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Ksju [112]
2 years ago
10

A newspaper article about an opinion poll says that "53% of Americans approve of the president's overall job performance." The p

oll is based on online survey interviews with 1,350 adults randomly chosen from a numbered list of one million responses from around the United States, excluding Alaska and Hawaii.
Part A: What is the population and sample in this poll? (3 points)

Part B: What type of sampling is used? (3 points)

Part C: Are there any sources of bias present? Explain. (4 points) (10 points)
Mathematics
1 answer:
Dvinal [7]2 years ago
7 0

Answer:

<u>Part A:</u> The population are the American adult citizens, excluding the ones from Alaska and Hawaii. The population is the people which the sample is trying to represent, as a hole.

The sample is a portion of this population, and in this case is represented by a randomly selected amount of people whose response to the interview has been selected.

<u>Part B:</u> The sample here has been selected in two steps. The <u>first</u> step is the one that we must pay attention to: the numbered list of one million responses from around the US (excluding Alaska and Hawaii). Because these responses were obtained by an online survey, the sample looks like a convenience sampling, as it depends on the availability and willingness from participants to take part of the study (is not compulsory for everyone, so not everyone is going to response, then there are people that is not going to be represented by). The <u>second</u> step is the random selection of a part of the previous responses. This last part will ensure that, the individuals that took part of the group that was interviewed, are well represented in the results.

<u>Part C:</u> As it was mentioned, there is a selection bias, because the information from the sample comes from a specific group of people that has certain features that may not represent all American adults citizens. For <u>example, the opinion of  those people who do not use internet</u>, will not be considered (and they may be a large number of persons). This situations weaken the conclusions obtained in the study, as they are not representative of the hole population.

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Suppose students' ages follow a skewed right distribution with a mean of 24 years old and a standard deviation of 3 years. If we
natta225 [31]

Answer: Option 'c' is correct.

Step-by-step explanation:

Since we have given that

Mean of students' age = 24 years

Standard deviation of students' age = 3 years

Sample size = number of students = 350

So, according to options,

a. The shape of the sampling distribution is approximately normal.

It is true as n >30, we will use normal.

b. The mean of the sampling distribution is approximately 24-years old.

It is true as it is given.

c. The standard deviation of the sampling distribution is equal to 5 years.

It is not true as it is given 3 years.

Hence, Option 'c' is correct.

8 0
2 years ago
A building engineer analyzes a concrete column with a circular cross section. The circumference of the column is 18pi meters. Wh
ella [17]

Answer

Circumference(C) of a circle is given by:

C = 2\pi r

where r is the radius and value of \pi = 3.14

As per the given statement:

A building engineer analyzes a concrete column with a circular cross section.

The circumference of the column is 18 \pi meters.

then;

18 \pi= 2\pi r

Divide both sides by 2 \pi we have;

9 = r

or

r = 9 meters

We have to find the area of the cross section of the column

Area of a circle is given by:

A = \pi r^2

then;

A = \pi \cdot 9^2 = 81 \pi meter square.

therefore, the area A of the cross section of the column is 81 \pi meter square.

3 0
2 years ago
Read 2 more answers
The average score of all golfers for a particular course has a mean of 64 and a standard deviation of 3. Suppose 36 golfers play
kondaur [170]

Answer:

the probability that the exceeded 65 = 0.3707

The average score of the 36 golfers exceeded 65

                        = 36 X 0.3707 = 13.3452

Step-by-step explanation:

<u>Step 1</u>:-

The average score of all golfers for a particular course has a mean of 64 and a standard deviation of 3.

mean (μ) = 64

standard deviation (σ) =3

by using normal distribution

given (μ) = 64 and  (σ) =3

i) when x =65

z = \frac{x-mean}{S.D} = \frac{65-64}{3} = 0.33 >0

P( X≥ 65) = P(z≥0.33)

              = 0.5 - A(z₁)

              = 0.5 - 0.1293 (see normal table)

             = 0.3707

The average score of the 36 golfers exceeded 65

                        = 36 X 0.3707 = 13.3452

                                               

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2 years ago
Tickets to a show cost $5.50 for adults and $4.25 for students. A family is purchasing 2 adult tickets and 3 student tickets. If
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Answer:

Step-by-step explanation:

The cost of an adult ticket to the show is $5.5

The cost of a student ticket to the show is $4.25

A family is purchasing 2 adult tickets and 3 student tickets. This means that the total cost of 2 adult tickets would be

2 × 5.5 = $11

It also means that the total cost of 3 student tickets would be

3 × 4.25 = $12.75

The total cost of 2 adult tickets and 3 student tickets would be

11 + 12.75 = $23.75

If the family pays $25, the exact amount of change they should receive is

25 - 23.75 = $1.25

5 0
3 years ago
Match the measurable quantity given by its formula with the appropriate unit(s)
lys-0071 [83]
Umm I think you have to edit the quest because I don’t understand it umm
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