Questions
A food safety guideline is that the mercury in fish should be below 1 part per...

A food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm). Listed below are the amounts of mercury​ (ppm) found in tuna sushi sampled at different stores in a major city. Construct a 95​% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna​ sushi? 0.61  0.78  0.10  0.88  1.27  0.56  0.91 What is the confidence interval estimate of the population mean mu​?

In: Math

The following should be performed using R and the R code included in your submission. To...

The following should be performed using R and the R code included in your submission.

To obtain first prize in a lottery, you need to correctly choose n different numbers from N and 1 number from 20, known as the supplementary. That is we first draw n numbers from 1:N without replacement and then 1 number from 1:20 in another draw. Suppose n=7 and N=35. Let X be the number of drawn numbers that match your selection, where the supplementary counts as 8, so that X=0,…,15. For a first prize X=15 i.e. all numbers are matched.

(a) Calculate probabilities P(X=x), x=0, 1, …, 7, without and with the supplementary. Plot the distribution function and the cumulative distribution function. Hint: Part of the answer involves the hypergeometric.

(b) Using R, generate 1,000,000 random numbers from this distribution and plot a histogram of the simulated data.

(c) Calculate the expected value, E(X), and the variance, σ2 (or Var(X)). Obtain the mean and the variance of the simulated data. Compare the estimates with the theoretical parameters.

(d) Assume that each week 10,000,000 entries are lodged, for a single draw. What is the value of � from the Poisson approximation to the number of entries with a first prize? Use the Poisson approximation for the following. What is the probability that there will be no entry with a first prize? What is the expected number of weeks until the first prize?

In: Math

The paint used to make lines on roads must reflect enough light to be clearly visible...

The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let μ denote the true average reflectometer reading for a new type of paint under consideration. A test of H0: μ = 20 versus Ha: μ > 20 will be based on a random sample of size n from a normal population distribution. What conclusion is appropriate in each of the following situations? (Round your P-values to three decimal places.)

(a)    n = 15, t = 3.3, α = 0.05
P-value =

State the conclusion in the problem context.

Reject the null hypothesis. There is not sufficient evidence to conclude that the new paint has a reflectometer reading higher than 20.

Reject the null hypothesis. There is sufficient evidence to conclude that the new paint has a reflectometer reading higher than 20.  

  Do not reject the null hypothesis. There is sufficient evidence to conclude that the new paint has a reflectometer reading higher than 20.

Do not reject the null hypothesis. There is not sufficient evidence to conclude that the new paint has a reflectometer reading higher than 20.

(b)    n = 9, t = 1.7, α = 0.01
P-value =

State the conclusion in the problem context.

Reject the null hypothesis. There is not sufficient evidence to conclude that the new paint has a reflectometer reading higher than 20.

Do not reject the null hypothesis. There is sufficient evidence to conclude that the new paint has a reflectometer reading higher than 20.    

Do not reject the null hypothesis. There is not sufficient evidence to conclude that the new paint has a reflectometer reading higher than 20.

Reject the null hypothesis. There is sufficient evidence to conclude that the new paint has a reflectometer reading higher than 20.

(c)    n = 29,

t = −0.3

P-value =

State the conclusion in the problem context.

Reject the null hypothesis. There is sufficient evidence to conclude that the new paint has a reflectometer reading higher than 20.

Reject the null hypothesis. There is not sufficient evidence to conclude that the new paint has a reflectometer reading higher than 20.  

Do not reject the null hypothesis. There is not sufficient evidence to conclude that the new paint has a reflectometer reading higher than 20.

Do not reject the null hypothesis. There is sufficient evidence to conclude that the new paint has a reflectometer reading higher than 20.

In: Math

All logs are to base e Size is in cubic centimeters, Age is in years, Weight...

