Experiment 1: Study the Relationship Between Volume and Temperature for a Sample of Air
Took a Erlenmeyer flask & placed it on the workbench; closed flask.
Added 1.50 atm of air to the flask.
Attached a gas syringe & thermometer to the Erlenmeyer flask; remember the Erlenmeyer flask volume is 150 mL.
The volume of air in the syringe = 75.0 mL
The total air volume = 225.0 mL
The temperature = 21.5 C
Ran the constant temperature bath to 0 °C; moved the Erlenmeyer flask into the constant temperature bath. Waited until the temperature stabilized at 0.0 °C.
The volume of air in the syringe = 58.6 mL
The total air volume = 208.6 mL
The temperature = 0.0 °C.
Ran the bath to 40 °C.
Waited until the temperature of the air stabilized at 40.0 °C.
The volume of air in the syringe = 89.1 mL
The total air volume = 239.1 mL
The temperature = 40 °C.
Ran the bath to 60 °C
The volume of air in the syringe = 104.4 mL
The total air volume = 254.4 mL
The temperature = 60 °C
Ran the bath to 80 °C
The volume of air in the syringe = 119.7 mL
The total air volume = 269.7 mL
The temperature = 80 °C
Ran the bath to 100 °C
The volume of air in the syringe = 135.0 mL
The total air volume = 285.0
What parameters were held constant in experiment 1?
A. pressure and number of moles
B. volume and pressure
C. temperature and volume
D. volume and number of moles
Given the data you collected in experiment 1, what is the relationship between volume & temperature?
A. The volume increases as the temperature increases.
B. The volume increases as the temperature decreases.
C. The volume changes randomly with temperature.
D. The volume remains constant as the temperature increases.
In: Chemistry
A retailer discovers that 3 jars from his last shipment of Spiffy peanut butter contained between 15.85 and 15.92 oz of peanut butter, despite the labeling indicating that each jar should contain 16 oz. of peanut butter. He is wondering if Spiffy is cheating its customers by filling its jars with less product than advertised. He decides to measure the weight of 50 jars from the shipment and use hypothesis testing to verify this.
(a) What are the null and alternative hypotheses for this experiment?
(b) Describe, in words, a Type I error for this experiment.
(c) Describe, in words, a Type II error for this experiment.
(d) Given the answer to (a), should the null hypothesis be rejected when the sample mean falls below or over a certain threshold? Should this threshold be below or above the value 16.0 oz?
(e) What is the distribution of X ̄, the sample mean?
(f) In his sample of 50 jars, the retailer finds an average weight of 15.84 oz and a sample standard deviation of 0.5 oz. He decides to use a significance level of 0.04. What is the conclusion from this hypothesis testing? Can you conclude that Spiffy is cheating its customers?
(g) What is the p-value? What is the meaning of this number?
(h) For what values of the sample mean would the null hypothesis be rejected?
(i) Calculate the probability of type II error if the true mean is 15.7 oz.
(j) Solve (f), (h) and (i) when the level of significance is 0.01. Is your new answer for (f) consistent with the p-value found in (g)? How is the probability of type II error affected when the probability of type I error changes?
In: Statistics and Probability
There is an urban legend that mothers have increased sensitivity
to and awareness of noises, in particular to that of children. A
social psychologist finds support for the legend in the literature
and now wants to confirm it. The psychologist designs a study where
he recruits women that think they are going to participate in a
sleep experiment where they will evaluate the comfort of different
mattresses overnight. While the women slept, the psychologist
introduces noises to test the minimum volume needed for the women
to be awakened by the noise. The wake-up decibel data are below.
What can be concluded with α = 0.10?
| mother | non-mother |
| 63 58 52 73 57 68 62 |
41 |
b)
Condition 1: (choose one)
1) non-mother 2) sleep experiment 3) decibels 4) women 5)
mother
Condition 2: (choose one)
1) non-mother 2) sleep experiment 3) decibels 4) women 5)
mother
c) Input the appropriate value(s) to make a
decision about H0.
(Hint: Make sure to write down the null and alternative hypotheses
to help solve the problem.)
p-value =_________________________ ; Decision: Reject H0
or Fail to reject H0
d) Using the SPSS results,
compute the corresponding effect size(s) and indicate
magnitude(s).
If not appropriate, input and/or select "na" below.
d =___________ ; ( choose one) 1.) trivial effect 2) small
effect 3) medium effect 4) large effect
r2 = ___________ ; ( Choose one) 1.) trivial
effect 2) small effect 3) medium effect 4) large effect
In: Statistics and Probability
Q3. (This question is based in R)
Now use the simulation ("X = rnorm(1000, mean = 10, sd = 2)", "Y = rnorm(1000, mean = 5, sd = 3)") to estimate the distribution of X+Y and create confidence intervals.
