Question

A random sample of 1200 car owners in a particular city found
384 car owners who received a speeding ticket this year. Find a 95%
confidence interval for the true percent of car owners in this city
who received a speeding ticket this year. **Express your
results to the nearest hundredth of a percent**.

Answer: ___________to ___________%

Answer #1

Given that, n = 1200 and x = 384

=> sample proportion = 384/1200 = 0.32

A 95% confidence level has significance level of 0.05 and critical value is,

The 95% confidence interval for the population proportion is,

**Answer : 29.36 to 34.64 %**

A random sample of 1800 car owners in a particular city found
324 car owners who received a speeding ticket this year. Find a 95%
confidence interval for the true percent of car owners in this city
who received a speeding ticket this year. Express your results to
the nearest hundredth of a percent.
Answer:
to %

HW 25
A random sample of 1300 home owners in a particular city found
143 home owners who had a swimming pool in their backyard. Find a
95% confidence interval for the true percent of home owners in this
city who have a swimming pool in their backyard. Express your
results to the nearest hundredth of a percent. . Answer: ___ to ___
%

A random sample of 1900 workers in a particular city found 494
workers who had full health insurance coverage. Find a 95%
confidence interval for the true percent of workers in this city
who have full health insurance coverage. Express your results to
the nearest hundredth of a percent.
Answer: _______ to________ %

A random sample of 1300 voters in a particular city found 273
voters who voted yes on proposition 200. Find a 95% confidence
interval for the true percent of voters in this city who voted yes
on proposition 200. Express your results to the nearest hundredth
of a percent.

A random sample of 2000 voters in a particular city found 820
voters who voted yes on proposition 200. Find a 95% confidence
interval for the true percent of voters in this city who voted yes
on proposition 200. Express your results to the nearest hundredth
of a percent.

A random sample of 700 voters in a particular city found 266
voters who voted yes on proposition 200. Find a 95% confidence
interval for the true percent of voters in this city who voted yes
on proposition 200. Express your results to the nearest hundredth
of a percent.

(1 point) A random sample of 800 workers in a particular city
found 296 workers who had full health insurance coverage. Find a
95% confidence interval for the true percent of workers in this
city who have full health insurance coverage. Express your
results to the nearest hundredth of a percent. .
Answer: _______to_________ %

Please show work so I
can understand how to solve thanks
A random sample of 800
adults in Flagstaff found 24 adults who had lost a job in the past
year. Find a 95% confidence interval for the true percent of adults
in Flagstaff who have lost a job in the past year. Express
your results to the nearest hundredth of a percent. .
Answer: to %

A random sample of 1200 Freshmen at UGA were asked if they had
applied for student football tickets. The proportion who answered
yes was 0.86. The standard error of this estimate is 0.010.
(a) Find the margin of error for a 95% confidence interval for
the population proportion who applied for football tickets.
(3 decimal places)
(b) Construct the 95% confidence interval.
Lower Limit
(3 decimal places)
Upper Limit
(3 decimal places)
(c) Since the sample was randomly selected, will...

A random sample of 56 credit card holders showed that
41 regularly paid their credit card bills on time.
1 Let P represent the proportion of all people who
regularly paid their credit card bills on time.
Find a point estimate P for P ( use decimal form and round to the
nearest hundredth).
a. What is the critical value Z used to find a 95%
confidence interval?
b. What is the margin of error EBP? (round
to the nearest...

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