Questions
The exponential distribution with rateλhas meanμ= 1/λ. Thus the method of moments estimator of λ is...

The exponential distribution with rateλhas meanμ= 1/λ. Thus the method of moments estimator of λ is 1/X. Use the following steps to verify that X is unbiased, but 1/X is biased. USING R:

a) Generate 10000 samples of size n= 5 from the standard exponential distribution (i.e.λ= 1) using rexp(50000) and arranging the 50000 random numbers in a matrix with 5 rows.

b) Use the apply() function to compute the 10000 sample means and store them in the object means. The 10000 estimators of λ can be stored in the object lambdas by lambdas = 1/means

c) Compute the sample mean of the object means, and sample mean of the object lambdas. What can you say about the bias of Xand of 1/X?

d) Repeat with a sample of size n= 10, using rexp(100000), and report your estimate of the bias of 1/X. Has the bias decreased?

Please answer with r code

In: Statistics and Probability

In the biathton event of the Olympic Games, a participant skis crosscountry and on four intermittent...

In the biathton event of the Olympic Games, a participant skis crosscountry and on four intermittent occasions stops at a rifle range and shoots a set of five shots. If the center of the target is hit, no penalty points are assessed. A particular individual has a history of hitting the center of the target 91% of the time. If 12 shots are taken, answer the following questions. Round your answers to 3 decimal places.

a. Find the probability that they will hit the target exactly 9 times. (for this part only you must demonstrate that you have used the binomial formula to receive credit for this problem).



b. Find the probability that they hit the target more than 10 times.



c. Find the probability that they wilt hit the target less than 8 times.


d. What is the mean of this distribution?

e. What is the standard deviation of this distribution.

In: Statistics and Probability

1. Use the R command X <- iris to assign Fishers’ iris dataset to the data...

1. Use the R command X <- iris to assign Fishers’ iris dataset to the data matrix X. Using the head(X) command summarize what each column of the dataset is measuring and represents. Assign Y as a new matrix of dimension 150 by 4 which has the values of X without the species label.

2. Compute and interpret (in summary English) each of the summary statistics X,S,R using R.

3. Visualize the dataset by making a scatterplot of Sepal Length vs. Sepal Width, a scatterplot of Petal Length vs. Petal Width. The pairs function is useful here. Use your plots and stats from #2 to comment on any evident correlations.

In: Statistics and Probability

Based on your data, is there a difference between the price of homes with only 1...

Based on your data, is there a difference between the price of homes with only 1 bathroom vs. homes with more than one bathroom? Test at the 1% level of significance. Regardless of your results, provide possible reasons why you might find no statistical difference between the two groups. Find the solution on SPSS

Can you please show me how to do this question on SPSS? you can take any random data but please show me how to do that?

In: Statistics and Probability

Is HDTV ownership related to quantity of purchases of other electronics? A Best Buy retail outlet...

Is HDTV ownership related to quantity of purchases of other electronics? A Best Buy retail outlet collected the following data for a random sample of its recent customers. At α = 0.10, is the frequency of in-store purchases independent of the number of large-screen HDTVs owned (defined as 50 inches or more)?

In-Store Purchases Last Month
HDTVs Owned None One More Than One Row Total
None 12 13 14 39
One 17 33 30 80
Two or More 18 45 65 128
Col Total 47 91 109 247



(b) Calculate the chi-square test statistic, degrees of freedom, and the p-value. (Round your test statistic value to 2 decimal places and the p-value to 4 decimal places.)

Test statistic
d.f.
p-value


(c) Find the critical value for chi-Square. Refer to the chi-square Appendix E table. (Round your answer to 2 decimal places.)

Critical value            

2)

Oxnard Kortholt, Ltd., employs 50 workers. Research question: At α = .05, do Oxnard employees differ significantly from the national percent distribution?

Health Care Visits National Percentage Oxnard Employees Frequency
No visits 15.6 3
1–3 visits 43.9 19
4–9 visits 25.1 13
10 or more visits 15.4 15
Total 100.0 50



)
Calculate the chi-square test statistic, degrees of freedom and the p-value. (Round your test statistic value to 3 decimal places and the p-value to 4 decimal places.)

Test statistic
d.f.
p-value


In: Statistics and Probability

If the average time to finish an IQ test is 85 minutes with standard deviation of...

If the average time to finish an IQ test is 85 minutes with standard deviation of 17 minutes and assuming the data are normally distributed,
1. what percentage of test takers finish less than 117 minutes?
2. what percentage of test takers finishes between one hour to hour and half?
3. find the time that separates the slowest 15% who finish the test?
4. find the time that separates the fastest 5% who finish the test?

In: Statistics and Probability

Ants on a Sandwich How many ants will climb on a piece of a peanut butter...