A) Form a set of Xs and Ys by repeating the individual experiment for B = 2000 times, where each experiment has n = 1000 samples. You may want to write a for loop and create two matrices "sample_X" and "sample_Y" to save those values.
B) Calculate the mean of X+Y for each experiment and save it to a vector which has a length of B, and plot a histogram of these means.
C) Now as we have a simulated sampling distribution of X+Y, calculate the 95% confidence interval for mean of X+Y (this can be done empirically).
D) In the above example, we have fixed the sample size n and number of experiments B. Next, we want to change B and n, and see how the confidence interval will change. Please write a function to create confidence intervals for any B and n.
E) Suppose the sample size n varies (100, 200, 300, .... , 1000) (fix B=2000) and the number of experiments B varies (1000, 2000, ... , 10000) (fix n=500). Plot your confidence intervals to compare the effect of changing the sample size n and changing the number of simulation replications B. What do you conclude? (Hint: Check function errbar() in Hmisc package for plot - library(Hmisc))
fix n, B varies
fix B, n variesIn: Statistics and Probability
Question 1: Determine whether the given description to an observational study or an experiment.
In a study of 424 children with particular disease, the subjects were monitored with an EEG while asleep.
Does the given description correspond to an observational study or an experiment?
A. The given description corresponds to an experiment.
B. The given description corresponds to an observational study.
C. The given description does not provide enough information to answer this question.
Question 2: Identify the type of sampling used (random, systematic, convenience, stratified, or cluster sampling) in the situation described below.
A man is selected by a marketing company to participate in a paid focus group. The company says that the man was selected because he was randomly chosen from all men in his tax bracket.
Which type of sampling did the marketing company use?
A. Systematic sampling, B. Random Sampling, C. Convenience Sampling, D. Cluster sampling, E. Stratified sampling
Question 3: Identify which type of sampling is used: random, systematic, convenience, stratified, or cluster.
A radio station asks its listeners to call in their opinion regarding the format of the morning show.
Which type of sampling is used?
A. Cluster, B. Stratified, C. Random, D. Systematic, E. Convenience
Question 4: Identify the type of sampling used (random, systematic, convenience, stratified, or cluster sampling) in the situation described below.
A woman experienced a tax credit. The tax department claimed that the woman was audited because everyone in four randomly selected districts was being audited.
Which type of sampling did the tax department use?
A. Cluster sampling
B. Systematic Sampling
C. Convenience Sampling
D. Random Sampling
E. Stratified Sampling
In: Statistics and Probability
Runiowa is a fashion shoe company that tries to manufacture much more durable heels in 2020. The management team of Runiowa suggests two rubber materials A and B and the research team of Runiowa is asked to design an experiment to gauge whether the rubber A is more durable than the rubber B. 300 people in the US aged between 18 and 65 were randomly chosen. The rubber A is allocated at random to the right shoe or the left shoe of each individual. Then, the rubber B has been assigned to the other. For example, if Mr. Nathaniel is one of 300 people randomly chosen, then the right heel of Mr. Nathaniel is randomly assigned to be made with the rubber A and then his left heel is to be made with the rubber B. The research team measures the amounts of heel wear both the rubber A (wA) and the rubber B (wB) in each individual and records the difference wA − wB of 300 individuals. Even though the individuals are heterogeneous with different heights and weights, those individual heterogeneities will not obscure the comparison of treatment groups by focusing on the paired differences of each individual. Also as long as the heel materials are randomly assigned for each individual, there has been no restrictions on shoe styles. Note that the age of subjects is ranging from 18 to 65. In this way, researchers compare treatments within blocks controlling heterogeneity of individuals. The research team also repeats this experiment design with 300 people in the US aged between 18 and 65 chosen at random.
Question:
Is there a conjecture?
What is the response variable?
What is the explanatory variable?
What levels of the factor(s) were used in the expereiment?
What are the treatments for this experiment?
What are the experimental units?
What is the control?
Hoe much replication was used?
How was randomization used?
In: Statistics and Probability
8. A pet food company has a business objective of expanding its product line beyond its current kidney- and shrimp-based cat foots. The company developed two new products, one based on chicken liver and the other based on salmon. The company conducted an experiment to compare the two new products with its two existing ones, as well as a generic beef-based product sold in supermarket chains.
