Questions
An article in the Journal of the American Medical Assocation described an experiment to investigate the...

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

It is known that the average American orders food from a restaurant in person or as...

  1. It is known that the average American orders food from a restaurant in person or as take out 4.2 times a week on average. Suppose we take a sample of 30 Americans and find that the average amount of meals ordered out is 3.2 per week with a sample standard deviation of 2.82 meals. We are interested in testing if the average amount of meals ordered out is actually lower than 4.2 now.
    1. Using the sample of 30 adults, what would the 95% confidence interval be for the population mean?
    2. What are the null and alternative hypotheses?
    3. What is the critical value at 95% confidence?
    4. Calculate the test statistic.
    5. Find the p-valu
    6. What conclusion would be made here at the 95% confidence level?
    7. Would my conclusion change if I changed alpha to .01? Show reasonin

In: Statistics and Probability

The Census Bureau gives this distribution for the number of people in American households in a...

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...

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...

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...

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...

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...

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...

"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 variable

    Xd

    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



    Explain what the p-value means for this problem.If

    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.If

    H0

    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.    If

    H0

    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.If

    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 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.)
    α =

    (ii) Decision:

    reject the null hypothesisdo not reject the null hypothesis    


    (iii) Reason for decision:

    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.


    (iv) Conclusion:

    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...

"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