Let the random variable X follow a normal distribution with a
mean of μ and a standard deviation of σ. Let 1 be the mean of a
sample of 36 observations randomly chosen from this population, and
2 be the mean of a sample of 25 observations randomly chosen from
the same population.
a) How are 1 and 2 distributed? Write down the form of the density
function and the corresponding parameters.
b) Evaluate the statement:
?(?−0.2?< ?̅1 < ?+0.2?)<?(?−0.2?< ?̅2 < ?+0.2?), as
to whether it is true or false.
In: Statistics and Probability
Use the following scenario to answer the questions below:
A campaign manager for a local politician is interested in the percentage of Leon County residents who intend to vote in the 2020 elections in November. Residents from each neighborhood in Leon County are randomly sampled. Using simple random sample techniques, larger neighborhoods have a larger number of residents randomly sampled and smaller neighborhoods have a smaller number of residents randomly sampled to get a representative sample.
Altogether 434 Leon County residents are sampled. The result is that 39% of the sampled Leon County residents intend to vote in the 2020 elections in November.
1. The result value 39% would be considered a statistic or parameter?
2. Would using a statistic or a parameter be more appropriate in this situation?
A). It does not matter as statistics and parameters have the same values.
B). A statistic because it would be more accurate than a parameter.
C). A statistic because it may be more easily obtained due to time constraints and available resources.
D). A parameter because it includes responses from everyone in the sample and will not vary from sample to sample.
3. A campaign manager for another candidate claims that the sample taken would not represent the population because the entire population was not surveyed. Which of the following provides the best evidence that the sample taken would be representative of the population?
A). A large enough sample of residents was randomly taken and proportional to the size of each neighborhood in Leon County.
B). 434 is a large enough sample of residents since more than 30 people were interviewed.
C). There is no evidence to say that the sample taken would be representative of the population.
D). 39% is a large enough percentage of Leon County residents who intend to vote in the 2018 midterm elections in November.
4. Suppose that someone at a local news station wants to consider the sampling variability and uses the results of the survey to calculate the margin of error for the result. The person at the station assumes that the normal distribution can be used in calculating the margin of error. Is the assumption of the normal distribution here correct?
A). Although a random sample was taken, the normal distribution is not appropriate since not enough residents were sampled.
B). The normal distribution is appropriate since a random sample was taken and there were enough residents who answered “Yes” as well as “No” to voting in the 2018 midterms in November.
C). The normal distribution is not appropriate since only a sample of all the residents of Leon County were surveyed.
D). The normal distribution is appropriate since a random sample was taken and there were more than 30 residents surveyed.
5. In the midterm primaries in August 2018, 37% of Leon County residents voted. Based on the information provided from the sample above, would it be surprising if 37% of all Leon county residents also intend to vote in the 2020 elections? State YES or NO and use statistical reasoning (inference) to support your position.
6. Using the information from the sample above, what is your conclusion regarding the proportion (or percentage) of Leon County residents who intend to vote in the 2020 elections in November? Be as detailed as possible in your explanation.
In: Statistics and Probability
7. In a survey sample of 83 respondents, about 30.1 percent of the sample work less than 40 hours per week. Calculate a 99 percent confidence interval for the proportion of persons who work less than 40 hours per week.
In: Statistics and Probability
In an effort to characterize the New Guinea crocodile (Crocodylus novaeguineae), measurements were taken of the dorsal cranial length (mm) (the length of the skull from the tip of the nose to the back of the cranial cap, denoted DCL) and the total length (cm) (denoted TL) of 50 harvested adult males. (Data on next page.)
1. Construct a histogram and a boxplot for each of the variables DCL and TL. Comment on the symmetry of the distribution of each variable.
2. Construct a scatterplot with DCL on the vertical axis and TL on the horizontal axis. Based on this plot, what can be said about the relationship between DCL and TL (i.e. if you vary DCL, what happens to TL)?
Data
TL DCL Observation
130 169 1
102 154 2
126 160 3
230 290 4
115 151 5
150 209 6
259 344 7
130 183 8
110 153 9
130 183 10
185 237 11
215 288 12
129 187 13
149 189 14
156 203 15
100 143 16
224 294 17
234 318 18
162 229 19
217 299 20
206 283 21
144 198 22
146 203 23
166 229 24
203 275 25
205 266 26
252 350 27
238 318 28
250 330 29
255 351 30
120 169 31
250 332 32
238 307 33
157 205 34
159 216 35
202 261 36
177 237 37
221 288 38
224 294 39
167 232 40
240 316 41
207 268 42
192 242 43
180 248 44
165 226 45
197 267 46
113 162 47
131 183 48
162 234 49
246 310 50
In: Statistics and Probability
Fiction Cruiseline offers three ways to exercise on their cruise ships. 73 of the 86 passengers participated in at least one method of exercise. 36 people went rock climbing, 44 people went ice skating, and 19 went to the fitness center. 14 people went rock climbing and ice skating, 11 people went rock climbing and to the fitness center, and 9 people went ice skating and to the fitness center. Draw a Venn Diagram for the three sets if necessary. Include how you found the number of ALL three activities.