For the experiment, a sample of 50 cats from the population at a local animal shelter was selected. Ten cats were randomly assigned to each of the five products being tested. Each of the cats was then presented with 3 ounces of the selected food in a dish at feeding time. The researchers defined the variable to be measured as the number of ounces of food that the cat consumed within a 5-minute period that began when the filled dish was presented to the cat. The results for this experiment are summarized in CatFood.
a. At the 0.05 level of significance, is there evidence of a differences in the mean amount of food eaten among the various products?
b. Does the result in (a) give you statistical permission to probe for individual differences between the food products?
| Kidney | Shrimp | Chicken Liver | Salmon | Beef |
| 3.00 | 0.25 | 2.61 | 0.63 | 2.97 |
| 3.00 | 0.34 | 3.00 | 1.37 | 2.85 |
| 1.47 | 0.16 | 3.00 | 3.00 | 0.44 |
| 2.62 | 0.72 | 2.43 | 0.54 | 0.90 |
| 1.51 | 3.00 | 3.00 | 3.00 | 3.00 |
| 3.00 | 3.00 | 3.00 | 2.55 | 3.00 |
| 3.00 | 0.29 | 2.85 | 2.93 | 0.48 |
| 3.00 | 3.00 | 0.18 | 2.73 | 0.45 |
| 0.74 | 1.99 | 0.98 | 3.00 | 3.00 |
| 2.21 | 2.30 | 1.51 | 0.96 | 0.05 |
In: Statistics and Probability
A pet food company has a business objective of expanding its product line beyond its current kidney- and shrimp-based cat foots. The company developed two new products, one based on chicken liver and the other based on salmon. The company conducted an experiment to compare the two new products with its two existing ones, as well as a generic beef-based product sold in supermarket chains.
For the experiment, a sample of 50 cats from the population at a local animal shelter was selected. Ten cats were randomly assigned to each of the five products being tested. Each of the cats was then presented with 3 ounces of the selected food in a dish at feeding time. The researchers defined the variable to be measured as the number of ounces of food that the cat consumed within a 10-minute period that began when the filled dish was presented to the cat. The results for this experiment are summarized in CatFood.
a. At the 0.05 level of significance, is there evidence of a differences in the mean amount of food eaten among the various products?
b. Does the result in (a) give you statistical permission to probe for individual differences between the food products?
Please show me how to do this in excel using the data analysis tab
| Kidney | Shrimp | Chicken Liver | Salmon | Beef |
| 2.37 | 2.26 | 2.29 | 1.79 | 2.09 |
| 2.62 | 2.69 | 2.23 | 2.33 | 1.87 |
| 2.31 | 2.25 | 2.41 | 1.96 | 1.67 |
| 2.47 | 2.45 | 2.68 | 2.05 | 1.64 |
| 2.59 | 2.34 | 2.25 | 2.26 | 2.16 |
| 2.62 | 2.37 | 2.17 | 2.24 | 1.75 |
| 2.34 | 2.22 | 2.37 | 1.96 | 1.18 |
| 2.47 | 2.56 | 2.26 | 1.58 | 1.92 |
| 2.45 | 2.36 | 2.45 | 2.18 | 1.32 |
| 2.32 | 2.59 | 2.57 | 1.93 | 1.94 |
In: Math
Synthesis of banana oil
this is my discussion and below it is the professors comments. Can you correct it for me?
Discussion:
The purpose of the experiment performed was to synthesize isopentyl acetate, or more commonly known as banana oil, from isopentyl alcohol and glacial acetic acid. Both liquids were combined and remained clear until the addition of concentrated sulfuric acid. The addition of an acid caused a reaction to occur that resulted in a polymer, which gave the mixture color. The liquid turned dark brown when the polymer formed. The polymer was removed via distillation.
After creating the ester, the biggest challenge was separating it from all the other chemicals in the flask, which include water, acetic acid, sulfuric acid, and isopentyl alcohol. The isopentyl acetate is not soluble in water; so two layers are created in the separatory funnel. While the product will not dissolve in water, both acetic acid and sulfuric acid are. Since the acids are also slightly in the organic layer, sodium bicarbonate is used to convert the acids into a salt, which is soluble in water, as seen in Figure 2. The sodium bicarbonate reaction also produces carbonic acid, which is unstable and will break down into water and CO2 as seen in Figure 3. The little amount of unreacted isopentyl alcohol is separated by removing the first ten drops of product during the distillation. This is possible because it has a lower boiling point then the ester that is produced.
This whole reaction was first done and called the Fischer esterification. The general reaction scheme can be seen above. In the experiment, isopentyl alcohol and acetic acid react in the presence of sulfuric acid to form isopentyl acetate. This reaction mechanism can be seen below in Figure 5.
Figure 5 – Fischer esterification mechanism
In the reaction above, Acetic acid takes a proton from concentrated sulfuric acid. Then it reacts with isopropentyl alcohol. Delocalization of electrons and protons in later steps leads to elimination of water molecule to form final product.
