Questions
Many consumers pay careful attention to stated nutritional contents on packaged foods when making purchases, so...

Many consumers pay careful attention to stated nutritional contents on packaged foods when making purchases, so it is important that the information on packages be accurate. The distribution of calorie content has been shown to be approximately normally distributed. A random sample of 12 frozen dinners of a certain type was selected and the calorie content of each one was determined to be 255, 244, 239, 242, 265, 245, 259, 248, 225, 226, 251, and 232. (a) Determine the sample mean and sample standard deviation for these 12 randomly selected frozen dinners. (b) The stated mean calorie content on the box is 240. Construct an appropriate test to see if the actual mean content differs from the stated value at an α = 0.05 significance level. Do not use the T-test feature of your calculator.

In: Math

Suppose We put five different dice into a hat. The dice have the following number of...

Suppose We put five different dice into a hat. The dice have the following number of side:4,6,8,12,20. When we choose a die from the hat, each of the five of the dice are equally likely to appear.

a) What is the probabilty that a “6” appears?

b) Now, suppose a “6” appears, what is the probability is was the 6-sided die that was chosen?

In: Math

In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor...

In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and the United Kingdom. They found that a substantially greater percentage of U.K. ads use humor.

(a) Suppose that a random sample of 387 television ads in the United Kingdom reveals that 130 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.K. television ads that use humor. (Round your answers to 3 decimal places.)   

pˆp^ =
The 95 percent confidence interval is [,].

(b) Suppose a random sample of 534 television ads in the United States reveals that 114 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.S. television ads that use humor. (Round your answers to 3 decimal places.)

pˆp^ =
The 95 percent confidence interval is [,].


(c) Do the confidence intervals you computed in parts a and b suggest that a greater percentage of U.K. ads use humor?


(Click to select)NoYes , the U.K. 95 percent confidence interval is (Click to select)abovenot above the maximum value
in the confidence interval for the U.S.

In: Math

Do streams with vegetated buffers (natural vegetation growth along the stream banks) have lower total phosphorus...

Do streams with vegetated buffers (natural vegetation growth along the stream banks) have lower total phosphorus concentrations [TP] than streams without vegetated stream buffers (and if so, by what magnitude)?   We have randomly sampled 20 streams in the piedmont of North Carolina (10 with buffers and ten without buffers) and measured TP concentrations (mg/L)

Table 1. Summary statistics of the natural log of total phosphorus concentrations

Mean (mg/L)

Standard Deviation (mg/L)

N

Buffered

1.5

0.30

10

Not Buffered

1.7

0.80

10

  1. Establish appropriate hypotheses for the analysis. What type of test should you conduct?
  2. Calculate the appropriate test statistic and p-value for your hypothesis test.   Interpret your p-value in a sentence.

c. What are the assumptions underlying this t-test?

In: Math

There are two suppliers of aluminum extrusions competing for your business. You are motivated to have...

There are two suppliers of aluminum extrusions competing for your business. You are motivated to have extrusions with good (i.e. high) yield strength and low variance. Each supplier provides you the following yield strength data (in ksi) from their respective processes:

X1= {76.5, 77.1, 76.1, 76.2, 75.9, 76.8, 77.0, 75.8, 76.6, 76.7}

X2= {78.2, 77.9, 77.2, 77.6, 76.2, 77.5, 77.2, 76.3, 77.3, 77.8}

At α = 0.05, whom will you choose? Statistically justify your answer.

Please solve by hand instead of using Excel.

In: Math

Joey Louzeshot is practicing his dart throwing skills. In the past, he hits the bullseye on...

Joey Louzeshot is practicing his dart throwing skills. In the past, he hits the bullseye on the target only one time for every 200 throws. To practice, he will throw darts on Sunday and Monday.

e) On Monday, Joey will practice by throwing the dart until hits the bullseye, then he will quit practicing. Let A be the number of attempts until he hits the bullseye. What are the distribution, parameter(s) and support of A?

f) What is the expected value and standard deviation of A?

In: Math

The average number of accidents at controlled intersections per year is 4.1. Is this average more...

The average number of accidents at controlled intersections per year is 4.1. Is this average more for intersections with cameras installed? The 43 randomly observed intersections with cameras installed had an average of 4.3 accidents per year and the standard deviation was 0.63. What can be concluded at the αα = 0.05 level of significance?

