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Ray Of Light [21]
3 years ago
9

A large school district is considering moving up the start date for the school year by two weeks. In order to determine if famil

ies would support this, surveys were mailed out to 1000 parents/guardians in the district. The survey said, "We believe that starting school two weeks earlier will be beneficial to the students in our District. Would you support moving up the start date for the school year by two weeks?" The School Board received 492 responses from the survey, 72% of which supported the proposal. -Describe a potential source of bias in the wording of the question. -Due to the bias described in part (a), is the sample proportion of 72% likely greater than or less than the actual proportion of families who support the proposal? Explain. -What type of sample was taken? -Less than half of those sent the letter returned a response. What type of bias is this?
Mathematics
1 answer:
Anvisha [2.4K]3 years ago
5 0

Answer:

(a) Confirmation bias

(b) Likely less than actual proportion of families who support the proposal

(c) Random sample

(d) Nonresponse bias

Step-by-step explanation:

In confirmation bias, the researcher already has and intends to prove an assumption about the effect the of a treatment before carrying out the survey, such that the survey or research is tries to dictate to the participants in the survey about the desired outcome

In the question, the belief of the researcher is expressed to the participants in a form of brief at the start of the survey which is attempting to influence the responses to the survey, such that the actual result should be less than what was observed.

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You interview four people. You find that person A is a married 24​-year-old white female with 2 pets and whose college GPA was 2
nadezda [96]

Answer:

The data file is included in attached excel file.

Step-by-step explanation:

There are seven variables in the data file for four persons interviewed. Person type, marital status, ethnic group and gender are categorical variables while age, number of pets and GPA are quantitative variables. Person type is classified as A,B,C and D. Marital status consists of category married and single. Age of persons lies between 20-50. Ethic group has three categories that are white, Asian and Hispanic. Gender of persons has two categories male and female. Number of pets for the persons interviewed lies in the range of 2 to 5. Last GPA variable ranges from 2.2 to 3.8.

Download xlsx
4 0
4 years ago
Which equation, when graphed, has x-intercepts at (-1,0) and (-5,0) and a y-intercept at (0, -30)?
Vladimir [108]

If a function has roots -1 and -5, it must be in the form

f(x)=a(x+1)(x+5)

We can fix the coefficient a by imposing the passing through (0,-30):

f(0)=-30=a(1)(5)\iff a=-6

So, the function is

f(x)=-6(x+1)(x+5)

7 0
3 years ago
Which posulate proves the two triangles are congruent?
san4es73 [151]

Answer:

D ASA

Step-by-step explanation:

ASA

5 0
2 years ago
The most common form of color blindness is an inability to distinguish red from green. However, this particular form of color bl
Alecsey [184]

Answer:

(a) The correct answer is P (CBM) = 0.79.

(b) The probability of selecting an American female who is not red-green color-blind is 0.996.

(c) The probability that neither are red-green color-blind is 0.9263.

(d) The probability that at least one of them is red-green color-blind is 0.0737.

Step-by-step explanation:

The variables CBM and CBW are denoted as the events that an American man or an American woman is colorblind, respectively.

It is provided that 79% of men and 0.4% of women are colorblind, i.e.

P (CBM) = 0.79

P (CBW) = 0.004

(a)

The probability of selecting an American male who is red-green color-blind is, 0.79.

Thus, the correct answer is P (CBM) = 0.79.

(b)

The probability of the complement of an event is the probability of that event not happening.

Then,

P(not CBW) = 1 - P(CBW)

                   = 1 - 0.004

                   = 0.996.

Thus, the probability of selecting an American female who is not red-green color-blind is 0.996.

(c)

The probability the woman is not colorblind is 0.996.

The probability that the man is  not color- blind is,

P(not CBM) = 1 - P(CBM)

                   = 1 - 0.004  

                   = 0.93.

The man and woman are selected independently.

Compute the probability that neither are red-green color-blind as follows:

P(\text{Neither is Colorblind}) = P(\text{not CBM}) \times  P(\text{not CBW})\\ = 0.93 \times  0.996 \\= 0.92628\\\approx 0.9263

Thus, the probability that neither are red-green color-blind is 0.9263.

(d)

It is provided that a one man and one woman are selected at random.

The event that “At least one is colorblind” is the complement of part (d) that “Neither is  Colorblind.”

Compute the probability that at least one of them is red-green color-blind as follows:

P (\text{At least one is Colorblind}) = 1 - P (\text{Neither is Colorblind})\\ = 1 - 0.9263 \\= 0.0737

Thus, the probability that at least one of them is red-green color-blind is 0.0737.

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Lina20 [59]

Answer:

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Step-by-step explanation:

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