Let X and Y have the following joint distribution:
X/Y | 0 | 1 | 2 |
0 | 5/50 | 8/50 | 1/50 |
2 | 10/50 | 1/50 | 5/50 |
4 | 10/50 | 10/50 | 0 |
Further, suppose σx = √(1664/625), σy = √(3111/2500)
a) Find Cov(X,Y)
b) Find p(X,Y)
c) Find Cov(1-X, 10+Y)
d) p(1-X, 10+Y), Hint: use c and find Var[1-X], Var[10+Y]
In: Math
3300 Econometics HW Set 1
DATE | Cons. | Disp.Icome |
2015-01-01 | $ 11,788.36 | $ 13,226.57 |
2015-04-01 | $ 11,887.54 | $ 13,327.81 |
2015-07-01 | $ 11,971.95 | $ 13,440.36 |
2015-10-01 | $ 12,039.65 | $ 13,471.39 |
2016-01-01 | $ 12,111.78 | $ 13,562.27 |
2016-04-01 | $ 12,214.10 | $ 13,541.45 |
2016-07-01 | $ 12,294.30 | $ 13,592.92 |
2016-10-01 | $ 12,372.73 | $ 13,685.36 |
2017-01-01 | $ 12,427.65 | $ 13,835.34 |
2017-04-01 | $ 12,515.86 | $ 13,909.77 |
2017-07-01 | $ 12,584.91 | $ 13,986.19 |
2017-10-01 | $ 12,706.37 | $ 14,065.92 |
2018-01-01 | $ 12,722.84 | $ 14,219.83 |
2018-04-01 | $ 12,842.02 | $ 14,306.61 |
2018-07-01 | $ 12,968.54 | $ 14,393.59 |
The data given in the data file in the Consumption file represent the real private consumption of the USA from Quarter I 2005 to III Quarter 2018.
Similarly, the Real Disposable Income is provided over the same time span.
Set up a regression that relates the dependent variable(Y) to the independent variable(X).
Derive Manually the coefficients of the regression. (Intercept(b1) and slope(b2)).
State the Regression equation.
Interpret the meaning of the slopes b2, in this problem.
Derive the Correlation Coefficient R^2
Derive the Standard Error of the regression
Derive the standard error of the Intercept (b1) and the standard error of the Slope (b2).
Derive the t values of the coefficients
Construct a 95% confidence interval for b1 and b2
Use a two tail α=5% level of significance, to test the confidence intervals for the slope(b2).
(Hint: All the formulas required to answer the questions are cited in chapters 2 and 3 of the textbooks. Use also the notes from the lectures).
In: Math
a) Prove, using the joint density function, and the definition of expectation of a function of two continuous random
variables (i.e., integration) that E (5X + 7Y ) = 5E (X ) + 7E (Y ).
b)
(h) Prove, using the joint density function and the definition of expectation of a function of two continuous random
variables (i.e., integration) that Var (5X + 7Y ) = 25Var (X ) + 49Var (Y ) + 70Cov (X; Y ).
In: Math
In 2010, Dr. Bob decided to gather research on the type of
disorders that present among his patients. His data collection
resulted in the following breakdown of patients by disorder: 54.9%
Schizophrenia; 21.1% Major Depression; 7.9% Obsessive-Compulsive
Disorder; 4.5% Anxiety Disorder; 2.9% Personality Disorder; 8.8%
Other. Information was collected from a random sample 0f 300
patients in 2018 to determine whether or not the data has changed
significantly. The sample data is given in the table below. At the
α=0.05 level of significance, test the claim that the disorder
breakdown of patients at Dr. Bob's hospital has not changed
significantly since 2010.
Which would be correct hypotheses for this test?
Type of disorder per patient in sample:
Disorder | Count |
---|---|
Schizophrenia | 145 |
Major Dispression | 73 |
Obsessive-Compulsive Disorder | 30 |
Anxiety Disorder | 12 |
Personality Disorder | 14 |
Other | 26 |
Test Statistic:
Give the P-value:
Which is the correct result:
Which would be the appropriate conclusion?
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Sample mean: x̄ = 48.74
Sample standard deviation: s = 32.5857
Size of your sample: n = 50
What is your Point Estimate? (round each answer to at least 4 decimals)
For a 95% confidence interval
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please provide me below
just i want
z tables , t tables , chi square tables
both right tailed ,left tailed .,two tailed
high quality of images only,other wise it wont help me, it leads thumbdown.
thankyou chegg
In: Math
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 14 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.22 gram.
(a)
Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.)
lower limitupper limitmargin of error
(b)
What conditions are necessary for your calculations? (Select all that apply.)
σ is knownσ is unknownn is largeuniform distribution of weightsnormal distribution of weights
(c)
Interpret your results in the context of this problem.
There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region. The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.
(d)
Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.10 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
hummingbirds
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The following table was derived from a study of HIV patients, and the data reflect the number of subjects classified by their primary HIV risk factor and gender. Test if there is a relationship between HIV risk factor and gender using a 5% level of significance:
Gender |
Total |
||
HIV Risk Factor |
Male |
Female |
|
IV drug user |
24 |
40 |
64 |
Homosexual |
32 |
18 |
50 |
Other |
15 |
25 |
40 |
71 |
83 |
154 |
What type of chi-square test will you use (goodness of fit or
test of independence)?
What are your hypotheses?
