Question

A random sample of 803 people was taken. 154 of the people in the sample favored Candidate A. Find 90% confidence interval for the true proportion of people who favors Candidate A. Enter in the lower limit of the confidence interval you found.

Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 154 / 803 = 0.192

1 - = 1 - 0.192 = 0.808

Z_{}
= 1.282

Margin of error = E = Z_{}
*
((
* (1 -
)) / n)

= 1.282 * (((0.192 * 0.808) / 803)

Margin of error = E = 0.018

A 90% lower confidence interval for population proportion p is ,

- E

0.192 - 0.018

0.174

**Lower limit: 0.174**

A random sample of 994 people was taken. 262 of the people in
the sample favored Candidate A. Find 90% confidence interval for
the true proportion of people who favors Candidate A. Enter in the
lower limit of the confidence interval you found.

In a random sample of 200 people, 154 said that they watched
educational television. Find the 90% confidence interval of the
true proportion of people who watched educational television. If
the television company wanted to publicize the proportion of
viewers, do you think it should use the 90% confidence
interval?

Educational Television: In a random sample of 200
people, 154 said that they watched educational
television. Find the 90% confidence interval of the true
proportion of people who watched educational television. Then write
in a z interval

A random sample of 100 people was taken. Eighty-five of
the people in the sample favored Candidate A. We are interested in
determining whether or not the proportion of the population in
favor of Candidate A is significantly more than 80%.
(1). The test statistic is
0.80
0.05
1.25
2.00
(2). The p-value is
a.
0.2112
b. 0.05
c. 0.025
d.
0.1056
(3). At 95% confidence, it can be
concluded that the proportion of the population in...

A random sample of 100 people was taken. Eighty-five of the
people in the sample favored Candidate A. We are interested in
determining whether or not the proportion of the population in
favor of Candidate A is significantly more than 80%. We consider
the following sets of hypothesis testing: (A) Ho: p > 0.8; Ha: p
<= 0.8 (B) Ho: p <= 0.8; Ha: p > 0.8 (C) Ho: p > 0.8;
Ha: p < 0.8 (D) Ho: p >...

The mean diastolic blood pressure for a random sample of 60
people was 84 millimeters of mercury. If the standard deviation of
individual blood pressure readings is known to be 10 millimeters of
mercury, find a 90% confidence interval for the true mean diastolic
blood pressure of all people. Then complete the table below.
What is the lower limit of the
90%
confidence interval?
What is the upper limit of the
90%
confidence interval?

In the run-up to the 2016 election, a random sample of 100
people was taken to determine a confidence interval for the
population proportion who would vote in favor of Proposition 61 to
legalize marijuana. The confidence interval was found to
be (.402 to .598).
What was the confidence level, (1-alpha)*100%, associated with
this confidence interval?
95%
90%
We would need to know the number in the sample who said they
would vote for Proposition 61. If we knew this number,
we...

The mean diastolic blood pressure for a random sample of
70
people was
94
millimeters of mercury. If the standard deviation of individual
blood pressure readings is known to be
8
millimeters of mercury, find a
90%
confidence interval for the true mean diastolic blood pressure
of all people. Then complete the table below.
Carry your intermediate computations to at least three decimal
places. Round your answers to one decimal place.
What is the lower limit of the
90%
confidence...

Let us assume that we randomly choose a sample of 144 people.
51% of people from our sample reported having voted for candidate A
in the 2004 election. Using this sample, find 95% confidence
interval for the population proportion of those who voted for this
candidate. Round to three digits.
Lower end of confidence interval:
Upper end of confidence interval:

Pinworm: In a random sample of 830 adults in
the U.S.A., it was found that 70 of those had a pinworm
infestation. You want to find the 90% confidence interval for the
proportion of all U.S. adults with pinworm.
(a) What is the point estimate for the proportion of all U.S.
adults with pinworm?
(b) What is the critical value of z (denoted
zα/2) for a 90% confidence
interval?
zα/2 =
(c) What is the margin of error (E) for...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 27 minutes ago

asked 30 minutes ago

asked 45 minutes ago

asked 48 minutes 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