Two samples, each with n = 8, produce an independent-measures t statistic of t = –2.15. Which of the following decisions is justified for a two-tailed test? (Hint: use the proper table to find the critical values).
In: Math
give an example of a discrete random variable X whose values are integers and such that E(X) = infinite. Prove that E(X) = infinite for your example. (hints: if you will be paid 2^k dollars for the kth head when you flip a fair coin., the expected value is infinite...) Or give other easy examples.
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The table represents the average protein intake in grams per day for people in different countries. Identify the following measures based on the given random sample of countries. Use the IQR method to determine which values are outliers
|
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Regarding problem R2 from chapter 26. Make a box representing a roulette wheel with 18 tickets that represent red and 20 that represent black or green. Draw 3800 times with replacement from this box and record the number of tickets drawn that are red. Repeat this process 10,000 times. What is the fraction of times (out of these 10,000 repeated trials) were the number of red tickets drawn at least 1,890? How does this compare to the P-value you got in the problem? Can you use pbinom( ) to compute this probability? Are these numbers different? Why?
"With a perfectly balanced roulette wheel, in the long run, red numbers should turn up 18 times in 38. To test its wheel, one casino records the results of 3800 plays finding 1890 reds numbers.Is that too many reds. Or chance variation?"
How do I compute this on R?
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It has been postulated that a student is more likely to smoke marijuana if both parents use alcohol and drugs. A paper published in Youth and Society (1979) reported that among 570 students whose parents use alcohol and drugs, 320 regularly smoke marijuana.
a. Is there any evidence that a student is more likely to smoke marijuana if both parents use alcohol and drugs?
b. Joann is planning for a new study on this problem. Suppose no prior information is available, what is the minimum sample size required such that the length of the 92.32% confidence interval for the true proportion of students smoke marijuana if both parents use alcohol and drugs is at most 0.075?
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A study by Bechtel et al., 2009, described in the Archives of Environmental & Occupational Health considered polycyclic aromatic hydrocarbons and immune system function in beef cattle. Some cattle were near major oil- and gas-producing areas of western Canada. The mean monthly exposure to PM1.0 (particulate matter that is <1μm in diameter) was approximately 7.2 μg/m3 with standard deviation 1.5. Assume that the monthly exposure is normally distributed.
a. What is the probability of a monthly exposure greater than 9 μg/m3?
b. What is the probability of a monthly exposure between 3 and 5 μg/m3?
c. What is the monthly exposure level that is exceeded with probability 0.05? d. What value of mean monthly exp
In: Math
Consider the following hypothesis test.
H0: μd ≤ 0
Ha: μd > 0
(a) The following data are from matched samples taken from two populations. Compute the difference value for each element. (Use Population 1 − Population 2.)
Element | Population | Difference | |
---|---|---|---|
1 | 2 | ||
1 | 21 | 21 | |
2 | 28 | 28 | |
3 | 18 | 16 | |
4 | 20 | 18 | |
5 | 26 | 25 |
(b) Compute d.
(c) Compute the standard deviation sd.
(d) Conduct a hypothesis test using α = 0.05.
Calculate the test statistic. (Round your answer to three decimal places.)
Calculate the p-value. (Round your answer to four decimal places.)
p-value =
What is your conclusion?
Reject H0. There is sufficient evidence to conclude that μd > 0.
Do not Reject H0. There is sufficient evidence to conclude that μd > 0.
Reject H0. There is insufficient evidence to conclude that μd > 0.
Do not reject H0. There is insufficient evidence to conclude that μd > 0.
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When σ is unknown and the sample is of size n ≥ 30, there are two methods for computing confidence intervals for μ.
Method 1: Use the Student's t distribution with
d.f. = n − 1.
This is the method used in the text. It is widely employed in
statistical studies. Also, most statistical software packages use
this method.
Method 2: When n ≥ 30, use the sample standard
deviation s as an estimate for σ, and then use
the standard normal distribution.
This method is based on the fact that for large samples, s
is a fairly good approximation for σ. Also, for large
n, the critical values for the Student's t
distribution approach those of the standard normal
distribution.
Consider a random sample of size n = 31, with sample
mean x = 44.4 and sample standard deviation s =
5.9.
(d) Now consider a sample size of 81. Compute 90%, 95%, and 99%
confidence intervals for μ using Method 1 with a Student's
t distribution. Round endpoints to two digits after the
decimal.
90% | 95% | 99% | |
lower limit | |||
upper limit |
(e) Compute 90%, 95%, and 99% confidence intervals for μ
using Method 2 with the standard normal distribution. Use
s as an estimate for σ. Round endpoints to two
digits after the decimal.
90% | 95% | 99% | |
lower limit | |||
upper limit |
In: Math
Two catalysts in a batch chemical process, are being considered for their effect on the output of a process reaction. Based on literature, data with different samples with catalyst 1 were found to result to the following yield:
78.53 |
78.17 |
81.05 |
81.77 |
78.13 |
81.24 |
81.84 |
79.63 |
78.59 |
79.74 |
79.84 |
On the other hand, you found the following for the yield using catalyst 2:
82.05 |
81.57 |
81.71 |
81.29 |
81.79 |
81.47 |
81.75 |
81.50 |
81.75 |
If you can assume that the distribution characterizing yield is approximately normal and the samples are independent:
Perform hypothesis testing using 97.5 % confidence to test whether the average yield using catalyst 2 is not equal to the average yield using catalyst 1
Report your findings with the same data and hypothesis, but now using significance testing (P-value approach) only for the last test (test on difference of means). Use the same confidence level as previously for any other test.