All logs are to base e

Size is in cubic centimeters, Age is in years, Weight is in pounds, Temperature is in degrees Fahrenheit, Height is in inches, Cost is in dollars, and Distance is in miles

The regression equation is
Log(Size) = 28.6 + 0.0292 Age - 1.124 Weight - 1.69 log(Temperature) + 1.02 log(Height) + 2.24 log(Cost) - 0.334 log(Distance)

Predictior Coef SE Coef T P
Constant 28.59 20.67 1.38 0.301
age 0.0292 0.083 12.27 0.006
weight -1.124 0.052 -21.40 0.001
Log(temp) -1.691 0.223 -7.58 0.002
Log(height) 1.0207 0.847 1.21 0.351
Log(cost) 2.239 0.42 5.06 0.004
Log(distance) -0.334 0.0112 -2.98 0.049

S = 64.1788   R-Sq = 86.1%   R-Sq(adj) = 72.1%

Answer the following question using three decimals.
If the effect is not statistically significant put in NA
Put in percentages without the percent sign, so put in 10 instead of 10%

1-

As age increases by 1 year the size increases by the percentage

2 -

As weight increases by one pound the size decreases by the percentage

3-

As temperature increases by 7% the size will decrease by the percentage

4-

As Height increases by 3% the size increases by the percentage

5- As cost increases by 5% the size increases by the percentage

In: Math

Time spent using​ e-mail per session is normally​ distributed, with mu equals 7 minutes and sigma...

Time spent using​ e-mail per session is normally​ distributed, with mu equals 7 minutes and sigma equals 2 minutes. Assume that the time spent per session is normally distributed. Complete parts​ (a) through​ (d). a. If you select a random sample of 25 ​sessions, what is the probability that the sample mean is between 6.8 and 7.2 ​minutes?

In: Math

Acme Outdoors Co. is introducing a new line of sport all-terrain vehicles (ATVs). Acme is considering...

Acme Outdoors Co. is introducing a new line of sport all-terrain vehicles (ATVs). Acme is considering program proposals from two competing companies: A & B. A's marketing plan is expected to generate high sales with a 75% probability and only a 25% likelihood of low sales.

An alternative marketing plan from company B could result in an initial 60% likelihood of high ATV sales and 40% probability of low sales. But company B's offer also contains a provision for an optional follow-on promotion if a low response is returned. If an optional follow-on ATV promotion is conducted by company B, there is a 70% chance of ultimately realizing high sales.

Draw an decision tree diagram with branches labeled and probabilities, do not solve.

In: Math

“Larry’s logs” is planning to change the supplier. For complete analysis he needs to compute the...

“Larry’s logs” is planning to change the supplier. For complete analysis he needs to compute the Multifactor Productivity (MFP) for each supplier. He expects to produce 240 units using old supplier, and 260 using the new supplier. He pays $10 per hour, and will use 300 labor hours with current supplier and 308 hours with the new supplier. Other costs will be consistent across both suppliers (materials are $10 per log), using a total of 100 logs; capital costs will be $350 and energy $150).

What is the total cost for current and new supplier?

What is the multifactor productivity for the current and new supplier?

What is percent change in multifactor productivity?

In: Math

1)    A lawn and garden retailer operates 4 stores in the DFW Metroplex. One of its...

1)    A lawn and garden retailer operates 4 stores in the DFW Metroplex. One of its most popular items is a lawn tractor. Weekly customer demand is distributed N(10,52) at each store. Each store replenishes its stock to 15 lawn tractors at the start of each week. Note: Assume that weekly demands at each store are independent.

a)   Consider just one store. What is the probability of a stockout in that store?

b)   What is the probability of a stockout in at least two of the four stores?

c)   Suppose that the four stores decide to pool their stock. Specifically, they decide to pool their weekly allocations (4 X 15 = 60) in a centrally located warehouse and draw from it as needed to satisfy their demand. How often will a store experience a stockout now?

In: Math

In a behavioral medicine study, Cease, Now, And Live-Longer(2017) found that there was a difference in...

In a behavioral medicine study, Cease, Now, And Live-Longer(2017) found that there was a difference in effectiveness of using rapid smoking as a treatment of smoking cessation as compared to a control method. They found that participants smoked less after receiving the rapid smoking treatment as compared to those participants in a control condition. Cease et al. (2017) had reviewed considerable research on other treatments for smoking cessation. For example, there is the use of nicotine replacement therapy and the use of pharmacotherapy like Chantix to treat smoking behavior. These methods have been shown to be effective in reducing or eliminating smoking behavior. There is a need to compare behavioral based smoking cessation treatments with other types of smoking cessation treatments like NRT and pharmacothearpy. To this end, the research term of Patch-It, Puff-It And Stop-it(2018) have designed a new study of compare the effectiveness of smoking cessation treatments.

