An experiment was conducted to test the effect of a new drug on a viral infection. After the infection was induced in 100 mice, the mice were randomly split into two groups of 50. The first group, the control group, received no treatment for the infection, and the second group received the drug. After a 30-day period, the proportions of survivors, p̂1 and p̂2, in the two groups were found to be 0.36 and 0.64, respectively.

(a) Is there sufficient evidence to indicate that the drug is effective in treating the viral infection? Use α = 0.05.

(b) Find the test statistic and rejection region. (Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused region.)

(c) Use a 95% confidence interval to estimate the actual difference (p₁ − p₂) in the survival rates for the treated versus the control groups. (Round your answers to two decimal places.)

Three couples and two individuals (eight total people) have been invited to an investment seminar and have agreed to attend.

- Suppose the probability that any particular couple or individual arrives late is .4 (a couple will travel together in the same vehicle, so either both people will be on time or else both people will arrive late).

- Assume that different couples and individuals are on time or late independently of one another.

- Calculate the probability that 6 or more people arrive at the seminar on time.

Hint: consider the disjoint cases: exactly 6 are on time, exactly 7 are on time, exactly 8 are on time, then think of all the combinations of couples and individuals that yield that amount.

Please dont copy the answer from another post and explain !!!

Two insulation thickness alternatives have been proposed for a process steam line subject to severe weather conditions. One alternative must be selected. Estimated savings in heat loss and installation cost are given below:

b.) Useful life of 2 cm Thickness alternative is 8 years and useful life of 5 cm thickness alternative is 7 years. Which thickness would you recommend for a MARR= 9% per year and negligible market (salvage) values using Present Worth Analysis and Annual Worth Analysis ? (Use repeatability assumption)

IN EXCEL PLZ!!

5

Which analysis tool that we studied (from the bulleted list below) should one use for each situation listed in a through d below? Also, state the null and alternative hypothesis for each situation.

a. Testing to see if a brand of tires marketed as lasting 60,000 miles does last 60,000 miles. Your sample size is 100 tires.

b. Testing to see if the failures recorded for these tires follow an exponential distribution.

c. Testing to see if samples of Firestone and Goodyear 60,000-mile tires have the same life. Your sample size is 15 Firestone and 20 Goodyear tires.

d. Testing two brands of tires to see if the age of the driver affects the number of gallons of gas used on a 500-mile trip for five different age groups of drivers.

Choose the best tool to apply to each test (a-d) from the list below.

Z-test / Normal distribution one sample versus threshold(s)

Z-test comparison of two samples T-Test one sample versus threshold(s)

T-test Two Sample assuming unequal variances

T-test Paired data (dependent samples)

Binomial test of one proportion versus threshold(s)

Binomial Test of two proportions

Chi-Square Contingency Table

Chi-Square Goodness of Fit

Human Genetics - What will your children be?

The purpose of this exercise is to give you some practice working with the major concepts of genetics using specific example of inherited traits present in humans. During this lab, we will observe patterns of inheritance for three general types of traits: autosomal traits controlled by two alleles (hair color, Rh factor, PTC tasting), an autosomal trait controlled by multiple alleles (blood type), and a sex-linked trait (color-blindness). Before we begin this exercise, we should introduce or review some of the terminology used when discussing genetics.

All of an organism's traits, whether they are visible traits such as hair and eye color or non-visible traits such as blood type, are controlled by genes. A gene is a specific region on a strand of DNA (deoxyribonucleic acid) that controls a specific trait. The 46 long strands of DNA in human cells, each containing many genes, are wound into chromosomes which become visible during mitosis (remember Ex.7). Most human cells are diploid, meaning they contain two sets of genetic information. Diploid human cells contain 23 pairs of chromosomes (46 chromosomes total). A pair of these chromosomes are referred to as homologous chromosomes because both chromosomes have genes that code for the same traits. One of the chromosomes in each pair was inherited from the mother and the other from the father.

While diploid cells have two copies of each gene, the copies may not be identical. For example, each of our cells has two genes that control hair color, but one gene may code for light hair and one may code for dark hair. The different versions of a gene are called alleles. The different alleles of a gene are often not expressed equally. For each trait, one allele is usually dominant and the other is recessive. A dominant allele is expressed whenever it is present, even if only one copy is present.
A recessive allele must be present in two copies to be expressed.