Calculate the probability for:
A randomly selected passenger did not go ice skating, given they did at least two activities
In: Statistics and Probability
After preliminary testing, it is determined that .01 of the residents of Selma have a certain virus. In Sangrian, a suburb of Selma (incidence of the virus is the same), the authorities test the 10,000 residents of the suburb. The test has the following characteristics. If you have the virus, the probability that the test is positive is .95, and if you don’t have the virus, the probability that the test is positive is .10. Suppose that 1,100 of the 10,000 residents test positive. Let X = the number among those 1,100 who really do have the virus.
Find E(X) and Variance(X). thank you :) !!
In: Statistics and Probability
I would like to see a proof of the Central Limit Theorem that applies to a simple probability dice scenerio, say rolling a 6 x amount of times. The goal is to help me understand the theorem with a simple example. Thanks!
In: Statistics and Probability
Suppose X1, . . . , Xn is a random sample from the Normal(μ, σ2) distribution, where μ is unknown but σ2 is known, and it is of interest to test H0: μ = μ0 versus H1: μ ̸= μ0 for some value μ0. The R code below plots the power curve of the test
Reject H0 iff |√n(X ̄n − μ0)/σ| > zα/2
for user-selected values of μ0, n, σ, and α. For a sequence of values of μ, the code computes the probability that the null hypothesis will be rejected according to the above test. In addition, for each value of μ in the sequence, a simulation is run: 100 data sets with sample size n are generated from the Normal(μ,σ2) distribution, and for each of the 100 data sets, it is recorded whether the null hypothesis was rejected. For each value of μ, the proportion of times the null hypothesis is rejected is recorded. This gets plotted as a dashed line.
mu.0 <- ???
n <- ???
sigma <- ???
alpha <- ???
mu.seq <- seq(mu.0 - 5,mu.0 + 5,length=50)
power.theoretical <- numeric()
power.empirical <- numeric()
for(j in 1:length(mu.seq))
{
power.theoretical[j] <- 1-(pnorm(qnorm(1-alpha/2)-sqrt(n)*(mu.seq[j] - mu.0)/sigma)
reject <- numeric()
for(s in 1:100)
{
}
-pnorm(-qnorm(1-alpha/2)-sqrt(n)*(mu.seq[j] - mu.0)/sigma))
x <- rnorm(n,mu.seq[j],sigma) x.bar <- mean(x)
reject[s] <- abs(sqrt(n)*(x.bar-mu.0)/sigma) > qnorm(1-alpha/2)
power.empirical[j] <- mean(reject) }
plot(mu.seq,power.theoretical,type="l",ylim=c(0,1),xlab="mu",ylab="power") lines(mu.seq, power.empirical,lty=2) abline(v=mu.0,lty=3) # vert line at null value abline(h=alpha,lty=3) # horiz line at size
(a) Putinμ0 =2,n=5,σ=2,andα=0.05andexecutethecode. Turnintheplot.
(b) Explain why the dashed line follows the solid line closely but not exactly.
(c) Interpret the height of the solid line at μ = 4.
(d) Interpret the height of the solid line at μ = 2.
(e) Interpret the height of the dashed line at μ = 2.
(f) What would be the effect on the height of the solid line at μ = 4 if i. the sample size n were increased?
ii. the standard deviation σ were increased? iii. the size α of the test were increased?
(g) What would be the effect on the height of the solid line at μ = 2 if i. the sample size n were increased?
ii. the standard deviation σ were increased? iii. the size α of the test were increased?
(h) What would be the effect on the dashed line of generating 500 data sets instead of only 100 data sets for the simulation at each value of μ?
In: Statistics and Probability
A new program of imagery training is used to improve the performance of basketball players shooting free-throw shots. The first group did an hour imagery practice, and then shot 30 free throw basket shots with the number of shots made recorded. A second group received no special practice, and also shot 30 free throw basket shots. The data are below. Did the imagery training make a difference? Set alpha = .05.
