The local news is conducting a poll of the residents’ opinions regarding a new traffic law that is being proposed. At the end of
1 answer:
You might be interested in
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%.
Answer:
March (3) to December (12)
Step-by-step explanation:
Given:
The graph above
Required:
Months where the number of Jerseys sold at least 150?
The above graph is a typical two dimensional graph.
A 2 dimensional graph has x and y axis.
The y axis (vertical) of this graph holds the number of jersey sold
While the x axis (horizontal) of this graph holds the number of month.
To check the month where at least 150 jerseys were sold, we mark the "150 mark" on the y axis then we trace it to the x axis.
For convenience sake, we'll make use of a line (see attachment).
Every month that falls under this line will not regarded as sales less than 150 (meaning that these months do not count as part of the answer)
From the line on the attachment, only the 3rd month to the 12th month satisfy this criteria.
Hence, the months at which number of sold jersey was atleast 150 is 3 to 12.
