An article in the Journal of the American Medical Assocation
described an experiment to investigate the effect of four
treatments on various body characteristics. In this double-blind
experiment, each of 59 female subjects age 65 or older was assigned
at random to one of the following four treatments:
1) P + P: placebo "growth hormone" and placebo "steroid"
2) P + S: placebo "growth hormone" and the steroid estradiol
3) G + P: growth hormone and placebo "steroid"
4) G + S: growth hormone and the steroid estradiol
The change in body fat mass was measured over the 26-week period
following the treatments. Using the information provided above and
in the partial table below, complete the ANOVA table (use 2
decimals for your answers).
Source of Variation | df | Sum of Squares | Mean Square | F |
Treatment | ||||
Error | 1.4 | X | ||
Total | 226.37 | X | X |
In: Statistics and Probability
In: Statistics and Probability
The Census Bureau gives this distribution for the number of people in American households in a certain year.
Family Size | Proportion |
---|---|
1 | 0.29 |
2 | 0.34 |
3 | 0.15 |
4 | 0.13 |
5 | 0.06 |
6 | 0.02 |
7 | 0.01 |
Note: In this table, 7 actually represents households of size 7 or greater. But for purposes of this exercise, assume that it means only households of size exactly 7.)
(a)
This is also the probability distribution for the size of randomly chosen households. The expected value of this distribution is the average number of people in a household. What is this expected value?
people
(b)
Suppose you take a random sample of 1,000 American households. About how many of these households will be of size 2?
households
About how many of these households will be of sizes 3 to 7?
households
(c)
Based on your calculations in part (b), how many people are represented in your sample of 1,000 households? (Hint: The number of individuals in your sample who live in households of size 7 is 7 times the number of households of size 7. Repeat this reasoning to determine the number of individuals in households of sizes 1 to 6. Add the results to get the total number of people represented in your sample.)
people
Calculate the probability distribution for the household size lived in by individual people. (Round your answers to four decimal places.)
Family Size | Proportion |
---|---|
1 | |
2 | |
3 | |
4 | |
5 | |
6 | |
7 |
Describe the shape of this distribution. What does this shape tell you about household structure? (Round your mean to one decimal place.)
The distribution is ---Select--- left skewed symmetric right skewed , with a median of and a mean of around people.
In: Statistics and Probability
In June 2008, the Journal of the American Academy of Dermatology reported on the use of an intermittent therapy for flare prevention and long-term disease control of eczema (red, inflamed, itchy skin). Eczema is a chronic skin disease that affects about 10% to 20% of infants and about 3% of adults and children in the United States. Researchers wanted to determine whether intermittent dosing with a specific topical inhibitor is effective in preventing eczema flare-ups in patients with stabilized eczema. Initially, 383 adults and children with moderate to severe eczema were treated with daily doses of the topical inhibitor. After 16 weeks, 197 were disease free and admitted into the maintenance phase of the study. These patients were randomized to a 3-times-weekly treatment with either the topical inhibitor or a placebo for 40 weeks. All tubes of ointments (topical inhibitor and placebo) were packaged in identical boxes and labeled “for investigational use only.” Then the packages were sealed with tamper-proof seals and shipped. Neither the researchers nor the patients knew who received the topical inhibitor or who received the placebo. During the maintenance phase researchers tracked the number of flare-free days for each patient.
Which of the following design features contributes to double
blinding?
A.Randomization of patients to one of two groups
B. Distinguishing the packaging for both treatments
C. The number of flare-free days for each patient
In: Statistics and Probability
Assume that there is no policy response to the increased saving by American families and consider the transition from the short-run to the medium-run (the medium-run is some intermediary or a transition state between the short-run and the long-run) and back to the long run. True or False. The US will see increased inflation during the transition back to long-run equilibrium. If true, explain why. If false, explain why not. (use the IS-LM and AS-AD logic to answer)
In: Economics
There is an American put option on a stock that expires in two months. The stock price is $55, and the standard deviation of the stock returns is 64 percent. The option has a strike price of $62, and the risk-free interest rate is an annual percentage rate of 4.4 percent. What is the price of the option? Use a two-state model with one-month steps. (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.)
What is Put value?
In: Finance
There is an American put option on a stock that expires in two months. The stock price is $100 and the standard deviation of the stock returns is 72 percent. The option has a strike price of $112 and the risk-free interest rate is an annual percentage rate of 5.6 percent.
What is the price of the option? Use a two-state model with one-month steps
In: Finance
There is an American put option on a stock that expires in two months. The stock price is $69 and the standard deviation of the stock returns is 59 percent. The option has a strike price of $78 and the risk-free interest rate is an annual percentage rate of 5.8 percent.
What is the price of the option? Use a two-state model with one-month steps. (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.)
In: Finance
"To Breakfast or Not to Breakfast?" by Richard Ayore
In the American society, birthdays are one of those days that
everyone looks forward to. People of different ages and peer groups
gather to mark the
18th, 20th, ,
birthdays. During this time, one looks back to see what he or
she has achieved for the past year and also focuses ahead for more
to come.
If, by any chance, I am invited to one of these parties, my
experience is always different. Instead of dancing around with my
friends while the music is booming, I get carried away by memories
of my family back home in Kenya. I remember the good times I had
with my brothers and sister while we did our daily routine.