H0:
HA:
Fill in the following table to calculate your test statistic:
IV Drug Use: |
Homosexual: |
Other: |
Total |
||
Male |
O = |
24 |
32 |
15 |
71 |
E = |
|||||
(O – E) = |
|||||
(O – E)2 / E = |
|||||
Female |
O = |
40 |
18 |
25 |
83 |
E = |
|||||
(O – E) = |
|||||
(O – E)2 / E = |
df =___________________
Critical value: ______________________
Conclusion: We _____________________ (reject / fail to reject) the
Null Hypothesis
Interpretation:
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Different varieties of the tropical flower Heliconia are fertilized by different species of hummingbirds. Over time, the lengths of the flowers and the form of the hummingbirds' beaks have evolved to match each other. Here are data on the lengths in millimeters of three varieties of these flowers on the island of Dominica. data332.dat
Do a complete analysis that includes description of the data and a significance test to compare the mean lengths of the flowers for the three species. (Round your answers for x to four decimal places, s to three decimal places, and s_(x^^\_) to three decimal places. Round your test statistic to two decimal places. Round your P-value to three decimal places.)
flower type n x^^\_ s s_(x^^\_)
H. bihai
H. caribaea red H.
caribaea yellow
F =
P =
variety length bihai 49.62 bihai 46.47 bihai 48.14 bihai 47.49 bihai 47.82 bihai 47.48 bihai 48.03 bihai 46.99 bihai 46.57 bihai 51.08 bihai 45.65 bihai 49.78 bihai 49.14 bihai 47.77 bihai 46.91 bihai 47.77 red 37.12 red 41.89 red 38.85 red 39.33 red 40.72 red 40.37 red 41.27 red 41.67 red 39.5 red 41.93 red 41.09 red 42.74 red 40.78 red 39.27 red 40.42 red 41.52 red 40.6 red 38.67 red 37.94 red 41.86 red 37.44 red 42.01 red 40.62 yellow 35.07 yellow 36.23 yellow 34.55 yellow 36.56 yellow 35.33 yellow 36.96 yellow 36.23 yellow 33.76 yellow 34.25 yellow 35.41 yellow 35.25 yellow 37.49 yellow 35.04 yellow 37.48 yellow 37.15
In: Math
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 14 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.20 gram. (a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.) lower limit upper limit margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) normal distribution of weights uniform distribution of weights n is large σ is known σ is unknown (c) Interpret your results in the context of this problem. The probability to the true average weight of Allen's hummingbirds is equal to the sample mean. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20. There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region. There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80. (d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.15 for the mean weights of the hummingbirds. (Round up to the nearest whole number.) ___hummingbirds
In: Math
In: Math
part 1
An independent measures study was conducted to determine whether a
new medication called "Byeblue" was being tested to see if it
lowered the level of depression patients experience. There were two
samples, one that took ByeBlue every day and one that took a
placebo every day. Each group had n = 30. What is the df value for
the t-statistic in this study?
a. 60
b. 58
c. 28
d. There is not enough information
part 2.
Engineers must consider the breadths of male heads when designing
helmets. The company researchers have determined that the
population of potential clientele have head breadths that are
normally distributed with a mean of 6.2-in and a standard deviation
of 0.8-in. Due to financial constraints, the helmets will be
designed to fit all men except those with head breadths that are in
the smallest 3.9% or largest 3.9%.
What is the minimum head breadth that will fit the clientele?
min =
What is the maximum head breadth that will fit the clientele?
max =
Do not round your answer.
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In how many ways can a five horse race end, allowing for the possibility that
two horses tie?
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A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a less than desirable battery life. They aim to compete with a new touch phone that is guaranteed to have a battery life more than two hours longer than the leading product. A recent sample of 65 units of the leading product provides a mean battery life of 5 hours and 39 minutes with a standard deviation of 92 minutes. A similar analysis of 51 units of the new product results in a mean battery life of 7 hours and 53 minutes and a standard deviation of 83 minutes. It is not reasonable to assume that the population variances of the two products are equal. Use Table 2. Sample 1 is from the population of new phones and Sample 2 is from the population of old phones. All times are converted into minutes. Let new products and leading products represent population 1 and population 2, respectively. a. Set up the hypotheses to test if the new product has a battery life more than two hours longer than the leading product. H0: μ1 − μ2 = 120; HA: μ1 − μ2 ≠ 120 H0: μ1 − μ2 ≥ 120; HA: μ1 − μ2 < 120 H0: μ1 − μ2 ≤ 120; HA: μ1 − μ2 > 120 b-1. Calculate the value of the test statistic. (Round all intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Test statistic b-2. Implement the test at the 5% significance level using the critical value approach. Do not reject H0; there is no evidence that the battery life of the new product is more than two hours longer than the leading product. Reject H0; there is no evidence that the battery life of the new product is more than two hours longer than the leading product. Do not reject H0; there is evidence that the battery life of the new product is more than two hours longer than the leading product. Reject H0; there is evidence that the battery life of the new product is more than two hours longer than the leading product.
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In a survey, 38% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so he randomly selected 150 pet owners and discovered that 53 of them spoke to their pet on the telephone. Does the vet have a right to be skeptical? use the confidence interval .1 level of significance. a) because np0(1-p0)= blank (=, not equal, greater, or less than) 10, the sample size is (blank- less or greater) 5% of the population size and the sample (blank-) the requirements for testing the hypothesis (blank- are or are not) satisfied. b)what are the null and alternative hypotheses? c)determine the test statistic. d)determine the critical values. e) does the vet have a right to be skeptical?
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