An engineering firm with a good track record is known to have a 40% success rate in
getting state-government construction contracts. In a recent year, the firm submitted bids
on eight construction projects to be funded by the state-government. The bids for
different projects are assessed independently of each other.
i) CHOOSE which of these probability distributions is most appropriate to describe a random variable X defined as "the number of approved state-government construction contracts bid by the engineering firm in the recent year". *
X~Poisson(8)
X~Po(3.2)
X~Binomial(8,0.4)
X~Negative Binomial(8,0.4)
X~Geometric(0.4)
ii) Using the random variable X in question 1(i), which of the following mathematical expressions indicates: the probability that the engineering firm will not get any state-government construction contracts that they have bid in the recent year? *
P(X=8)
P(X > 1)
1 - P(X=0)
P(X is at most 0)
iii) Hence, which of the following answers is correct for the probability that the firm will not get any state-government construction contracts that they have bid in the recent year? *
0.0168
0.0408
0.6866
0.3134
0.9832
Y~Hypergeometric(8,2,5)
Y~Negative Binomial(2, 0.0408)
Y~Geometric(0.6)
Y~Binomial(8, 0.6)
Y~Negative Binomial(2, 0.0168)
Y~Negative Binomial(2, 0.6)

A study was conducted to explore the effects of ethanol on sleep time. Fifteen rats were randomized to one of three treatments. Treatment 1 got only water (control). Treatment 2 got 1g of ethanol per kg of body weight, and treatment 3 got 2g/kg of ethanol. The amount of REM sleep in a 24hr period was recorded, in minutes. Data are below:

Treatment 1: 63, 54, 69, 50, 72

Treatment 2: 45, 60, 40, 56

Treatment 3: 31, 40, 45, 25, 23, 28

a) Create 95% CIs for all pairwise comparisons of means using the Tukey method. Do this by hand and show your work. You may use R to check your answers. Summarize your results using letter codes. What do you conclude?

b) What would have been the multiplier if Fisher’s LSD method had been used? Bonferroni? Use a t table. You may check your answer in R. You do not need to make the complete CIs for this question, just give the values of the multipliers.

In 2010, the United Nations claimed that there was a higher rate of illiteracy in men than in women from the country of Qatar. A humanitarian organization went to Qatar to conduct a random sample. The results revealed that 45 out of 234 men and 42 out of 251 women were classified as illiterate on the same measurement test. Do these results indicate that the United Nations' findings were correct?

1. Test an appropriate hypothesis and state your conclusion.

2. Find a 95% confidence interval for the difference in the proportions of illiteracy in men and women from Qatar. Interpret your interval.

Using the binomial distribution formula, find the possibility that a family that has 13 children has 10 girls.

On average, 36 babies are born in Vanderbilt Children’s Hospital every day.

(a) Let X be the number of babies that will be born in Whoville in the next hour. What type of distribution does X have? Circle the correct answer. binomial hypergeometric negative binomial Poisson normal exponential

(b) A baby was just born in Vanderbilt Children’s Hospital. Let Y be the number of hours until the next baby is born. What type of distribution does Y have? Circle the correct answer. binomial hypergeometric negative binomial Poisson normal exponential

Inference - type 2 error

3. A soda vending machine is scheduled to ship an average of nine (9) ounces of soda per glass, with a standard deviation of one ounce (1) ounce. The machine manufacturer wishes to set the control limit so that for a sample of 36 soft drinks, 5% of the sample averages is higher than the upper control limit and 5% of the sample averages, lower than the limit of lower control
a. In what values should the control limits be programmed?
b. What is the probability that if the population average changes to 8.9, the change will not be detected?
C. What is the probability that if the population average changes to 9.3, the change will not be detected?

1. A jar contains 100 gummy bears: 50 ordinary candies, and 50 cannabis edibles. Of the ordinary candies, 40 are red and 10 are green. Of the cannabis edibles, 20 are red, and 30 are green. A gummy bear is drawn at random from the jar, in such a way that each gummy bear is equally likely to be the one drawn. What is the probability that the drawn gummy bear is a cannabis edible, conditional on it being red?

(a) The rent prices for a studio apartment in Dublin follows a normal distribution with mean 1315€ and standard deviation 147€. A postdoctoral researcher working at Trinity is looking for a studio apartment. Compute the probability that he finds one within the price range 900€ - 1200€. (correct to 3 decimal places rounded down).

(b) consider again rent prices in Dublin. Find the value x for the price of a studio apartment such that the probability of paying less of x is 0.95. (correct to 3 decimal places rounded down.

#1)

A sample of 12 joint specimens of a particular type gave a sample mean proportional limit stress of 8.57 MPa and a sample standard deviation of 0.78 MPa.

