In a study of crime, the FBI found that 13.2% of all Americans had been victims of crime during a 1-year period. This result was
1 answer:
Answer: The lower bound of confidence interval would be 0.116.
Step-by-step explanation:
Since we have given that
p = 13.2%= 0.132
n = 1105
At 90% confidence,
z = 1.645
So, Margin of error would be
So, the lower bound of the confidence interval would be
Hence, the lower bound of confidence interval would be 0.116.
You might be interested in
Answer:
9 feet
Step-by-step explanation:
Given:
The border of the garden is a right angled triangle.
Two lengths are given as 12 ft and 15 ft.
Let the length of the shortest timber be 'x' feet.
Now, in a right angled triangle, the longest length is called the hypotenuse.
As 15 feet is the largest length, it is the hypotenuse of the triangle. Now, applying Pythagoras theorem, we get:
The negative value is neglected as length can never be negative.
Therefore, the length of the shortest timber is 9 feet.
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%.