Group 1: 15, 17, 20, 25 26, 27
Group 2: 5, 6, 10, 15, 18, 20
A) You must use all five steps in hypothesis testing:
B) Solve for and evaluate the effect size of this study using Cohen's D.
C) Create a 90% confidence interval for this problem.
In: Statistics and Probability
A researcher wants to determine whether high school students who attend an SAT preparation course score significantly different on the SAT than students who do not attend the preparation course. For those who do not attend the course, the population mean is 1050 (μ = 1050). The 16 students who attend the preparation course average 1150 on the SAT, with a sample standard deviation of 300. On the basis of these data, can the researcher conclude that the preparation course has a significant difference on SAT scores? Set alpha equal to .05.
Q1: The appropriate statistical procedure for this example would be a
A. z-test
B. t-test
Q2: The most appropriate null hypothesis (in words) would be
A. There is no statistical difference in SAT scores when comparing students who took the SAT prep course with the general population of students who did not take the SAT prep course.
B. There is a statistical difference in SAT scores when comparing students who took the SAT prep course with the general population of students who did not take the SAT prep course.
C. The students who took the SAT prep course did not score significantly higher on the SAT when compared to the general population of students who did not take the SAT prep course.
D. The students who took the SAT prep course did score significantly higher on the SAT when compared to the general population of students who did not take the SAT prep course.
Q3: The most appropriate null hypothesis (in symbols) would be
A. μSATprep = 1050
B. μSATprep = 1150
C. μSATprep 1050
D. μSATprep 1050
Q4: Based on your evaluation of the null in and your conclusion, as a researcher you would be more concerned with a
A. Type I statistical error
B. Type II statistical error
Calculate the 99% confidence interval. Steps:
Q5: The mean you will use for this calculation is
A. 1050
B. 1150
Q6: What is the new critical value you will use for this calculation?
Q7: As you know, two values will be required to complete the following equation:
__________ __________
Q8: Which of the following is a more accurate interpretation of the confidence interval you just computed?
A. We are 99% confident that the scores fall in the interval _____ to _____.
B. We are 99% confident that the average score on the SAT by the students who took the prep course falls in the interval _____ to _____.
C. We are 99% confident that the example above has correct values.
D. We are 99% confident that the difference in SAT scores between the students who took the prep course and the students who did not falls in the interval _____ to _____.
In: Statistics and Probability
A printing shop that produces advertising specialties produces paper cubes of various sizes, of which the 3.5 inch cube is the most popular. The cubes are cut from a stack of paper on cutting presses. The two sides of the cube are determined by the distance of stops on the press from the cutting knife and remain fairly constant, but the height of the cube varies depending on the number of sheets included in a “lift” by the operator. The lift height does not remain constant within an operator or between operators. The difficulty is in judging, without taking much time, what thickness of lift will give the correct height when it is pressed by the knife and cut. The humidity in the atmosphere also contributes to this difficulty, because the paper swells when humidity is high. The operators tend to err on the safe side, by lifting a thicker stack of paper than necessary.
The company management believes the cubes are being made much taller than the target, thus giving away excess paper and causing loss to the company. They have received advice from a consultant that they could install a paper-counting machine, which will give the correct lift containing exactly the same number of sheets each time a lift is made. This, however, will entail a huge capital investment. To see if the capital investment would be justifiable, the company management wants to assess the current loss in paper because of the variability of the cube heights from the target.
Data were collected by measuring the heights of 20 groups of five cubes and are provided in the table below. Estimate the loss incurred because of the cubes being taller than 3.5 inches. A cube that is exactly 3.5 inches in height weighs 1.2 lb. The company produces 3 million cubes per year, and the cost of paper is $64 per hundred-weight (100lb.).
Note that the current population of cube heights has a distribution (assume this to be normal) with an average and standard deviation, and the target population of the cubes is also a distribution with an average of 3.5 in. and a standard deviation to be determined. (You can not make every cube exactly 3.5 in. in height.) The target standard deviation can be smaller than the current standard deviation, especially if the current process is subject to some assignable causes.