(a) Calculate and interpret a 95% lower confidence bound for the true average proportional limit stress of all such joints. (Round your answer to two decimal places.)
_______ MPa

Interpret this bound. (pick one)

o With 95% confidence, we can say that the value of the true mean proportional limit stress of all such joints is centered around this value.

o With 95% confidence, we can say that the value of the true mean proportional limit stress of all such joints is less than this value.   

o With 95% confidence, we can say that the value of the true mean proportional limit stress of all such joints is greater than this value.

What, if any, assumptions did you make about the distribution of proportional limit stress? (pick one)

o We must assume that the sample observations were taken from a normally distributed population.

o We must assume that the sample observations were taken from a uniformly distributed population.    

o We do not need to make any assumptions.

o We must assume that the sample observations were taken from a chi-square distributed population.

(b) Calculate and interpret a 95% lower prediction bound for proportional limit stress of a single joint of this type. (Round your answer to two decimal places.)
_______ MPa

Interpret this bound. (pick one)

o If this bound is calculated for sample after sample, in the long run, 95% of these bounds will provide a lower bound for the corresponding future values of the proportional limit stress of a single joint of this type.

o If this bound is calculated for sample after sample, in the long run 95% of these bounds will provide a higher bound for the corresponding future values of the proportional limit stress of a single joint of this type.    

o If this bound is calculated for sample after sample, in the long run 95% of these bounds will be centered around this value for the corresponding future values of the proportional limit stress of a single joint of this type.

You may need to use the appropriate table in the Appendix of Tables to answer this question.

#2) An article reported that for a sample of 40 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO₂ level (ppm) was 654.16, and the sample standard deviation was 165.4.

(a) Calculate and interpret a 95% (two-sided) confidence interval for true average CO₂ level in the population of all homes from which the sample was selected. (Round your answers to two decimal places.)

( ____ , ____ ) ppm

Interpret the resulting interval. (pick one)

o We are 95% confident that this interval does not contain the true population mean.

o We are 95% confident that the true population mean lies below this interval.    

o We are 95% confident that this interval contains the true population mean.

o We are 95% confident that the true population mean lies above this interval.

(b) Suppose the investigators had made a rough guess of 170 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 56 ppm for a confidence level of 95%? (Round your answer up to the nearest whole number.)
______ kitchens

#3) It was reported that in a survey of 4713 American youngsters aged 6 to 19, 15% were seriously overweight (a body mass index of at least 30; this index is a measure of weight relative to height). Calculate a confidence interval using a 99% confidence level for the proportion of all American youngsters who are seriously overweight. (Round your answers to three decimal places.)

( _____ , _____ )

Please help with these questions. Thanks!

An auditor for a government agency was assigned the task of evaluating reimbursement for office visits to physicians paid by Medicare. The audit was conducted on a sample of 75 reimbursements, with the following results:

• In 12 of the office visits, there was an incorrect amount of reimbursement.
• The amount of reimbursement was xbar = $93.70, s = $34.55.

1. At the 0.05 level of significance, is there evidence that the population mean reimbursement was less than $100?
2. At the 0.05 level of significance, is there evidence that the proportion of incorrect reimbursements in the population was greater than 0.10?
3. Discuss the underlying assumptions of the test used in (a).
4. What is your answer to (a) if the sample mean equals $90?
5. What is your answer to (b) if 15 office visits had incorrect reimbursements?

A simple random sample with

n = 56

provided a sample mean of 29.5 and a sample standard deviation of 4.4.

A. Develop a 90% confidence interval for the population mean.

B. Develop a 95% confidence interval for the population mean.

C. Develop a 99% confidence interval for the population mean.

1. By hand solve the problem.

The life in hours of a thermocouple used in furnace is known to be approximately normally distributed, with standard deviation σ=15 hours . A random sample of 15 thermo-couples resulted in the following data:

1. Using Z₀. Is there evidence to support the claim that the mean is 190 hours (α=.05) ?
2. What is the P-value ?
3. Construct a 99% two-sided CI on the mean life.
4. Use the CI found in part (c) to test the hypothesis of part a

Three independent group students took an exam. The data presented below represent the students’ test scores.

Group 1 Group 2 Group 3

95 79 65

98 76 68

80 60 70

77 88 87

99 69 66

90 66 72

Based on the post hoc (Tukey) test, which two groups have a significant difference at the 0.05 level ?

a. group 1 and group 2

b. group 1 and group 3

c. group 2 and group 3

d. both a and b

8) Errors: Type I and Type II are errors that are possible even when a hypothesis test is done correctly. A hypothesis test is based on probabilities (p-values) This means there is always a probability of drawing the wrong conclusion even when done correctly. Please review the following:

a.) What are type I and type II errors?

b.) Be able to discuss what a type I or type II error is in a given scenario

c.) What is the relationship between alpha, beta and the sample size?

4.) When would you use the normal distribution? T-distribution? Chi-Square distribution?