When we write out a genetics problem, we often use letters to denote the different alleles for a particular trait. An uppercase letter denotes a dominant allele, and a lowercase letter denotes a recessive allele. As we will see later, other notations may also be used to indicate which alleles are present. If a person has two copies of the same allele (AA or aa), they are considered homozygous for that trait. If they possess two different alleles (Aa), they are considered heterozygous for that trait. When we refer to the actual genetic makeup of an individual, the alleles that are present, this is called the genotype. If we refer to the outward appearance of the individual, or what trait is expressed, then this is referred to as the phenotype of that individual.

Next, we will consider the specific traits that we will be working with in lab today.
Autosomal traits controlled by two alleles: Genes for these traits are carried on two homologous autosomal (non-sex) chromosomes. There are only two alleles or versions for each gene present in the population.

Ability to taste PTC: PTC (phenylthiocarbimide) is a chemical that tastes bitter to some people and is tasteless to others. The ability to taste PTC is an inherited trait, and the ability to taste PTC is dominant (T) over the non-taster (t) condition.

Hair Color: In this case, we will consider only dark hair or light hair. Dark hair (B) is dominant over light hair (b).

Rh factor: The Rh factor is an antigen found on red blood cells. It is named after the Rhesus monkey where it was discovered. People who possess this antigen on their red blood cells are considered Rh+ (dominant), while those without the antigen are Rh-(recessive). The Rh factor is of concern if an Rh- mother is carrying an Rh+ child, since the mother's body may form antibodies to the child's blood, resulting in numerous health problems.

Autosomal traits controlled by multiple alleles: Genes for these traits are also carried on homologous autosomal chromosomes, but there are more than two alleles (versions) of the gene present in the population. However, only two of these alleles can be
present in any one individual.

Blood type: A person's blood type or phenotype (A, B, AB, O) is determined by the presence of certain antigens on the surface of their red blood cells. Control for the inheritance of blood types is based on three different alleles: IA, IB, and IO. Again, only two of these alleles will be present in any one individual. The IA allele codes for the "A" antigen, the IB allele codes for the "B" antigen. The presence of the IO allele results in no antigen on the surface of the red blood cell. A person's blood type (phenotype) is the result of the combination of these alleles. The IA and IB alleles are co-dominant and are equally expressed if both are present. The IA and IB are both dominant to the IO allele. The only way for a person to have type O blood is to have both recessive alleles. Fill in the table at the right with the proper phenotype (blood type) for each possible genotype.

Sex chromosomes and sex-linked traits: Sex in humans is determined by the presence or absence of certain whole chromosomes. Females have two X chromosomes and males have an X and a Y chromosome. Genes for sex-linked traits are usually carried on the X chromosome and are absent on the homologous Y chromosome. As a result, males only have a single copy of any gene located on the X chromosome and they will express whichever allele is present, even if it is normally a recessive allele.

Color-blindness: The gene that controls color vision is located on the X chromosome and has two alleles: normal (X) and color-blind (X°). The normal condition is dominant over the color-blind condition. Females that are phenotypically normal, but heterozygous for color-blindness (X X°) are often referred to as carriers

Now that you have become familiar with the basic terms of genetics and the specific traits that we will observe today, you are ready to put this knowledge to use. To do this, we will play a "family" game. During the course of the lab, each of you will determine your genetic makeup, find a mate, and determine the combination of traits that would be possible in your offspring. Here is how it works:

Step 1: Names - These will be used to determine your mates. Men will draw a last name from the beaker marked males and women will draw a last name from the beaker marked female. If we do not have equal numbers of men and women, some people may have multiple mates.

Step 2: Determine sex & sex-linked traits - Men will draw only ONE X chromosome and ONE Y chromosome. Women will draw TWO X chromosomes. The X chromosome will have an allele for normal vision or color blindness.

Step 3: Determine autosomal traits - In order to find out your traits, you will draw TWO chromosomes with alleles for each of the following traits: hair color, ability to taste PTC, Rh factor, and blood type.

Step 4: Take a social break and find your mate. When you find someone with the same last name, have a seat beside your new mate.

Step 5: Once you have become acquainted with your mate;-) arrange your alleles in front of you in the order shown below for Jane Smith.

