Question

Use this formula to find a confidence interval which is xbar +/-E where E is = zc* s / sqrt n On the first one we will have a 95% confidence interval with z=1.96 and n=100 Pick a number for xbar between 60-100. Pick a number for s between 4-8. Find the confidence interval given xbar +/- E Note to find the square root of n use n^.5 where you are raising to the .5 or 1/2 power which is a square root. You can also use the square root key. Here is a short cut to use excel!!! Let's say you pick xbar = 80 and s=6. Since you have a 95% confidence interval alpha = 100%-95% =5% or .05 as a decimal n is 100. This is a z score so we use =confidence.norm(alpha, sd, n) to find E In this case you will type this in an excel cell to find E and then take the xbar + E and xbar - E =confidence.norm(.05,6,100) then hit enter If you had a t test where n< 30 then do then use the following =confidence.t(alpha,sd,n) to find E. For a proportion you will use the formula to find E. Here is an example First find phat which is p with a ^ over the p. It is equal to p/n so if p=40 and n=100 you have .4 qhat = 1- phat = .6 The confidence interval is phat +/- E where E= zc * (phat * qhat/n)^.5 So if you have a=.05 the zc = 1.96 so you have in excel for E =1.96*(.4*.6 /100)^.5 Then take (.4-E, .4+E) Go to the original problem and Instead of 100 let n=400 Show both intervals and the numbers you used. Why do you think they changed?

Answer #1

Let Zc=1.96 and n=100 ,

The 95% confidence interval is given by:

The 95% confidence interval is

###

Let n=100

p=40 then phat=40/100=0.40

qhat=1-phat

=1-0.4

=0.60

Zc=1.96

The 95% confidence interval for the proportion is given by:

The 95% confidence interval for the proportion is

Let n=400

p=40 then phat=40/400=0.10

qhat=1-0.10=0.90

The 95% confidence interval for the proportion is given by:

The 95% confidence interval for the proportion is

Using techniques from an earlier section, we can find a
confidence interval for μd. Consider a
random sample of n matched data pairs A,
B. Let d = B − A be a random
variable representing the difference between the values in a
matched data pair. Compute the sample mean
d
of the differences and the sample standard deviation
sd. If d has a normal distribution or
is mound-shaped, or if n ≥ 30, then a confidence
interval for μd...

Use the given degree of confidence and sample data to find a
confidence interval for the population standard deviation σ. Assume
that the population has a normal distribution. Round the confidence
interval limits to the same number of decimal places as the sample
standard deviation.
Weights of men: 90% confidence; n = 14, xbar = 156.9 lbs, s =
11.1 lb
A) 8.5lbs < σ < 16.5 B) 8.2lbs < σ < 15.6 lbs C)
8.7lbs < σ < 14.3lbs...

Using techniques from an earlier section, we can find a
confidence interval for μd. Consider a
random sample of n matched data pairs A,
B. Let d = B − A be a random
variable representing the difference between the values in a
matched data pair. Compute the sample mean d of the
differences and the sample standard deviation
sd. If d has a normal distribution or
is mound-shaped, or if n ≥ 30, then a confidence
interval for μd...

Using techniques from an earlier section, we can find a
confidence interval for μd. Consider a random sample of n matched
data pairs A, B. Let d = B − A be a random variable representing
the difference between the values in a matched data pair. Compute
the sample mean d of the differences and the sample standard
deviation sd. If d has a normal distribution or is mound-shaped, or
if n ≥ 30, then a confidence interval for μd...

Find an example of a confidence interval for a proportion in the
media or scholarly literature (do not use a statistics textbook or
website/article that is teaching or demonstrating statistics to
find the example).
At the very least it must include either the lower and upper
bounds or a point estimate with a margin of error. Make sure you
have a Proportion confidence interval and NOT a CI for the mean,
odds ratio, hazard ratio, or relative risk as these...

#1 Use the confidence level and sample data to find a confidence
interval for estimating the population μ. Find the margin of error
E and round your answer to the nearest tenth. Test scores: n = 93,
LaTeX: \bar{x}x ¯ = 88.5, σ = 6; 99% confidence
#2 A sociologist develops a test to measure attitudes towards
public transportation, and n = 82 randomly selected subjects are
given the test. Their mean score is LaTeX: \bar{x}x ¯ = 75.9 and...

Write the equation used for…
Binomial Probability Formula:
Confidence interval for a proportion:
Confidence interval for a mean:
Test statistic for testing proportions:
Test statistic for testing means:
Symbols and Equations
Write the correct symbol for each
term
Sample mean:
Sample proportion:
Sample size:
Sample standard deviation:
Population proportion:
Population mean:
Population standard deviation:
Confidence Intervals *Week
10
On my recent trip to space I observed the heights of 50 aliens.
I found the average height to be 48 inches...

Chapter 6, Section 1-CI, Exercise 011
Use the normal distribution to find a confidence interval for a
proportion p given the relevant sample results. Give the
best point estimate for p, the margin of error, and the
confidence interval. Assume the results come from a random
sample.
A 95% confidence interval for p given that p^=0.34 and
n=475.
Round your answer for the best point estimate to two decimal
places, and your answers for the margin of error and the...

Find the indicated confidence interval. Assume the standard
error comes from a bootstrap distribution that is approximately
normally distributed.
a) A 95% confidence interval for a proportion p if the sample
has n=100 with p^=0.37, and the standard error is SE=0.05.
b) A 95% confidence interval for a mean μ if the sample has n=60
with x¯=70 and s=12, and the standard error is
c) A 90% confidence interval for a mean μ if the sample has n=20
with x¯=22.5...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 34 minutes ago

asked 54 minutes ago

asked 58 minutes ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago

asked 3 hours ago

asked 3 hours ago

asked 3 hours ago

asked 3 hours ago

asked 4 hours ago

asked 4 hours ago