  1. For this study, we should use Select an answer z-test for a population proportion t-test for a population mean
  2. The null and alternative hypotheses would be:

H0:H0:  ? μ p  ? > = ≠ <       

H1:H1:  ? μ p  ? < = > ≠    

  1. The test statistic ? z t  =  (please show your answer to 3 decimal places.)
  2. The p-value =  (Please show your answer to 4 decimal places.)
  3. The p-value is ? > ≤  αα
  4. Based on this, we should Select an answer accept reject fail to reject  the null hypothesis.
  5. Thus, the final conclusion is that ...
    • The data suggest that the population mean is not significantly more than 4.1 at αα = 0.05, so there is statistically insignificant evidence to conclude that the population mean number of accidents per year at intersections with cameras installed is more than 4.1 accidents.
    • The data suggest that the sample mean is not significantly more than 4.1 at αα = 0.05, so there is statistically insignificant evidence to conclude that the sample mean number of accidents per year at intersections with cameras installed is more than 4.3 accidents.
    • The data suggest that the populaton mean is significantly more than 4.1 at αα = 0.05, so there is statistically significant evidence to conclude that the population mean number of accidents per year at intersections with cameras installed is more than 4.1 accidents.
  6. Interpret the p-value in the context of the study.
    • If the population mean number of accidents per year at intersections with cameras installed is 4.1 and if another 43 intersections with cameras installed are observed then there would be a 2.17496463% chance that the population mean number of accidents per year at intersections with cameras installed would be greater than 4.1.
    • If the population mean number of accidents per year at intersections with cameras installed is 4.1 and if another 43 intersections with cameras installed are observed then there would be a 2.17496463% chance that the sample mean for these 43 intersections with cameras installed would be greater than 4.3.
    • There is a 2.17496463% chance that the population mean number of accidents per year at intersections with cameras installed is greater than 4.1 .
    • There is a 2.17496463% chance of a Type I error.
  7. Interpret the level of significance in the context of the study.
    • There is a 5% chance that the population mean number of accidents per year at intersections with cameras installed is more than 4.1.
    • If the population mean number of accidents per year at intersections with cameras installed is 4.1 and if another 43 intersections with cameras installed are observed then there would be a 5% chance that we would end up falsely concluding that the population mean number of accidents per year at intersections with cameras installed is more than 4.1.
    • If the population population mean number of accidents per year at intersections with cameras installed is more than 4.1 and if another 43 intersections with cameras installed are observed then there would be a 5% chance that we would end up falsely concluding that the population mean number of accidents per year at intersections with cameras installed is equal to 4.1.
    • There is a 5% chance that you will get in a car accident, so please wear a seat belt.

In: Math

mean iron for males = 17.9 mg std devmales = 10.9 mg mean iron for women...

mean iron for males = 17.9 mg

std devmales = 10.9 mg

mean iron for women = 13.7 mg

stdev women = 8.9 mg

Using the information in Review Exercises 14 and 15, and assuming independent random samples of size 100 and 120 for women and mean, respectively, find the probability that the difference in sample mean iron level is greater than 5 mg

Using the information in Review Exercises 14 and 15, and assuming independent random samples of size 100 and 120 for women and mean, respectively, find the probability that the difference in sample mean iron level is greater than 5 mg

In: Math

A recent survey was conducted to determine how people consume their news. According to this​ survey,...

A recent survey was conducted to determine how people consume their news. According to this​ survey, 60​% of men preferred getting their news from television. The survey also indicates that 42​% of the sample consisted of males. ​ Also, 42​% of females prefer getting their news from television. Use this information to answer the following questions.

a. Consider that you have 30000 ​people, given the information​ above, how many of them are​ male?
b. Again considering that you have 30000 ​people, how many of them are​ female?
c. Out of the males in your​ sample, how many of them prefer getting their news on​ television?
d. Out of the males in your​ sample, how many of them prefer getting their news​ on-line?
e. Out of the females in your​ sample, how many of them prefer getting their news on​ television?
f. Out of the females in your​ sample, how many of them prefer getting their news​ on-line?
g. How many people in your sample prefer getting their news from​ television?
h. How many people in your sample prefer getting their news from​ on-line?
nothing
i. Given that a person prefers getting their news on television​, what is the probability that the person is male ​(round to 3 decimal​ places)?

PLEASE SHOW ALL WORK

In: Math

Assume that a set of test scores is normally distributed with a mean of 120 and...

Assume that a set of test scores is normally distributed with a mean of 120 and a standard deviaton of 5. Use the 68-95-99.7 rule to the find the followng quantities.

a. The percentage of scores less than 120 is ____% (round to one decimal place as needed).

b. The percentage of scores greater than 125 is ______% (round to one decimal place as needed).

c. The percentage of scores between 110 and 125 is ____% (round to one decimal place as needed).

In: Math

Based on data from a statistical abstract, only about 15% of senior citizens (65 years old...

Based on data from a statistical abstract, only about 15% of senior citizens (65 years old or older) get the flu each year. However, about 30% of the people under 65 years old get the flu each year. In the general population, there are 13.5% senior citizens (65 years old or older). (Round your answers to three decimal places.)

(a) What is the probability that a person selected at random from the general population is senior citizen who will get the flu this season?
(b) What is the probability that a person selected at random from the general population is a person under age 65 who will get the flu this year?

