A social worker is involved in a study about family structure. She o.pdf

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A social worker is involved in a study about family structure. She obtains information regarding the number of children per family for a certain community from the census data. Identify the variable of interest, determine whether it is discrete or continuous and list some possible values. Solution.

A social psychologist interested in cultural differences compared women of two ethnic groups on a Role Approval Index on which high scores indicated high degrees of approval of one\'s own social role. The results were as follows. Ethnic Group A: N=15, M=55, S2=6.5 Ethnic Group B: N=23, M=51, S2=4.5 If the standard deviation of the distribution of the difference between means is .76, what is the t score? Solution d.

A small town has been hit hard by the flu this winter. It is estimat.pdf

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A skeptical paranormal researcher claims the proportion of Americans.pdf

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A skeptical paranormal researcher claims the proportion of Americans that have seen a UFO, p, is less than 1 in every one thousand. Identify the type 2 error for the test. Please show all work on how you got this answer. Solution Type II error is when you do not reject the null hypothesis even though it is false. Ho: p = 1/1000 H1: p < 1/1000 (the researches claim) In this example the type II error would be finding that the proportion of Americans that have seen a UFO is 1/1000 or higher when in fact, the true proportion is less than 1 in 1000..

A skeptical paranormal researcher claims that the proportion of Amer.pdf

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A situation in which the Fed holds interest rates constant (at a low.pdf

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A simple random sample of 1000 American adults found that the averag.pdf

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A simple random sample of 1000 American adults found that the average number of hours spent watching television during a typical week was 13.8. A simple random sample of 500 Canadians yielded an average of 12.5 hours per week of television viewing. Assume that the American and Canadian distributions for weekly television, viewing times have the same standard deviations. The sampling variability associated with the sample means (described above) is A.) smaller for the sample of Canadians, since the population of Canada is less than half that of the United States. B.) larger for the sample of Canadians because the sample size is smaller. C.) smaller for the sample of Canadians, because their population is smaller. D.) larger for the sample of Canadians, since Canadian citizens are more widely dispersed throughout their country than American citizens are in the United States. Hence, Canadians have more variable views. Solution b. remember that sample standard deviation is given by sigma/sqrt(n).

A simple random sample of 100 observations was taken from a large po.pdf

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A shipment of 11 printers contains 2 that are defective. Find the pr.pdf

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A set of waveforms applied to two D flip-flops is shown in Figure 4-56. These waveforms are applied to the flip-flops shown along with the values of their timing parameters. (a) List the time(s) at which there are timing violations in signal D1 for flip-flop 2. (b) List the time(s) at which there are timing violations in signal D2 for flip-flop 2. Solution flip flop 1 is a positive edge triggered flop. The inputs should be seen at the rising edge of the clock for signal D1, at the fourth rising edge i.e t = 28ns, violation occurs. D1 is rising at the same edge, it should be stable for Ts = 1ns before and Th =0.5ns after the edge. 2) flipflop 2 is a negative edge triggered. At second negative edge i.e t = 16ns and third negative edge i.e t = 24ns, D2 signal viloates the timing requirements of flop2.

A significant advantage of internal funding is the fact that retaine.pdf

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A significant advantage of internal funding is the fact that retained earnings do not incur flotation cost for the company; this advantage comes at the cost of reduced capital gains for the firm’s shareholders. True False Solution A significant advantage of internal funding is the fact that retained earnings do not incur flotation cost for the company; this advantage comes at the cost of reduced capital gains for the firm’s shareholders. TRUE UNLIKE OTHER SOURCES OF FINANCEING THE RETAINED EARNNINGS DOES NOOT REQUIRE ANY ISSUE RELATED COST..

A short quiz has two truefalse questions and one multiple choice qu.pdf

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A short quiz has two true/false questions and one multiple choice question with four choices. A student guesses at each question. Assuming the choices are all equally likely, what is the probability that the student gets all three correct? Solution The events are independent, so P(A and B) = P(A)*P(B) --- P(2 correct on truefalse) = (1/2)(1/2) = 1/4 P(1 correct mult-choice) = 1/4 ---- P(this and that) = (1/4)(1/4) = 1/16.

A share of common stock has just paid a dividend of $2.00 (D0 = 2.00.pdf

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A set of 50 data values has a mean of 27 and a variance of 16. I. .pdf

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A set of 50 data values has a mean of 27 and a variance of 16. I. Find the standard score (z) for a data value = 19. II. Find the probability of a data value < 19. III. Find the probability of a data value > 19. Show all work. Solution I am assuming X ~ N(27, 16) to find the probability of greater than 19, just take the fact that the probabilities add to 1 and 1 - the above probability is the probability that x is greater than 19.

A set of 50 data values has a mean of 21 and a variance of 16. I. .pdf

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A set of 50 data values has a mean of 21 and a variance of 16. I. Find the standard score (z) for a data value = 31. II. Find the probability of a data value > 31. III. Find the probability of a data value < 31. Show all work. Solution If the variance is 4, sigma is the sqrt(4) = 2 z = (x - mu) / sigma z = (31 - 28) / 2 z = 1.5 So from there, look at a z-table. P(z>1.5) = 1 - .9332 = .0668 P(z<1.5) = .9332.

A set of 50 data values has a mean of 16 and a variance of 4. I. F.pdf

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A set of 50 data values has a mean of 16 and a variance of 4. I. Find the standard score (z) for a data value = 12. II. Find the probability of a data value > 12. Solution n=50 =16 = population mean 2 = 4, so standard devation =2 A) z = (x - )/ = (12 - 16)/2 = 2 B) P(x > 12) = P( (x - )/ > (12-16)/2 ) = P( z > 2) = .0228 I did not use the standard error of the sampling distribution [ (x - )/(/n) ] because that is only used when dealing with a sample. The problems clearly state \"data value\" which I understood as it being a single observation rather than the sample mean..

A set of 10 data values has a mean and a median of $1425 and ranges .pdf

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