In: Math
****Return to problem 5.3.2 and answer the questions posed, using the t distribution rather than the standard normal distribution. In other words, the problem was originally posed assuming that the population’s standard deviation σ is known. Now answer the questions assuming the standard deviation is estimated from the sample itself. What are the degrees of freedom?***
5.3.2 The study cited in Exercise 5.3.1 reported an estimated mean serum cholesterol level of 183 for women aged 20–29 years. The estimated standard deviation was approximately 37. Use these estimates as the mean m and standard deviation s for the U.S. population. If a simple random sample of size 60 is drawn from this population, find the probability that the sample mean serum cholesterol level will be:
(a) Between 170 and 195 (b) Below 175 (c) Greater than 190
The real question is bolded the other is just the info needed to answer the question.
In: Math
The random variable x is greater than or equal to 50, with a mean of 70, and a variance of 36. What is the probability for x less than or equal to 50? What is the probability for x between 80 and 60?
.
.
.
The salary of employees is normally distributed with mean of $50,000 annually. Standard deviation of $10,000. What percentage of the employees make between $45,000-$60,000?
In: Math
Given Hypothesis: Ho: population mean μo = 200 Ha: population mean μo < 200 This is Left-Tailed test. Population Standard Deviation σ = 50. Given Significance Level is 0.10 (10%). Critical z-value for 0.10 significance level and Left-Tailed test is (-1.28). Rejection Region will be to the left of z= - 1.28. -3…………..-2………….-1…………0……….1 Choose your sample mean, x̅ (any integer between 180 and 195) and your sample size, n, ( any integer between 16 and 36). Perform the steps below and reach your conclusion: reject Ho or do not reject Ho. Steps to follow: 1. Calculate your test statistics z-value: z = ( x̅ - 200) / (50/√n) 2. Compare your z value with the critical value -1.28. 3. If your z-value is less than -1.28 (falls in Rejection Region) then reject Ho. If your z-value is greater than -1.28 (does not fall in Rejection Region) then do not reject Ho. Show all three steps; just the answer will not be graded.
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The following table shows the percentage of on-time arrivals, the number of mishandled baggage reports per 1,000 passengers, and the number of customer complaints per 1,000 passengers for 10 airlines.
Airline | On-Time
Arrivals (%) |
Mishandled
Baggage per 1,000 Passengers |
Customer
Complaints per 1,000 Passengers |
---|---|---|---|
Airline 1 | 83.3 | 0.79 | 1.50 |
Airline 2 | 79.1 | 1.91 | 0.80 |
Airline 3 | 86.9 | 1.64 | 0.95 |
Airline 4 | 86.5 | 2.13 | 0.76 |
Airline 5 | 87.2 | 3.04 | 0.47 |
Airline 6 | 78.2 | 2.21 | 1.11 |
Airline 7 | 83.1 | 3.18 | 0.23 |
Airline 8 | 85.7 | 2.19 | 1.70 |
Airline 9 | 77.1 | 2.89 | 1.77 |
Airline 10 | 77.2 | 3.79 | 4.29 |
(a)
If you randomly choose an Airline 4 flight, what is the probability that this individual flight has an on-time arrival?
(b)
If you randomly choose one of the 10 airlines for a follow-up study on airline quality ratings, what is the probability that you will choose an airline with less than two mishandled baggage reports per 1,000 passengers?
(c)
If you randomly choose 1 of the 10 airlines for a follow-up study on airline quality ratings, what is the probability that you will choose an airline with more than one customer complaint per 1,000 passengers?
(d)
What is the probability that a randomly selected Airline 3 flight will not arrive on time?
In: Math
You will be performing an analysis on female heights, given of set of 30 heights that were randomly obtained. For this project, it is necessary to know that the average height for women is assumed to be 65 inches with a standard deviation of 3.5 inches. You will use these numbers in some of your calculations.
a. As you know, a random sample of 30 women’s heights was
obtained. Describe the sampling distribution. Use a full sentence
here to describe the sampling distribution. HINT: See the Chapter 8
Lecture Notes Key or the Chapter 8 video lecture if you are stuck.
You will not use the raw data yet.
b. In StatCrunch or Excel, find and state the mean of the 30
women’s heights that were provided (round to two decimal places).
Then (using the values from step 3), calculate the probability that
a random sample of 30 women’s heights would result in a mean that
was the value you found or more. Use this probability in a full
sentence.
c. For this step, please work under the assumption that we do not
know the population mean and standard deviation (and in fact, if we
are running this test, it means that we are not sure of these
values). Construct a 95% confidence interval for the average height
of a female. Interpret this confidence interval. HINT: You can use
the “with data” option
Height (in Inches) | Name |
72.44 | Emma |
67.53 | Olivia |
66.71 | Ava |
62.02 | Isabella |
73.89 | Sophia |
65.95 | Mia |
65.83 | Charlotte |
64.15 | Amelia |
65.39 | Evelyn |
59.68 | Abigail |
64.24 | Harper |
66.60 | Emily |
65.40 | Elizabeth |
64.72 | Avery |
67.11 | Sofia |
61.97 | Ella |
62.83 | Madison |
67.20 | Scarlett |
66.62 | Victoria |
68.78 | Aria |
66.13 | Grace |
64.47 | Chloe |
66.64 | Camila |
62.39 | Penelope |
63.90 | Riley |
62.97 | Layla |
59.31 | Lillian |
66.14 | Nora |
67.54 | Zoey |
63.45 | Mila |
In: Math
Assume that women's heights are normally distributed with a mean given by mu equals 62.1 in, and a standard deviation given by sigma equals 2.7 in. (a) If 1 woman is randomly selected, find the probability that her height is less than 63 in. (b) If 32 women are randomly selected, find the probability that they have a mean height less than 63 in.
In: Math