Ants on a Sandwich How many ants will climb on a piece of a peanut butter sandwich left on the ground near an ant hill? To study this, a student in Australia left a piece of a sandwich for several minutes, then covered it with a jar and counted the number of ants. He did this eight times, and the results are shown in Table 3.10. (In fact, he also conducted an experiment to see if there is a difference in number of ants based on the sandwich filling. The details of that experiment are given in Chapter 8, and the full dataset is in SandwichAnts.)42

Table3.10 Number of ants on a sandwich Number of ants 43 59 22 25 36 47 19 21

(a) Find the mean and standard deviation of the sample.

(b) Describe how we could use eight slips of paper to create one bootstrap statistic. Be specific.

(c) What do we expect to be the shape and center of the bootstrap distribution?

(d) What is the population parameter of interest? What is the best estimate for that parameter?

(e) A bootstrap distribution of 5000 bootstrap statistics gives a standard error of 4.85. Use the standard error to find and interpret a 95% confidence interval for the parameter defined in part (d).

In: Statistics and Probability

Sign Runs Test/G Consider the following sequence of 1 and 6: 6 111 66 111 66...

Sign Runs Test/G

Consider the following sequence of 1 and 6:

6 111 66 111 66 11 6 11 666 11 6 1111 666 11 66

With a 0.01 significance level, we wish to test the claim that the above sequence was produced in a random manner. Answer each of the following questions


(a) The null hypothesis H0 is given by

A. n1=n2
B. ρ=0
C. β=0
D. The data are in an order that is not random
E. The data are in a random order
F. Median=0
G. G=0
H. r=0
I. None of the above.

(b) The null hypothesis H1 is given by

A. ρ≠0
B. G≠0
C. n1≠n2
D. The data are in an order that is not random
E. The data are in a random order
F. r≠0
G. Median ≠0
H. β≠0
I. None of the above.

(c) The number of runs G is

A. 27
B. 30
C. 15
D. 17
E. 21
F. 19
G. None of the above.

(d) What kind of test should you conduct

A. Both sign and goodness of fit tests
B. Sign test
C. Independence Test
D. Goodness of fit test
E. Runs test for randomness and the test statisc is G
F. Runs test for randomness and the test statistic is a z-score
G. None of the above.

(e) What is/are the critical(s) value(s)

A. The smallest one is -1.96 and the largest one is 1.96
B. The smallest one is 11 and the largest one is 24
C. The only critical value is 11
D. The negative one is -1.645 and the positive one is 1.645
E. The smallest one is -2.575 and the largest one is 2.575
F. The only critical value is 24
G. The only critical value is 23
H. The only critical value is -2.575
I. None of the above.

(f) The test statistic is

A. z=1.65 with n1=15 and n2=18.
B. G=15 with n1=15 and n2=18.
C. z=1.49 with μG=19.37 and σG=4.80.
D. z=1.56 with n1=18 and n2=15.
E. z=−.84 with μG=17.36 and σG=2.8.
F. z=1.56 with n1=15 and n2=18.
G. z=−.49 with μG=19.37 and σG=2.8.
H. G=15 with n1=18 and n2=15.
I. z=−.49 with G=17.36 and σG=4.80.
J. None of the above.

(g) The conclusion:

A. We reject H0 and then there is enough evidence to support the calim that the above sequence was in a random order
B. We fail to reject H0 and then there is enough evidence to support the calim that the above sequence was in a random order
C. We reject H0 and then there isn't enough evidence to support the calim that the above sequence was in a random order
D. We fail to reject H0 and then there isn't enough evidence to reject the calim that the above sequence was in a random order.
E. We fail to reject H0 and then there isn't enough evidence to support the calim that the above sequence was in a random order
F. We fail to reject H0 and then there is enough evidence to reject the calim that the above sequence was in a random order
G. We reject H0 and then there is enough evidence to reject the calim that the above sequence was in a random order
H. We reject H0 and then there isn't enough evidence to reject the calim that the above sequence was in a random order
I. None of the above.

In: Statistics and Probability

Consider the following time series. t 1 2 3 4 5 yt 6 10 8 13...

Consider the following time series. t 1 2 3 4 5 yt 6 10 8 13 15 (a) Choose the correct time series plot. (i) (ii) (iii) (iv) What type of pattern exists in the data? (b) Use simple linear regression analysis to find the parameters for the line that minimizes MSE for this time series. If required, round your answers to two decimal places. y-intercept, b0 = 4.1 Slope, b1 = 2.1 MSE = ???? (c) What is the forecast for t = 6? If required, round your answer to one decimal place. 16.7 I know how to do it all but the MSE.

In: Statistics and Probability

Motorola used the normal distribution to determine the probability of defects and the number of defects...

Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of  ounces.

a. The process standard deviation is , and the process control is set at plus or minus  standard deviation . Units with weights less than  or greater than  ounces will be classified as defects. What is the probability of a defect (to 4 decimals)?

In a production run of  parts, how many defects would be found (round to the nearest whole number)?

b. Through process design improvements, the process standard deviation can be reduced to . Assume the process control remains the same, with weights less than  or greater than  ounces being classified as defects. What is the probability of a defect (round to 4 decimals; if necessary)?

In a production run of  parts, how many defects would be found (to the nearest whole number)?

c. What is the advantage of reducing process variation, thereby causing a problem limits to be at a greater number of standard deviations from the mean?

- Select your answer -It can substantially reduce the number of defectsIt may slightly reduce the number of defectsIt has no effect on the number of defectsItem 5

In: Statistics and Probability

A store owner claims the average age of her customers is 30 years. She took a...

A store owner claims the average age of her customers is 30 years. She took a survey of 33 randomly selected customers and found the average age to be 32.8 years with a standard error of 1.821. Carry out a hypothesis test to determine if her claim is valid.

(a) Which hypotheses should be tested?

H0: p = 30 vs. Ha: p ≠ 30

H0: μ = 30 vs. Ha: μ ≠ 30  

  H0: μ = 30 vs. Ha: μ > 30

H0: μ = 32.8 vs. Ha: μ ≠ 32.8

(b) Find the test statistic:   (Use 4 decimals.)

(c) What is the P-value?   (Use 4 decimals.)

(d) What should the store owner conclude, for α = 0.05?

Reject the initial claim of 30 years. There is insufficient evidence the mean customer age is different than 30.

Do not reject the initial claim of 30 years. There is insufficient evidence the mean customer age is different than 30.    

Do not reject the initial claim of 30 years. There is sufficient evidence the mean customer age is different than 30.

Reject the initial claim of 30 years. There is sufficient evidence the mean customer age is different than 30.

In: Statistics and Probability

Enter answers rounded to 5 decimal places. 1. In pharmacologic research a variety of clinical chemistry...

Enter answers rounded to 5 decimal places.

1. In pharmacologic research a variety of clinical chemistry measurements are routinely monitored closely for evidence of side effects of the medication under study. Suppose typical blood glucose levels are normally distributed with mean 90 mg/dL and standard deviation 38 mg/dL.

1a. If the normal range is 65-120 mg/dL, that what percentage of values will fall into the normal range?

1b. In some studies only values that are at least 1.5 times as high as the upper limit of normal are identified as abnormal. What percentage of values would fall in this range?

1c. What is the probability that two people will have a test of 1.5 times as high as the upper limit of normal? (Assume the tests are indpendent)

1d. What blood glucose levels separate the lower and upper 5% of blood glucose values? (round your answer to the nearest tenth) ,

In: Statistics and Probability

Suppose x is a normally distributed random variable with muμequals=3434 and sigmaσequals=44. Find a value x...

Suppose x is a normally distributed random variable with

muμequals=3434

and

sigmaσequals=44.

Find a value

x 0x0

of the random variable x.

a.

​P(xgreater than or equals≥x 0x0​)equals=.5

b.

​P(xless than<x 0x0​)equals=.025

c.

​P(xgreater than>x 0x0​)equals=.10

d.

​P(xgreater than>x 0x0​)equals=.95

In: Statistics and Probability

The ages of a group of 50 women are approximately normally distributed with a mean of...

The ages of a group of 50 women are approximately normally distributed with a mean of 51 years and a standard deviation of 6 years. One woman is randomly selected from the​ group, and her age is observed.

a. Find the probability that her age will fall between 56 and 61 years.

b. Find the probability that her age will fall between 48 and 51 years.

c. Find the probability that her age will be less than 35 years

. d. Find the probability that her age will exceed 40 years.

In: Statistics and Probability

The following table shows ceremonial ranking and type of pottery sherd for a random sample of...

The following table shows ceremonial ranking and type of pottery sherd for a random sample of 434 sherds at an archaeological location.

Ceremonial Ranking Cooking Jar Sherds Decorated Jar Sherds (Noncooking) Row Total

A 89 46 135

B 93 52 145

C 79 75 154

Column Total 261 173 434

Use a chi-square test to determine if ceremonial ranking and pottery type are independent at the 0.05 level of significance.

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

Since the P-value > α, we fail to reject the null hypothesis.

Since the P-value > α, we reject the null hypothesis.

Since the P-value ≤ α, we reject the null hypothesis.

Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is sufficient evidence to conclude that ceremonial ranking and pottery type are not independent.

At the 5% level of significance, there is insufficient evidence to conclude that ceremonial ranking and pottery type are not independent.

In: Statistics and Probability