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April 16, 2014

April 16, 2014

**Recent Homework Questions About Statistics**

Post a New Question | Current Questions

**STATISTICS**

Program evaluation differs from policy analysis with respect to _____. a. the time dimension within the policy process b. the cost c. the time d. the role of the stakeholder
*Monday, December 9, 2013 at 9:00pm*

**Statistics**

True or False If a sample of at least 30 scores is randomly selected from a normal population, the sample mean will be equal to the population mean.
*Monday, December 9, 2013 at 7:48pm*

**Statistics**

A sample of n = 9 scores is obtained from a population with ƒÝ = 70 and ƒã = 18. If the sample mean is M = 76, then what is the z-score for the sample mean? a. z = 0.33 b. z = 0.50 c. z = 1.00 d. z = 3.00
*Monday, December 9, 2013 at 7:32pm*

**Statistics**

Z = (score-mean)/SD You need the values of the mean and standard deviation. Unless you are dealing with a distribution of means, n does not matter.
*Monday, December 9, 2013 at 12:41pm*

**Statistics**

1.00 - .49 - .25 = ?
*Monday, December 9, 2013 at 12:35pm*

**Psychological Statistics**

What research situations have you chosen? IQ compared to GPA? (r = ?) IQ compared to letter grade in a particular course? (X^2 = ?) I'll leave the rationals up to you.
*Monday, December 9, 2013 at 12:32pm*

**statistics**

Arranged in order of value as you have them, with 20 scores, between the 17th and 18th scores would be the 90th percentile (defining the top 10%). Indicate your specific subject in the "School Subject" box, so those with expertise in the area will respond to the ...
*Monday, December 9, 2013 at 12:23pm*

**public speaking**

Paul is delivering a speech on the hunger epidemic in the world. His speech could be helped with the use of: A. facts and statistics. B. narratives. C. examples. D. All of the above Answer: D. All of the above
*Monday, December 9, 2013 at 8:27am*

**STATISTICS HELP**

A student wishes to test the accuracy of the measuring jug she is using in an experiment. The label on the jug claims that it measures volume to an accuracy of plus or minus 5 cm3. The density of water is about 0.999 g cm−3, which means that 100 cm3 of water should have ...
*Monday, December 9, 2013 at 8:17am*

**typo - Statistics**

near the end it should say: So now we have 4 cases hobbits believing in neither one = .65 hobbits believing in both = .09 hobbits believing in only elves = .19 hobbits believing in only ents = .07 add them up to get 1
*Sunday, December 8, 2013 at 11:48pm*

**Statistics**

conditional probability: prob (ents given elves) = prob(endts | elves) = prob(elves AND ents)/prob (elves) prob(ents | elves) = .09/.28 = .3214 prob (2hobbits believing in elves) = (.28)(.28) = .0794 They do add up to 1 if you add up ALL possible cases, you added up only 3 A ...
*Sunday, December 8, 2013 at 11:44pm*

**Statistics**

Suppose that you visit Shire, and find that 28% of Hobbits believe in the existence of Elves, 16% believe in the existence of Ents, and 9% believe in the existence of both Elves and Ents. You come across a Hobbit: i) What is the probability that he will believe in Ents given ...
*Sunday, December 8, 2013 at 10:54pm*

**Statistics**

.0288
*Sunday, December 8, 2013 at 5:12pm*

**Psychological Statistics**

Post a behavioral research situation that could use a Pearson coefficient research study and a chi square research study. Present the rationale for each selection. Be very specific in your presentation.
*Sunday, December 8, 2013 at 4:33pm*

**Statistics**

The probability of Joe is at work is 49%. The probability Joe is at home is 25%. What is the probability Jow is neither at Work nor at home? A) 39% B)74% C)52% D)26% I get the answer at about 38%,but doesn't round to 39%, can anyone help, I'm stuck.
*Sunday, December 8, 2013 at 4:02pm*

**Statistics**

How would you find the z-score for the following X=86 and the n=17.
*Sunday, December 8, 2013 at 1:28pm*

**Statistics**

Although you indicate n = 17, you only have 16 scores listed. Find the mean first = sum of scores/number of scores Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance. Standard...
*Sunday, December 8, 2013 at 12:48pm*

**Business Math and Statistics**

1000(1+.02)^4 = 1082.43
*Sunday, December 8, 2013 at 7:38am*

**Business Math and Statistics**

You deposit $1,000 for 4 years at an interest rate of 2.0%. If the interest is compounded annually, how much money do you have after 4 years?
*Sunday, December 8, 2013 at 1:33am*

