Question

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. If the dean wanted to estimate the proportion of all students receiving financial aid to within 3% with 99% reliability, how many students would need to be sampled?

Please show al work on how you got the answer.

Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 118/200 = 0.59

= 0.59

1 - = 1-0.59 = 0.41

margin of error = E = 0.03

At 99% confidence level

= 1-0.99% =1-0.99 =0.01

/2
=0.01/ 2= 0.005

Z/2
= Z_{0.005 = 2.576}

Z/2 = 2.576

sample size = n = (Z_{
/ 2} / E )^{2} *
* (1 -
)

= (2.576/0.03)^{2} *0.59*0.41

= 1784

sample size = n = 1784

Answer =1784 student .

A university dean is interested in determining the proportion of
students who receive some sort of financial aid. Rather than
examine the records for all students, the dean randomly selects 200
students and finds that 118 of them are receiving financial aid. If
the dean wanted to estimate the proportion of all students
receiving financial aid to within 1% with 98% reliability, how many
students would need to be sampled?
Select one: A. 132 B. 5637 C. 13,133 D. 3177

A university dean is interested in determining the proportion of
students who receive some sort of financial aid. Rather than
examine the records for all students, the dean randomly selects 200
students and finds that 118 of them are receiving financial aid.
Use a 98% confidence interval to estimate the true proportion of
students on financial aid.

A university dean is interested in determining the proportion of
students who receive some sort of financial aid. Rather than
examine the records for all students, the dean randomly selects 200
students and finds that 122 of them are receiving financial aid.
Use a 95% confidence interval to estimate the true proportion of
students who receive financial aid.

1.) A university dean is interested in determining the
proportion of students who receive some sort of financial aid.
Rather than examine the records for all students, the dean
randomly selects 200 students and finds that 118 of them are
receiving financial aid. Use a 95% confidence interval to estimate
the true proportion of students who receive financial aid.
a.) 0.59+/-0.068
b.) 0.59+/-0.005
c.) 0.59+/-0.002
d.)0.59+/-0.474
2.)
Fill in the blanks. The owner of Get-A-Away Travel has recently
surveyed a...

1. A university Dean is interested in determining the proportion
of students who receive
some sort of financial aid. Rather than examine the records for all
students, the dean randomly selects
200 students and finds that 118
of them are receiving financial aid. By hand, and using formulas
for the
“Margin of Error” (i.e. not calculator confidence interval), find a
98% confidence interval to estimate
the true proportion of students on financial aid. (Use 3 decimal
places for proportions.)
2....

A university dean states that the proportion of students who
receive some sort of financial aid has increased compared to 55% of
students who received financial aid over the last several years.
The dean randomly selects 200 students and finds that 118 of them
are receiving aid.
Form the hypothesis to test the dean’s statement.
Conduct this test and explain if you support the dean’s
statement using 5% significance level for this test.

A dean at a large state university (student population exceeds
40,000) is interested in determining the proportion of students who
receive some sort of financial aid. Rather than examine the records
for all students, the dean randomly selects 200 students and finds
that 118 of them are receiving financial aid.
A. State the conditions that should be
met to reliably estimate the proportion of students on financial
aid using a confidence interval and explain whether or not they
are met....

Question: The probability that a standard
normal variable, Z, is below 1.96 is 0.4750. (T/F)
Question-After an
extensive advertising campaign, the manager of a company wants to
estimate the proportion of potential customers that recognize a new
product. She samples 120 potential consumers and finds that 54
recognize this product. She uses this sample information to obtain
a 95 percent confidence interval that goes from 0.36 to 0.54.
True or False: 95 percent of the people will recognize the
product...

The Dean of Students at a large university wanted to estimate
the proportion of students who are willing to report cheating by
fellow students.
So, her staff surveyed 172 students currently enrolled in the
introduction to biology class. The students were asked, “Imagine
that you witness
two students cheating on a quiz. Would you tell the professor?”
19 of the surveyed students responded “yes.”
Using these data, calculate a 95% confidence interval for the
proportion of all students at the...

1. A psychology professor at a university is interested in
determining the proportion of undergraduate students at her
university who self-identify as early-risers, night-owls, or
neither. She decides to perform an observational study to
investigate this question, so she randomly selects 500
undergraduate students from a university with 11088 undergraduate
students enrolled to whom she poses her question. 274 respond that
they would call themselves night-owls, 124 replied that they are
early-risers, and 102 respond that they are neither. Is...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago