths. Use x=89 (for 89​0,000 registered​ boats) with a 99​% confidence level.

Boats (tens of thousands)   Manatees
67    54
68    37
66    34
73    49
74    40
70    59
76    56
82    67
83    84
83    80
90    82
91    94
95   74
93   69
97   78
99   93
97   73
98   91
98   97
91   81
90   87
90   82
89   73
90   7

Find the indicated prediction interval.

___manatees < y < manatees

​(Round to three decimal places as​ needed.)

A dairy scientist is testing a new feed additive. She chooses 13 cows at random from a large population. She randomly assigns n_old = 8 to the old diet and n_new = 5 to a new diet including the additive. The cows are housed in 13 widely separated pens. After two weeks, she milks each cow and records the milk produced in pounds:

Old Diet: 43, 51, 44, 47, 38, 46, 40, 35 New Diet: 47, 75, 85, 100, 58

Let μnew and μold be the population mean milk productions for the new and old diets, respectively. She wishes to test H0 : μnew − μold = 0 against HA : μnew − μold ̸= 0 using α = 0.05.

(a) Graph the data as you see fit. Why did you choose the graph(s) that you did and what does it (do they) tell you?

(b) Choose a test appropriate for the hypotheses and justify your choice based on your answer to part (a). Then perform the test by computing a p-value, and making a reject or not reject decision. Do this without R and show your work. (Also do it with R, if you wish, to check your work). Finally, state your conclusion in the context of the problem.

n a 2008​ survey, people were asked their opinions on astrology​ - whether it was very​ scientific, somewhat​ scientific, or not at all scientific. Of 1438 who​ responded, 76 said astrology was very scientific. a. Find the proportion of people in the survey who believe astrology is very scientific. b. Find a​ 95% confidence interval for the population proportion with this belief. c. Suppose a TV news anchor said that​ 5% of people in the general population think astrology is very scientific. Would you say that is​ plausible? Explain your answer. a. The proportion of people in the survey who believe astrology is very scientific is . 0529. ​(Round to four decimal places as​ needed.) b. Construct the​ 95% confidence interval for the population proportion with the belief that astrology is very scientific. left parenthesis nothing comma nothing right parenthesis ​(Round to three decimal places as​ needed.) Enter your answer in the edit fields and then click Check Answer.

Consider two random variables X and Y, with Y = (a+bX)

- Find E(Y)

- Find Cov(X,Y)

- Find Corr(X,Y)

(9). The National Health Statistics Reports dated Oct. 22, 2008, stated that for a sample size of 277 18-year-old American males, the sample mean waist circumference was 86.3 cm. A somewhat complicated method was used to estimate various population percentiles, resulting in the following values.

5^th   10^th   25^th  50^th 75^th 90^th 95^th

69.6 70.9 75.2 81.3 95.4 107.1 116.4

(a) Is it plausible that the waist size distribution is at least approximately normal? Explain your reasoning.

1. Since the mean and median are substantially different, and the difference in the distance between the median and the upper quartile and the distance between the median and the lower quartile is relatively large, it seems plausible that waist size is at least approximately normal.
2. Since the mean and median are nearly identical, and the distance between the median and the upper quartile and the distance between the median and the lower quartile are almost the same, it does not seem plausible that waist size is at least approximately normal.
3. Since the mean and median are nearly identical, and the distance between the median and the upper quartile and the distance between the median and the lower quartile are nearly the same, it seems plausible that waist size is at least approximately normal.
4. Since the mean and median are substantially different, and the distance between the median and the upper quartile and the distance between the median and the lower quartile are nearly the same, it seems plausible that waist size is at least approximately normal.
5. Since the mean and median are substantially different, and the difference in the distance between the median and the upper quartile and the distance between the median and the lower quartile is relatively large, it does not seem plausible that waist size is at least approximately normal.

(b)   Make a conjecture on the shape of the population distribution.

1. The upper percentiles stretch much farther than the lower percentiles. Therefore, we might suspect a right-skewed distribution.
2. The lower percentiles stretch much farther than the upper percentiles. Therefore, we might suspect a right-skewed distribution.     
3. The upper percentiles stretch much farther than the lower percentiles. Therefore, we might suspect a left-skewed distribution.
4. The lower percentiles stretch much farther than the upper percentiles. Therefore, we might suspect a left-skewed distribution.
5. It is plausible that waist size is at least approximately normal.