**Statistics**

Reiny Thank you for your help on this homework.
*Saturday, December 7, 2013 at 9:11pm*

**Statistics**

Find the Standard Deviation for this distribution of scores. Then find the z score for X=86. N=17 98, 92, 86, 84 , 76, 74 , 72 , 72, 72 , 70, 70, 66, 60 ,54, 32,30
*Saturday, December 7, 2013 at 9:02pm*

**statistics**

Find the mean first = sum of scores/number of scores Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance. Standard deviation = square root of variance 95% = mean ± 1.96 ...
*Saturday, December 7, 2013 at 7:43pm*

**Statistics**

Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you. See later post. Again, what are the total number of plants?
*Saturday, December 7, 2013 at 7:23pm*

**STATISTICS**

What you want is the probability of getting 1, 2 or 3 hermaphrodite plants. However, what is the total number of plants? If there are only three, then the answer = 1 (certainty). If the events are independent, the probability of both/all events occurring is determined by ...
*Saturday, December 7, 2013 at 7:08pm*

**STATISTICS**

Two-thirds of papaya plants availabe are hermaphrodite and are self-pollinating. One-third of plants are female and need to be paired withh an hermaphrodite plant in order to be pollinated. Unless the plants are pollinated, they will not bear fruit. If you select three papaya ...
*Saturday, December 7, 2013 at 6:06pm*

**Healthcare Statistic**

Given the following null hypothesis, give an example of a Type II error. H0: There is no difference in the level of understanding of this chapter between students who have previously taken a statistics course and students who have not previously taken a statistics course.
*Saturday, December 7, 2013 at 5:50pm*

**Statistics**

Two-thirds of papaya plants availabe are hermaphrodite and are self-pollinating. One-third of plants are female and need to be paired withh an hermaphrodite plant in order to be pollinated. Unless the plants are pollinated, they will not bear fruit. If you select three papaya ...
*Saturday, December 7, 2013 at 4:21pm*

**Statistics**

i dont know
*Saturday, December 7, 2013 at 2:55am*

**Easy Statistics**

prob(Joe not at work) = .51 prob(Joe not at home) = .75 prob(joe not at work and not at home) = (.51)(.75) = .3825 = 38% Don't know how they got 39%
*Friday, December 6, 2013 at 9:44pm*

**Easy Statistics**

The probability Joe is at work is 49%,. The probability Joe is at home is 25%. What is the probability Joe is neither at work nor at home? a) 39% b)74% c)52% d) 26% *I personally think it is a or c. I am just needing some help thanks!
*Friday, December 6, 2013 at 9:22pm*

**statistics**

Where are the scores
*Friday, December 6, 2013 at 4:10pm*

**statistics**

Nine students took the SAT. Their scores are listed below. Later on, they read a book on test preparation and retook the SAT. Their new scores are listed below. Construct a 95% confidence interval for µ d (the true mean difference in scores). Assume that the distribution...
*Friday, December 6, 2013 at 3:39pm*

**Statistics**

This is a little clearer. The larger portion (body) for positive Z scores is to the left. For negative Z scores, it is to the right. Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/...
*Friday, December 6, 2013 at 12:53pm*

**Statistics**

We cannot draw on these posts. The Z score is your score in terms of standard deviations from the mean. There would be a tail on both sides. Are you looking for the smaller portion? https://www.google.com/search?client=safari&rls=en&q=normal+distribution+standard...
*Friday, December 6, 2013 at 12:49pm*

**Statistics**

From your limited data, there is no variability measure for t-test. I wouldn't even know how to figure the mean. You might want to use chi-square test for whatever you are hypothesizing. To use the Chi-square (X^2) method: X^2 = ∑ (O-E)^2/E, where O = observed ...
*Friday, December 6, 2013 at 12:36pm*

**statistics**

The espresso project you are heading up will do well with probability 0.80. Given that it does well, you believe that the herbal tea project is likely to do well with probability 0.70.However, if the espresso project does not do well, then the herbal tea project has only a 25...
*Friday, December 6, 2013 at 9:13am*

**Statistics**

Draw a vertical line through a normal distribution for each of the following z-score locations. Determine whether the body is on the right or left side od the line and find the proportion in the body. a. z= 2.20 b. z= 1.60 c. z= -1.50 d. z= -0.70
*Friday, December 6, 2013 at 12:47am*

**Statistics**

Draw a vertical line through s normal distribution for each of the following z-score locations. Determine whether the tail is on the right or left side of the line and find the proportion in the tail. a. z=2.00 b. z=0.60 c. z= -1.30 d. z= -0.30
*Friday, December 6, 2013 at 12:43am*

**Statistics**

z = (90-100)/20 = -.5 z table = 0.3085 z = (110-100)/20 = .5 z table = 0.6915 0.6915-0.3085 = 0.3830 D
*Thursday, December 5, 2013 at 8:30pm*

**Statistics**

A normal distribution has a mean of ƒÝ = 100 with ƒã = 20. If one score is randomly selected from this distribution, what is the probability that the score will have a value between X = 90 and X = 110? a. 0.6915 b. 0.3085 c. 0.1915 d. 0.3830
*Thursday, December 5, 2013 at 8:16pm*

**Statistics (?)**

What is your question?
*Thursday, December 5, 2013 at 4:39pm*

**Statistics**

Next time, give your response, so we can evaluate it. True
*Thursday, December 5, 2013 at 4:22pm*

**Statistics**

True
*Thursday, December 5, 2013 at 4:21pm*

**Statistics**

True
*Thursday, December 5, 2013 at 4:21pm*

**Statistics**

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability [(1-.95)/2] = Z = 1.96. You divide by 2, because you are considering both above and below. 95% = mean ± 1.96 SEm The ...
*Thursday, December 5, 2013 at 4:17pm*

**Statistics**

Mean = 4.33 SE 3 1- a = .683 a= 0.317 Za/2 = Z0.1585 = 1 z table invNorm(0.1585) = 1 mean -+ Z *SE 4.33 -1(3), 4.33+1*3 (1.33, 7.33) You can follow concept and you will get rest of your answer.
*Thursday, December 5, 2013 at 1:07pm*

**Statistics**

Somebody be so kind and brake this down to me, please it's for a test I need the help.
*Thursday, December 5, 2013 at 11:56am*

**Statistics**

Given the following information, determine the 68.3 percent, 95.5 percent, and 99.7 percent confidence intervals. overbar above X equals 4.33 comma SE sub m equals 3 I've looked at the table as recommended. I just do not understand this at all. I'm really lost on this ...
*Thursday, December 5, 2013 at 11:54am*

**Statistics**

true
*Thursday, December 5, 2013 at 9:54am*

**Statistics**

true, but your answer should be reduced to 1/6
*Thursday, December 5, 2013 at 9:52am*

**Statistics**

True or False For any normal distribution, the proportion in the tail beyond z = 2.00 is p = 0.0228.
*Thursday, December 5, 2013 at 7:01am*

**Statistics**

True or False A jar contains 15 red marbles and 75 blue marbles. If you randomly select a marble from this jar, the probability of obtaining a red marble is p = 15/90.
*Thursday, December 5, 2013 at 7:00am*

**Statistics**

True or False One requirement of a random sample is that every individual in the population has an equal chance of being selected.
*Thursday, December 5, 2013 at 6:59am*

**Statistics**

True or False The value for a probability can never be less than zero, unless you have made a computational error.
*Thursday, December 5, 2013 at 6:56am*

**Statistics**

True or False All probabilities can be expressed as decimal values ranging from 0 to 1.00.
*Thursday, December 5, 2013 at 6:54am*

**AP Statistics**

A standard deck of cards consisting of 52 cards, 13 in each of 4 different suits, is shuffled, and 4 cards are drawn without replacement. What is the probability that all four cards are of a different suit?
*Wednesday, December 4, 2013 at 8:24pm*

**statistics**

joij
*Wednesday, December 4, 2013 at 5:33pm*

**statistics**

I have just a couple questions I have trouble understanding... can someone please help me? The data below indicate the rankings of a set of employees according to class theory and on-the-job practice evaluations: Theory 1 7 2 10 4 8 5 3 6 9 Practice 2 8 1 7 3 9 6 5 4 10 What ...
*Wednesday, December 4, 2013 at 2:22pm*

**Statistics**

Your program plans to acquire a total of 600 end items costing $535 million over a five-year period. The first production contract is to be awarded in FY04. The number of items to be procured (not delivered) each year and their estimated cost (in then-year dollars) is shown ...
*Tuesday, December 3, 2013 at 10:03pm*

**Statistics**

You might want to round it to the nearest person. It rounds to 17, but if you want all the records to be covered, you might round it up to 18. This would give the FTEs a little slack.
*Tuesday, December 3, 2013 at 1:00pm*

**statistics**

(100*10)/(7*60) = ?
*Tuesday, December 3, 2013 at 12:41pm*

**statistics (?)**

What is your question? If you want to know Kelsey's score: Z = (score-mean)/SD -1.85 = (score-72)/4
*Tuesday, December 3, 2013 at 12:38pm*

**statistics**

poor Kelsey.
*Tuesday, December 3, 2013 at 5:30am*

**statistics**

The average for the statistics exam was 72 and the standard deviation was 4. Kelsey was told by the instructor that she scored 1.85 standard deviations below the mean.
*Tuesday, December 3, 2013 at 4:23am*

**statistics**

The HIM department at community hospital will experience a 15 percent increase in the number of discharges coded per day as a result of opening a cardic clinic in the facility. The 15 percent increase is projected to be 100 additional records per day. The standard time to code...
*Tuesday, December 3, 2013 at 4:05am*

**statistics**

mn counting rule
*Tuesday, December 3, 2013 at 3:43am*

**Statistics**

A vending machine is designed to dispense a mean of 7.6 oz. of coffee into an 8-oz cup. If the standard deviation of the amount of coffee dispensed is 0.4 oz. and the amount is normally distributed, find the percent of times that the machine will: a.) Dispense from 7.4 oz. to ...
*Tuesday, December 3, 2013 at 12:02am*

**Statistics**

Dear PsyDAG, Thanks for your help, I suppose the answer to be 17.2 FETs?
*Monday, December 2, 2013 at 9:51pm*

**Statistics**

Dear PsyDAG, Thanks for your help, I suppose the answer to be 17.2 FETs?
*Monday, December 2, 2013 at 9:51pm*

**statistics**

Mean=9 Variance=792 Standard Deviation=28.14
*Monday, December 2, 2013 at 8:54pm*

**statistics**

whether it is one-tailed or two tailed? and if one tailed...I do agree with you now that this is the left tail test. However I am not sure how to handle negative value on t. Here is what I did. I used formula for hypothesis test with mu unknow, t=(xbar-mu)/(sd/sqr n)=(26-25.02...
*Monday, December 2, 2013 at 8:23pm*

**elementary statistics**

There are different ways you can do this kind of problem, but the formula below might be one of the easier ways: s/[1 + (1.645/√2n)] ..to.. s/[1 - (1.645/√2n)] ...where s = standard deviation, 1.645 represents the 90% confidence interval using a z-table, and n = ...
*Monday, December 2, 2013 at 5:42pm*

**Statistics**

1. D Chi Square distributions are positively skewed (skewed to the right). As the degrees of freedom increases, the Chi Square distribution will approach a normal symmetrical distribution. Smaller degrees of freedom will skew more to the right. 2. A Ho: The variables are ...
*Monday, December 2, 2013 at 5:28pm*

**Statistics 2**

Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you. A
*Monday, December 2, 2013 at 12:06pm*

**Statistics**

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. 13/52 * 12/51 * 11/50 * 10/49 * (13*3)/48 = ?
*Monday, December 2, 2013 at 12:05pm*

**Statistics**

8600/(100*5) = ?
*Monday, December 2, 2013 at 11:32am*

**Statistics**

community physicians clinic is a large clinic with 85 physicians. they treat about 8,600 patients each week. coders are expected to code 100 clinic records each day. how many Full Time Employees are needed to code these records? ( assume a five day work week)
*Monday, December 2, 2013 at 7:41am*

**STATISTICS **

To calculate the mean for grouped data, which of the following steps must be completed? a. multiply the number of subjects within each category by the value of that category b. divide the results of the multiplication c. multiply by the number of subjects d. multiply each ...
*Monday, December 2, 2013 at 5:32am*

**Statistics**

Yes. 5% of 300 = 15
*Sunday, December 1, 2013 at 10:28pm*

**Statistics**

smith is a weld inspector at a shipyard. he knows from keeping track of good and substandard welds that for the afternoon shift 5% of all welds done will be substandard. If Smith checks 300 of the 7500 welds completed that shift, would it be unusual for Smith to find 30 or ...
*Sunday, December 1, 2013 at 10:21pm*

**Business Statistics**

Using the telephone numbers listed in your local directory as your population, randomly obtain 20 samples of size 3. From each telephone number identified as a source, take the fourth, fifth, and sixth digits. a. Calculate the mean of the 20 samples b. Draw a histogram showing...
*Sunday, December 1, 2013 at 10:14pm*

**Statistics**

We have a population of all adult women. We take a random sample from this population and create a confidence interval. If we take many, many, many different samples from this population we will obtain many, many, many different sample means and confidence intervals created ...
*Sunday, December 1, 2013 at 9:01pm*

**Statistics**

1. The Chi-square distribution, used in the Chi-square test of independence, varies in shape by degrees of freedom. What does the Chi-square distribution look like for 4 degrees of freedom. A) Unimodal and symmetric. B) Bimodal and symmetric. C) Unimodal and skewed to the left...
*Sunday, December 1, 2013 at 8:14pm*

**Statistics **

Explaining confidence. A student reads that a 95% confidence interval for the mean ideal weight given by adult American women is 140 ± 1.4 pounds. Asked to explain the meaning of this interval, the student says, “95% of all adult American women would say that ...
*Sunday, December 1, 2013 at 8:09pm*

**Statistics**

Suppose 5 cards are drawn, without replacement, from a standard bridge deck of 52 cards. Find the probability of drawing 4 clubs and 1 non- club.
*Sunday, December 1, 2013 at 4:26pm*

**Statistics 1**

(10+3 +4+ 25 + 1+4+ 4 + 12+ 6 +8 +10)/11 87/11 = 7.9 Answer B
*Sunday, December 1, 2013 at 2:37pm*

**Statistics 2**

To calculate the mean for grouped data, which of the following steps must be completed? a. multiply the number of subjects within each category by the value of that category b. divide the results of the multiplication c. multiply by the number of subjects d. multiply each ...
*Sunday, December 1, 2013 at 2:05pm*

**Statistics 1**

Professor White asked his students to put the number of hours each studied for the final exam on a piece of paper and turn that paper in with the exam. The pieces of paper indicate that students studied 10, 3, 4, 25, 1, 4, 4, 12, 6, 8, and 10 hours. What is the mean number of ...
*Sunday, December 1, 2013 at 2:03pm*

**child care**

This is probably true, although it's just a guess on my part. "Reliable statistics on the prevalence of shaken baby syndrome do not exist. Estimates in the United States approach 50,000 cases each year." http://www.healthline.com/galecontent/shaken-baby-...
*Sunday, December 1, 2013 at 1:48pm*

**statistics**

It is the Z test (google it). Where is your confusion centered?
*Sunday, December 1, 2013 at 12:23pm*

**Math.stat help :(**

Ho: mean1 = mean2 Ha: mean1 ≠ mean2 Since these are just samples rather than the whole population, µ is an inappropriate designation. µ is parameter rather than a statistic. Z = (mean1 - mean2)/standard error (SE) of difference between means SEdiff = √(...
*Sunday, December 1, 2013 at 12:19pm*

**Statistics**

Z = (score-mean)/SD Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
*Sunday, December 1, 2013 at 12:09pm*

**Probability and Statistics**

Suppose that the antenna lengths of woodlice are approximately normally distributed with a mean of 0.2 inches and a standard deviation of 0.05 inches. What proportion of woodlice have antenna lengths that are more than 0.23 inches? Round your answer to at least four decimal ...
*Sunday, December 1, 2013 at 11:37am*

**Statistics**

1-p(x = 0) + p(x = 1) 1-[(11C0)(.015)^0(.985)^11 + 11C1 *0.015^1 *.985^10)] 1-[.84684 + .14185] 1-.9887 = 0.0113
*Saturday, November 30, 2013 at 11:18pm*

**elementary statistics**

A statistics professor is used to having a variance in his class grades of no more than 100. He feels that his current group of students is different, and so he examines a random sample of midterm grades (listed below). At 0.05 alpha level can it be conculded that the variance...
*Saturday, November 30, 2013 at 10:30pm*

**Statistics**

Of the parts produced by a particular machine, 1.5% are defective. If a random sample of 11 parts produced by this machine contains 2 or more defective parts, the machine is shut down for repairs. Find the probability that the machine will be shut down for repairs based on ...
*Saturday, November 30, 2013 at 9:55pm*

**Probability and Statistics**

you're right
*Saturday, November 30, 2013 at 9:45pm*

**Probability and Statistics**

β = P(Type II Error) =P(6≤X ≤13|p=0.5) (0.5)^x(0.5)^15-x ≈ 0.8454 β = P(Type II Error) =P(6≤X ≤13|p=0.7 0.7^x (0.3)^15-x ≈
*Saturday, November 30, 2013 at 9:42pm*

**Probability and Statistics**

The proportion of adults living in a small town who are college graduates is estimated to be p = 0.6. To test this hypothesis, a random sample of 15 adults is selected. If the number of college graduates in the sample is anywhere from 6 to 12, we shall not reject the null ...
*Saturday, November 30, 2013 at 8:06pm*

**elementary statistics**

A statistics professor is used to having a variance in his class grades of no more than 100. He feels that his current group of students is different, and so he examines a random sample of midterm grades (listed below). At 0.05 alpha level can it be conculded that the variance...
*Saturday, November 30, 2013 at 7:44pm*

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