(c) Repeat parts (a) and (b) for a community that has 88% senior citizens. A: B:

(d) Repeat parts (a) and (b) for a community that has 51% senior citizens.A: B:

Let z be a random variable with a standard normal distribution. Find the indicated probability. (Round your answer to four decimal places.)

In: Math

Romans Food Market, located in Saratoga, New York, carries a variety of specialty foods from around...

Romans Food Market, located in Saratoga, New York, carries a variety of specialty foods from around the world. Two of the store’s leading products use the Romans Food Market name: Romans Regular Coffee and Romans DeCaf Coffee. These coffees are blends of Brazilian Natural and Colombian Mild coffee beans, which are purchased from a distributor located in New York City. Because Romans purchases large quantities, the coffee beans may be purchased on an as-needed basis for a price 10% higher than the market price the distributor pays for the beans. The current market price is $0.47 per pound for Brazilian Natural and $0.62 per pound for Colombian Mild. The compositions of each coffee blend are as follows:

Blend
Bean Regular DeCaf
Brazilian Natural 75% 40%
Colombian Mild 25% 60%

Romans sells the Regular blend for $3.60 per pound and the DeCaf blend for $4.40 per pound. Romans would like to place an order for the Brazilian and Colombian coffee beans that will enable the production of 1000 pounds of Romans Regular coffee and 500 pounds of Romans DeCaf coffee. The production cost is $0.80 per pound for the Regular blend. Because of the extra steps required to produce DeCaf, the production cost for the DeCaf blend is $1.05 per pound. Packaging costs for both products are $0.25 per pound. Formulate a linear programming model that can be used to determine the pounds of Brazilian Natural and Colombian Mild that will maximize the total contribution to profit.

Let BR = pounds of Brazilian beans purchased to produce Regular
BD = pounds of Brazilian beans purchased to produce DeCaf
CR = pounds of Colombian beans purchased to produce Regular
CD = pounds of Colombian beans purchased to produce DeCaf

If required, round your answers to three decimal places. For subtractive or negative numbers use a minus sign even if there is a plus sign before the blank. (Example: -300)

Max BR + BD + CR + CD
s.t.
Regular blend BR + CR =
DeCaf blend BD + CD =
Regular production BR CR =
DeCaf production BD + CD =
BR, BD, CR, CD ≥ 0

What is the optimal solution and what is the contribution to profit? If required, round your answer to the nearest whole number.

Optimal solution:

BR =
BD =
CR =
CD =

If required, round your answer to the nearest cent.

Value of the optimal solution = $  

In: Math

Can you please tell me what more information I need to provide to solve below problem?...

Can you please tell me what more information I need to provide to solve below problem?

1. Find out below using R

Hints: Use iris dataset from R (built in data set in R)

a) Create a new data frame called virginica.versicolor (that only contains these two species)

the command I used:

virginica.versicolor <- iris[iris$Species %in% c("versicolor", "virginica"), ]

b) What is your null hypothesis regarding sepal lengths for the two species (virginica.versicolor) ? And what is your alternate hypothesis?

c) Describe your hypotheses in terms of your test statistic: what would be the t under the null hypothesis, H0, and what would be the statement about t under your alternate hypothesis Ha?

d) Would you do a one- or non-(i.e., two-sided) directional test? Why?

e) Conduct a Student’s t-test using the formula format as follows:

t.test(sepal.length ~ species, data = virginica.versicolor, var.equal = T).

f) Explain what the three different sections do within the t.test() function.

g) Did your function run a one- or non-directional test?

h) What is your t-value? Based on the results of your t-test, what is your conclusion and why?

In: Math

After a million measurements of thing x, we find a sample mean of 60.29 and standard...

After a million measurements of thing x, we find a sample mean of 60.29 and standard deviation of 1.46. What chance, in percent (0-100) does the next measurement have of being outside 3 standard deviations from the mean? Do not include the percent sign.

After a million measurements of thing x, we find a sample mean of 50.25 and standard deviation of 1.92. What chance, in percent (0-100) does the next measurement have of being outside 2 standard deviations from the mean? Do not include the percent sign.

In: Math

Consider the quarterly electricity production for years 1-4: Year 1 2 3 4 Q1 99 120...

Consider the quarterly electricity production for years 1-4:
Year 1 2 3 4
Q1 99 120 139 160
Q2 88 108 127 148
Q3 93 111 131 150
Q4 111 130 152 170
(a) Estimate the trend using a centered moving average. PLEASE PROVIDE ANSWER A AND PLOT THE TREND  ON A SCATTER PLOT
(b) Using a classical additive decomposition, calculate the seasonal component.
(c) Explain how you handled the end points.
Note: Explain all the steps and computations

In: Math