Sean is conducting an experiment with an isotope of a particular element that has a half life of 4 hours. He begins with a sampl
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Answer:
(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].
(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval
(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.
Step-by-step explanation:
We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.
Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;
P.Q. = ~ N(0,1)
where, = sample proportion of Americans who decide to not go to college = 48%
n = sample of American adults = 331
p = population proportion of Americans who decide to not go to
college because they cannot afford it
<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>
<em />
<u>So, 90% confidence interval for the population proportion, p is ;</u>
P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level
of significance are -1.645 & 1.645}
P(-1.645 < < 1.645) = 0.90
P( <
<
) = 0.90
P( < p <
) = 0.90
<u>90% confidence interval for p</u> = [ ,
]
= [ ,
]
= [0.4348, 0.5252]
(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].
(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.
3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.
So, the margin of error =
= 54.79
n =
n = 3001.88 ≈ 3002
Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.
I found a similar problem to your problem here, which is shown in the attached picture. So, from the picture, we have to find the equation for the red line. All we have to do is find two points of the line. That would be: Point 1(2,0) and Point 2(-2,3). The general equation would be:
y - y₁ = (y₂-y₁)/(x₂ - x₁) * (x - x₁)
Substituting the coordinates to the equation,
y - 0 = (3-0)/(-2 - 2) * (x - 2)
y = -3(x -2)/4
Rearranging,
<em>4y = -3x + 6 or 4y + 3x = 6</em>
y - y₁ = (y₂-y₁)/(x₂ - x₁) * (x - x₁)
Substituting the coordinates to the equation,
y - 0 = (3-0)/(-2 - 2) * (x - 2)
y = -3(x -2)/4
Rearranging,
<em>4y = -3x + 6 or 4y + 3x = 6</em>
Answer:
The joint probility density will be c = 1/36
The joint probability distribution will be c = 1/15
Step-by-step explanation:
For step by step explanation see the pucture attached