During the distillation, the temperature was recorded for the boiling point of isopentyl acetate. It was found to be at 129 degrees Celsius. This is low compared to the literature value of 142 degrees Celsius. This difference is believed to be due to the different typed of thermometers used to record the boiling point. The literature value most likely used a mercury thermometer, while the experiment performed used an alcohol thermometer. Alcohol thermometers are not as accurate as mercury thermometers, but have to be used due a law banning mercury thermometers in the state of New Jersey. Because of this, not much can be said about the purity of the substance obtained.
While an NMR was not taking during the experiment, nor was an IR spectrum, attached is both an NMR example and an IR example of isopentyl acetate. On the NMR sheet, all hydrogen peaks are labeled on a drawing of the molecule, and all the information is written in the chart. On the IR sheet, peaks above 1500 cm-1 are labeled for what they are believed to be. While the pictures were found from a reputable source, they were not from the experiment performed so one cannot be perfectly sure that the desired substance was obtained.
The last topic for the reaction was the product loss. Most organic reactions are not perfect and have relatively low yields. The 42 percent yield was most likely due to side reactions occurring in the flask. Since there were several chemicals in the flask together, side reactions are bound to occur, creating a lower yield. For example, a polymer was formed which changed the color of the flask. This was not intended to happen so this reaction lowered the yield of the experiment.
professors comments.
You need to explain what substances should have been in the flask at the end of refluxing. Show a flow
diagram of the purification process. You did an extraction, a drying and a distillation.
Alcohol thermometers are quite accurate in the range they are designed to function. This type of
alcohol thermometer was not designed to work around 142 degrees C.
Treat the NMR and IR spectra as if they were obtained by you for your product. You should have tables
in your Results Section for IR, NMR and GC. You should discuss these results in more detail in the
discussion. What do we learn from each measurement.
Give more details of the many causes for obtaining less than 100% Yield.
Discuss whether this experiment is a good example of Green Chemistry. Does the Atom efficiency make
it a good candidate. Does the Reaction efficiency make it a good example of Green chemistry. What
about other things.
In: Chemistry
Account Type: SAVING
Account Owner: Customer B
Interest Rate: 1.1
Balance: 500.0
Account Type: CHECKING
Account Owner: Customer A
Interest Rate: 1.2
Balance: 200.0
Cannot withdraw more than account balance
Account Type: SAVING
Account Owner: Customer B
Interest Rate: 1.1
Balance: 700.0
Account Type: CHECKING
Account Owner: Customer A
Interest Rate: 1.2
Balance: 200.0
solution provided:
package csci1011.csci1011.lab8;
import java.text.NumberFormat;
import java.text.DecimalFormat;
class Account {
enum ACCOUNT_TYPE
{
CHECKING,SAVINGS;
}
//create five variable
private ACCOUNT_TYPE aType;
private String accountOwner;
private double interestRate;
private double Balance;
//setter and getter methods
private ACCOUNT_TYPE getaType() {
return aType;
}
private void setaType(ACCOUNT_TYPE aType) {
this.aType = aType;
}
private String getAccountOwner() {
return accountOwner;
}
private void setAccountOwner(String accountOwner) {
this.accountOwner = accountOwner;
}
private double getInterestRate() {
return interestRate;
}
private void setInterestRate(double interestRate) {
this.interestRate = interestRate;
}
private double getBalance() {
return Balance;
}
private void setBalance(double Balance) {
this.Balance = Balance;
}
void initialize(ACCOUNT_TYPE atype,String name,double interestRate,double balance)
{
setaType(atype);
setAccountOwner(name);
setInterestRate(interestRate);
setBalance(balance);
}
void display()
{
System.out.println(" Account Type : "+aType.name());
System.out.println("Account Owner : "+getAccountOwner());
System.out.println("Interest Rate : "+getInterestRate());
System.out.println("Balance : "+getBalance());
}
//deposit to the account
void deposit(double amount)
{
double bal=getBalance();
setBalance(bal+amount);
System.out.println("Successfull Deposit ! ");
}
//withdraw from account
void withdraw(double amount)
{
//if balance is 0 or amount is greater than balance
if(getBalance()<=0 || amount>getBalance())
{
System.out.println("You don't have enough money to withdraw");
}
else
{
double bal=getBalance();
setBalance(bal-amount);
System.out.println("Successfully withdrawn ");
}
}
public static void main(String args[])
{
//two objects
Account ac=new Account();
Account ac1=new Account();
//initialize with Savings
ac.initialize(ACCOUNT_TYPE.SAVINGS, "Customer B ", 1.1, 500);
//initialize with Checking
ac1.initialize(ACCOUNT_TYPE.CHECKING, "Customer A ", 1.2, 200);
//display
ac.display();
ac1.display();
//deposit into account
ac.deposit(20);
//withdraw from account
ac.withdraw(60);
//display
ac.display();
}
}
In: Computer Science