Genotype XX

Bb
IA IO Rh+Rh- TT

Phenotype
Female with normal vision Dark hair
Blood type A
Rh+
PTC Taster

Step 6: Fill in the table below with the information for you and your mate. If needed, review the information on pages 1-2

MY RESULTS

my phenotypes: SEX/VISION = xBxb (female normal vision), BLOOD TYPE=IAIA (A), +/- BLOOD TYPE = Rh-Rh- (-), PTC TASTING = Tt (taster), HAIR COLOR = BB (dark)

Mates phenotypes: SEX/VISION = xbY (colorblind male), BLOOD TYPE= IAIO (A), +/- BLOOD TYPE = Rh- Rh- (-), PTC TASTING = tt (non taster), HAIR COLOR= Bb (dark)

Mates Phenotypes:

Step 7: You are now ready to determine all the possible gametes that each of you could contribute to your offspring. Remember, gametes are formed by meiosis, which is reduction division. Each gamete will only get one allele from each pair of homologous chromosomes.

The number of possible gametes is equal to 2 (because we are dealing with 2 alleles) raised to the ?n?

power, where ?n? equals the number of pairs of heterozygous alleles. For example, looking at Jane Smith=s genotype we can see that she is heterozygous for three traits: hair color (Bb), blood type (IAIO), and Rh factor (+-). According to our formula, she could have (23) = 8 different gametes. How many possible gametes could you have? _____ How many possible gametes could your mate have?_____

Step 8: Each person should now determines their own gametes and writes them below. The method illustrated below using Jane Smith=s genotype will allow you to systematically find all possible gametes for you and your mate. Each column represents one possible gamete. Jane could have eight different gametes (23 = 8), so there should be eight columns shown below when we are finished. You may have more or less than eight columns depending on the number of heterozygous pairs of alleles that you have. Choose one trait that Jane (or you) is heterozygous for (ex., Rh factor) and give the possible gametes. (Hint: there should be 2, one for each allele).

Use the space below to write out all of the possible gamete combinations for you and your partner.

The total number of different looking children that you and your mate could produce can be determined by multiplying your number of possible gametes times your mate's number of possible gametes. Considering only these traits, how many different children could you and your mate produce?_________

Look at your gamete possibilities and genotypes and those of your mate to answer the following questions. Use simple Punnett squares to determine the possible outcomes of the crosses where needed.

1. Will any of your children be color-blind?

2. What are the possible hair colors of your children?

3. Will any of your children be non-tasters?

4. What are the possible blood types of your children?

5. Will any of your children be blood type O-?

6. Will any of your children be AB+?

7. Could you possibly produce a child who would have dark hair, could taste PTC, and have AB- blood?

8. Would you bet on the sex of your next child? Why or why not?

9. If you were a betting person, are there any traits, concerning your child, on which you would bet? Explain.

10. Today we can determine the genetic makeup of individuals and determine whether they carry traits that are potentially harmful such as Huntington=s disease. Knowing this, would you participate in a genetic typing for certain various characteristics before you get married? Why or why not?

1.An individual has $15,000 invested in a stock with a beta of 0.5 and another $50,000 invested in a stock with a beta of 1.6. If these are the only two investments in her portfolio, what is her portfolio's beta? Do not round intermediate calculations. Round your answer to two decimal places.

Suppose that the five measured SampleRTT values are
130 ms, 106 ms, 150 ms, 140 ms, and 115 ms.

1) Compute the EstimatedRTT after each of these SampleRTT values is obtained, using a value of α = 0.125 and assuming that the value of EstimatedRTT was 120 ms just before the first of these five samples were obtained.

2) Compute also the DevRTT after each sample is obtained, assuming a value of β = 0.25 and assuming the value of DevRTT was 10 ms just before the first of these five samples was obtained.

3) Last, compute the TCP TimeoutInterval after each of these samples is obtained.

4) Will there be a timeout event for any of the packages? Explain. (hint you can use Excel in this one like we did in class)

1. Calculate the future value of a single amount. You were left $3500 from your grandmother's will and have decided to invest it. You have a couple of investment options in mind and are curious how much $3500 will be worth at the end of years 1 to 5, assuming a constant interest rate of 8%.
2. Calculate the doubling period. You're also curious when your money will double, assuming the same 8% interest rate and want to use the "Rule of 69" for the most accurate prediction.
3. Calculate the present value of a single future amount. Your grandmother was a finance professor at Champlain College and, for some reason, always challenged you to "think critically" and "figure out the math" for just about every financial decision you made in your life...and certainly when it came to asking her for money. In her will she left you one last tricky set of options - two additional options for the cash, one of which was to receive the $3500 three years from now. She asked you to calculate the present value of the future $3500 using an ROI of 8% and make a decision on whether you want the cash three years from now, or not.
4. Calculate the future value of an annuity. The final option your grandmother presented to you was to receive a regular annuity payment $700 over 5 years and she wanted you to calculate the annuity's future value with an ROI of 8%. Do you chose this option? Why or why not?