Patch-it et al. (2018) designed a study that used three conditions to evaluate the effectiveness of smoking cessation treatments. They used the rapid smoking condition that was used in the Cease et al. (2017) study. In the rapid smoking condition, the participants deeply inhaled on a cigarette every six seconds that was repeated for six cigarettes smoked. There were no rest periods between cigarettes. In another condition nicotine replacements therapy was used. The researchers used a patch with 14 mg of nicotine to be worn for 16 hours a day during the treatment phase of the study. A control condition was used where participants puff a sham cigarette at the same rate and duration as the rapid smoking condition.

Patch-it et al. (2018) recruited a large pool of college student smokers, who had reported to smoke for at least one year and expressed a desire to quit completely. From this pool of potential participants, 30 smokers were randomly selected to participate. With informed consent, the smokers agreed to participate in the study. Ten smokers(10) were randomly assigned to the rapid smoking condition (RAPID SMOKING). The two treatment conditions and control condition lasted for two weeks(14 Days).
A measure of cigarette use was taken at two months after the treatment program was ended. These days were recorded as the average daily number (rounded to a whole number) of reported cigarettes (CIGARETTES) smoked during a period of seven days. These data were obtained from daily logs submitted by the participants.

CONTROL 10 16 20 16 19 15 15 14 16 11

RAPID SMOKING 13 4 12 11 6 6 11 12 12 12

PATCH 14 MG 10 4 4 7 7 4 2 8 10 10

2. State the statistical hypothesis to be used in the data analysis (notation and words).

In: Math

a) What is an exponential distribution (include an APA citation)? ___________________________ b) When would you use...


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)

In: Math

1. a. A researcher wishes to estimate the proportion of adults who have​ high-speed Internet access....

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?

In: Math

The health of the bear population in Yellowstone National Park is monitored by periodic measurements taken...

The health of the bear population in Yellowstone National Park is monitored by periodic measurements taken from anesthetized bears. A sample of 38 bears has a mean weight of 189.6 lb.

At α = .05, can it be concluded that the average weight of a bear in Yellowstone National Park is different from 187 lb? Note that the standard deviation of the weight of a bear is known to be 8.2 lb.

(a) Find the value of the test statistic for the above hypothesis.
(b) Find the critical value.
(c) Find the p-value.
(d) What is the correct way to draw a conclusion regarding the above hypothesis test?
(A) If the answer in (b) is greater than the answer in (c) then we cannot conclude at the 5% significance
level that the average weight of a bear in Yellowstone National Park is different from 187 lb.
(B) If the answer in (b) is greater than the answer in (c) then we conclude at the 5% significance
level that the average weight of a bear in Yellowstone National Park is different from 187 lb.
(C) If the answer in (a) is greater than the answer in (b) then we conclude at the 5% significance
level that the average weight of a bear in Yellowstone National Park is different from 187 lb.
(D) If the answer in (a) is greater than the answer in (c) then we conclude at the 5% significance
level that the average weight of a bear in Yellowstone National Park is different from 187 lb.
(E) If the answer in (c) is less than 0.05 then we cannot conclude at the 5% significance
level that the average weight of a bear in Yellowstone National Park is different from 187 lb.
(F) If the answer in (c) is greater than 0.05 then we conclude at the 5% significance
level that the average weight of a bear in Yellowstone National Park is different from 187 lb.
(G) If the answer in (a) is greater than the answer in (b) then we cannot conclude at the 5% significance
level that the average weight of a bear in Yellowstone National Park is different from 187 lb.
(H) If the answer in (a) is greater than the answer in (c) then we cannot conclude at the 5% significance
level that the average weight of a bear in Yellowstone National Park is different from 187 lb.

In: Math

Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping...

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 μx = 49 and standard error σx = $8.    
The sampling distribution of x is approximately normal with mean μx = 49 and standard error σx = $0.13.
The sampling distribution of x is approximately normal with mean μx = 49 and standard error σx = $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.    

In: Math

The weights for newborn babies is approximately normally distributed with a mean of 5lbs and a...

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: Math

In 2018 and 2019, Health Canada commissioned two national surveys on the use of cannabis among...

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.

In: Math