Estimate the current loss in paper because of the cubes being too tall. You first may have to determine the attainable variability before estimating the loss. If any of the information you need is missing, make suitable assumptions, and state them clearly.
|
3.61 |
3.59 |
3.53 |
3.63 |
3.63 |
3.57 |
3.61 |
3.52 |
3.47 |
3.53 |
3.61 |
3.65 |
3.30 |
3.52 |
3.60 |
|
3.59 |
3.59 |
3.58 |
3.58 |
3.57 |
3.60 |
3.64 |
3.44 |
3.59 |
3.61 |
3.49 |
3.65 |
3.52 |
3.50 |
3.53 |
|
3.61 |
3.58 |
3.58 |
3.64 |
3.57 |
3.59 |
3.54 |
3.52 |
3.60 |
3.58 |
3.44 |
3.54 |
3.60 |
3.51 |
3.63 |
|
3.62 |
3.63 |
3.52 |
3.63 |
3.53 |
3.55 |
3.64 |
3.60 |
3.59 |
3.63 |
3.59 |
3.54 |
3.57 |
3.59 |
3.63 |
|
3.60 |
3.62 |
3.60 |
3.62 |
3.57 |
3.54 |
3.58 |
3.49 |
3.60 |
3.55 |
3.53 |
3.62 |
3.51 |
3.49 |
3.63 |
|
3.63 |
3.65 |
3.60 |
3.61 |
3.63 |
|
3.66 |
3.64 |
3.63 |
3.62 |
3.61 |
|
3.68 |
3.65 |
3.63 |
3.64 |
3.63 |
|
3.63 |
3.64 |
3.61 |
3.61 |
3.63 |
|
3.64 |
3.61 |
3.63 |
3.61 |
3.62 |
In: Statistics and Probability
6.42 Let X1,..., Xn be an i.i.d. sequence of Uniform (0,1) random variables. Let M = max(X1,...,Xn).
(a) Find the density function of M. (b) Find E[M] and V[M].
In: Statistics and Probability
You may need to use the appropriate appendix table or technology to answer this question.
Consider the following hypothesis test.
| H0: μ = 15 |
| Ha: μ ≠ 15 |
A sample of 50 provided a sample mean of 14.05. The population standard deviation is 3.
(a)
Find the value of the test statistic. (Round your answer to two decimal places.)
(b)
Find the p-value. (Round your answer to four decimal places.)
p-value =
(c)
At
α = 0.05,
state your conclusion.
Reject H0. There is sufficient evidence to conclude that μ ≠ 15.Reject H0. There is insufficient evidence to conclude that μ ≠ 15. Do not reject H0. There is sufficient evidence to conclude that μ ≠ 15.Do not reject H0. There is insufficient evidence to conclude that μ ≠ 15.
(d)
State the critical values for the rejection rule. (Round your answers to two decimal places. If the test is one-tailed, enter NONE for the unused tail.)
test statistic≤test statistic≥
State your conclusion.
Reject H0. There is sufficient evidence to conclude that μ ≠ 15.Reject H0. There is insufficient evidence to conclude that μ ≠ 15. Do not reject H0. There is sufficient evidence to conclude that μ ≠ 15.Do not reject H0. There is insufficient evidence to conclude that μ ≠ 15.
In: Statistics and Probability
Consider the following hypothesis test.
H0: μ ≤ 12
Ha: μ > 12
A sample of 25 provided a sample mean
x = 14
and a sample standard deviation
s = 4.57.
(a)
Compute the value of the test statistic. (Round your answer to three decimal places.)
(b)
Use the t distribution table to compute a range for the p-value.
p-value > 0.2000.100 < p-value < 0.200 0.050 < p-value < 0.1000.025 < p-value < 0.0500.010 < p-value < 0.025p-value < 0.010
(c)
At
α = 0.05,
what is your conclusion?
Do not reject H0. There is sufficient evidence to conclude that μ > 12.Do not reject H0. There is insufficient evidence to conclude that μ > 12. Reject H0. There is insufficient evidence to conclude that μ > 12.Reject H0. There is sufficient evidence to conclude that μ > 12.
(d)
What is the rejection rule using the critical value? (If the test is one-tailed, enter NONE for the unused tail. Round your answer to three decimal places.)
test statistic≤test statistic≥
What is your conclusion?
Do not reject H0. There is sufficient evidence to conclude that μ > 12.Do not reject H0. There is insufficient evidence to conclude that μ > 12. Reject H0. There is insufficient evidence to conclude that μ > 12.Reject H0. There is sufficient evidence to conclude that μ > 12.
In: Statistics and Probability
Question Four
An academic conference held this past January consistent of 250 participants including undergraduate students, masters students, PhD students, and, professors from business faculties, engineering faculties, and nursing faculties to discuss various ways of increasing the learning desires of students. The following table gives us the breakdown of participants by levels of education and faculties.
Business Engineering Nursing Totals
Undergraduate students 30 15 35 80
Masters students 40 38 12 90
PhD students 10 12 8 30
Professors 20 15 15 50
Totals 100 80 70 250
_____
In: Statistics and Probability