Every morning, I remember we went to the shamba (garden) to weed
our crops. I remember one day arguing with my brother as to why he
always remained behind just to join us an hour later. In his
defense, he said that he preferred waiting for breakfast before he
came to weed. He said, "This is why I always work more hours than
you guys!"
And so, to prove him wrong or right, we decided to give it a try.
One day we went to work as usual without breakfast, and recorded
the time we could work before getting tired and stopping. On the
next day, we all ate breakfast before going to work. We recorded
how long we worked again before getting tired and stopping. Of
interest was our mean increase in work time. Though not sure, my
brother insisted that it was more than two hours. Using the data in
the table below, solve our problem. (Use
α = 0.05)
Work hours with breakfast | Work hours without breakfast |
---|---|
8 | 6 |
6 | 4 |
8 | 4 |
5 | 4 |
9 | 7 |
8 | 7 |
10 | 7 |
7 | 5 |
6 | 6 |
9 | 5 |
NOTE: If you are using a Student's t-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
Part (a)
State the null hypothesis.H0: μd > 0
H0: μd ≠ 0
H0: μd < 0
H0: μd ≥ 0
H0: μd = 0
Part (b)
State the alternative hypothesis.Ha: μd < 0
Ha: μd ≤ 0
Ha: μd > 0
Ha: μd ≥ 0
Ha: μd ≠ 0
Part (c)
In words, state what your random variableXd
represents.Xd
represents the average work times of the 10 days.Xd
represents the average difference in work times on days when eating breakfast and on days when not eating breakfast.Xd
represents the total difference in work times on days when eating breakfast and on days when not eating breakfast.Xd
represents the difference in the average work times on days when eating breakfast and on days when not eating breakfast.Part (d)
State the distribution to use for the test. (Enter your answer in the form z or tdf where df is the degrees of freedom.)Part (e)
What is the test statistic? (If using the z
distribution round your answer to two decimal places, and if using
the t distribution round your answer to three decimal
places.)
---Select--- z t =
Part (f)
What is the p-value?p-value < 0.0100.010 < p-value < 0.050 0.050 < p-value < 0.100p-value > 0.100
H0
is false, then there is a chance equal to the p-value that the sample average difference between work times on days when eating breakfast and on days when not eating breakfast is less than 2.1.IfH0
is true, then there is a chance equal to the p-value that the sample average difference between work times on days when eating breakfast and on days when not eating breakfast is less than 2.1. IfH0
is true, then there is a chance equal to the p-value that the sample average difference between work times on days when eating breakfast and on days when not eating breakfast is at least 2.1.IfH0
is false, then there is a chance equal to the p-value that the sample average difference between work times on days when eating breakfast and on days when not eating breakfast is at least 2.1.Part (g)
Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value.Part (h)
Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.(i) Alpha (Enter an exact number as an integer, fraction, or decimal.)reject the null hypothesisdo not reject the null hypothesis
Since p-value < α, we reject the null hypothesis.Since p-value > α, we do not reject the null hypothesis. Since p-value > α, we reject the null hypothesis.Since p-value < α, we do not reject the null hypothesis.
There is sufficient evidence to conclude that the mean difference in work times on days when eating breakfast and on days when not eating breakfast has increased.There is not sufficient evidence to conclude that the mean difference in work times on days when eating breakfast and on days when not eating breakfast has increased.
Part (i)
Explain how you determined which distribution to use.The t-distribution will be used because the samples are dependent.The standard normal distribution will be used because the samples involve the difference in proportions. The standard normal distribution will be used because the samples are independent and the population standard deviation is known.The t-distribution will be used because the samples are independent and the population standard deviation is not known.
In: Statistics and Probability
"To Breakfast or Not to Breakfast?" by Richard Ayore
In the American society, birthdays are one of those days that
everyone looks forward to. People of different ages and peer groups
gather to mark the
18th, 20th, ,
birthdays. During this time, one looks back to see what he or
she has achieved for the past year and also focuses ahead for more
to come.
If, by any chance, I am invited to one of these parties, my
experience is always different. Instead of dancing around with my
friends while the music is booming, I get carried away by memories
of my family back home in Kenya. I remember the good times I had
with my brothers and sister while we did our daily routine.
Every morning, I remember we went to the shamba (garden) to weed
our crops. I remember one day arguing with my brother as to why he
always remained behind just to join us an hour later. In his
defense, he said that he preferred waiting for breakfast before he
came to weed. He said, "This is why I always work more hours than
you guys!"
And so, to prove him wrong or right, we decided to give it a try.
One day we went to work as usual without breakfast, and recorded
the time we could work before getting tired and stopping. On the
next day, we all ate breakfast before going to work. We recorded
how long we worked again before getting tired and stopping. Of
interest was our mean increase in work time. Though not sure, my
brother insisted that it was more than two hours. Using the data in
the table below, solve our problem. (Use
α = 0.05)
State the distribution to use for the test. (Enter your answer in the form z or tdf where df is the degrees of freedom.)
Part (e)
What is the test statistic? (If using the z distribution round your answer to two decimal places, and if using the t distribution round your answer to three decimal places.)
Part (f)
What is the p-value?
Work hours with breakfast | Work hours without breakfast |
---|---|
8 | 6 |
6 | 5 |
10 | 6 |
5 | 4 |
9 | 7 |
8 | 7 |
10 | 7 |
7 | 5 |
6 | 6 |
9 |
5 |
In: Statistics and Probability