(C)    Suppose that the population mean waist size is 85 cm and that the population standard deviation is 15 cm. How likely is it that a random sample of 277 individuals will result in a sample mean waist size of at least 86.3 cm? (Round your answers to four decimal places.)

________________________________________________________________

(d)    Referring back to (C), suppose now that the population mean waist size in 82 cm. Now what is the (approximate) probability that the sample mean will be at least 86.3 cm? (Round your answers to three decimal places.)

________________________________________________________________

(e)     In light of this calculation, do you think that 82 cm is a reasonable value for μ?

1. No 82 cm is not a reasonable value for μ since if the population mean waist size is 82 cm, there would be almost no chance of observing a sample mean waist size of 86.3 cm (or higher) in a random sample of 277 men.
2. Yes 82 cm is a reasonable value for μ since if the population mean waist size is 82 cm, there would be almost no chance of observing a sample mean waist size of 86.3 cm (or higher) in a random sample of 277 men.    
3. Yes 82 cm is a reasonable value for μ since it is almost the same as 50^th percentile 81.3.
4. Yes 82 cm is a reasonable value for μ since if the population mean waist size is 82 cm, there is a reasonably large chance of observing a sample mean waist size of 86.3 cm (or higher) in a random sample of 277 men.
5. No 82 cm is not a reasonable value for μ since if the population mean waist size is 82 cm, there is a reasonably large chance of observing a sample mean waist size of 86.3 cm (or higher) in a random sample of 277 men.

You and a friend, along with an eccentric rich probabilist, are observing a Poisson process whose arrival rate is λ = .5 per hour. The probabilist offers to pay you $100 if there is at least one arrival between noon and 2pm, and also offers to pay your friend $100 if there is at least one arrival between 1pm and 3pm.

a. What is the probability that either you or your friend, or both, gets $100?

b. What is the probability that one of you wins $100, but not both?

Consider a Poisson process with arrival rate λ per minute. Given that there were three arrivals in the first 2 minutes, find the probability that there were k arrivals in the first minute; do this for k = 0, 1, 2, and 3.

Given that P(A) = .4, P(A ∩ B) = .1, and P((A ∪ B) c ) = .2, find P(B).

13. Using traditional methods, it takes 94 hours to receive a basic flying license. A new license training method using Computer Aided Instruction (CAI) has been proposed. A researcher used the technique with 210 students and observed that they had a mean of 95 hours. Assume the standard deviation is known to be 5. A level of significance of 0.05 will be used to determine if the technique performs differently than the traditional method. Is there sufficient evidence to support the claim that the technique performs differently than the traditional method?

What is the conclusion?

A. There is not sufficient evidence to support the claim that the technique performs differently than the traditional method.

B. There is sufficient evidence to support the claim that the technique performs differently than the traditional method.

14. Using traditional methods it takes 90 hours to receive an advanced driving license. A new training technique using Computer Aided Instruction (CAI) has been proposed. A researcher believes the new technique may reduce training time and decides to perform a hypothesis test. After performing the test on 190 students, the researcher decides to reject the null hypothesis at a 0.02 level of significance.

What is the conclusion?

A. There is sufficient evidence at the 0.020 level of significance that the new technique reduces training time.

B. There is not sufficient evidence at the 0.02 level of significance that the new technique reduces training time.

Suppose 56% of the population has a college degree. If a random sample of size 503 is selected, what is the probability that the proportion of persons with a college degree will be greater than 54%? Round your answer to four decimal places.

In one of PLE’s manufacturing facilities, a drill press that has three drill bits is used to fabricate metal parts. Drill bits break occasionally and need to be replaced. The present policy is to replace a drill bit when it breaks or can no longer be used. The operations manager is considering a different policy in which all three drill bits are replaced when any one bit breaks or needs replacement. The rationale is that this would reduce downtime. It costs $200 each time the drill press must be shut down. A drill bit costs $85, and the variable cost of replacing a drill bit is $14 per bit. The company that supplies the drill bits has historical evidence that the reliability of a single drill bit is describes by a Poisson probability distribution with the mean time between failures is an exponential distribution with mean μ = 1 / λ = 1 / 0.01 = 100 hours. (Professor Cursio: see below.) The operations manager at PLE would like to compare the cost of the two replacement policies. Develop spreadsheet models to determine the total cost for each policy over 1,000 hours and make a recommendation. Explain and summarize your findings in a report