Question

A sample size of **26** is taken from a normal
distribution and a confidence interval is constructed. The sample
mean is **61** and the population standard deviation
is σ = **18**.

Using a **99****%** confidence level,
find the ** margin of error**,

Answer #1

a.) Given a normal distribution with σ = 0.380. Find the
required sample size for a 95% confidence level (estimating the
mean), given a margin-of-error of 6%.
b.) Given the sample results taken from a normal population
distribution: mean = 4.65, σ = 0.32, and n = 17.
For a 99% confidence interval, find the margin-of-error for the
population mean. (use 2 decimal places)
c.) Given the sample results taken from a normal population
distribution: mean = 1.25, σ =...

A sample mean, sample size, population standard deviation,
and confidence level are provided. Use this information to complete
parts (a) through (c) below.
x=54, n=14, σ=5, confidence level=99%
A. Use the one-mean z-interval procedure to find a confidence
interval for the mean of the population from which the sample was
drawn.
The confidence interval is from ___to___
B. Obtain the margin of error by taking half the length of the
confidence interval.
What is the length of the confidence interval?...

Given the sample results taken from a normal population
distribution: mean = 4.65, σ = 0.32, and n = 13. Find the
margin-of-error and the 95% confidence interval for the population
mean. (use 2 decimal places)

Given the sample results taken from a normal population
distribution: mean = 4.65, σ = 0.32, and n = 19.
For a 90% confidence interval, find the margin-of-error for the
population mean. (use 2 decimal places)

A sample mean, sample size, population standard deviation,
and confidence level are provided. Use this information to complete
parts (a) through (c) below.
x =2323, n=3434, σ=55, confidence level=95%
a.
Use the one-mean z-interval procedure to find a confidence
interval for the mean of the population from which the sample was
drawn.
The confidence interval is from__ to__
b.Obtain the margin of error by taking half the length of the
confidence interval.
What is the length of the confidence interval?...

A sample of n = 16 is to be taken from a distribution
that can reasonably be assumed
to be Normal with a standard deviation σ of 100. The sample mean
comes out to be 110.
1. The standard error of the mean, that is, the standard deviation
of the sample mean,
is σx¯ = σ/√
n. What is its numerical value?
2. The 97.5 percentile, 1.96, of the standard Normal distribution
is used for a 95% confi-
dence interval....

(S 9.2) Recall that a confidence interval for the sample mean
can be calculated using the interval
x¯?tn?1?sn??????x¯+tn?1?sn???
Thus, the margin of error is
tn?1?sn???
We can recover the margin of error from an interval constructed
on the calculator using algebra.
Suppose a random sample of size 16 was taken from a normally
distributed population, and the sample standard deviation was
calculated to be s = 6.3. We'll assume the sample mean is
10 for convenience.
a) Calculate the margin...

a. Use the one-mean t-interval procedure with the sample mean,
sample size, sample standard deviation, and confidence level
given below to find a confidence interval for the mean of the
population from which the sample was drawn.
b. Obtain the margin of error by taking half the length of the
confidence interval. c. Obtain the margin of error by using the
formula t Subscript alpha divided by 2 Baseline times StartFraction
s Over StartRoot n EndRoot EndFraction .
x=30 n=25...

QUESTION PART A: For a confidence level of 80% with a sample
size of 26, find the critical t value.
QUESTION PART B: Assume that a sample is used to estimate a
population mean μμ. Find the margin of error M.E. that
corresponds to a sample of size 13 with a mean of 46.5 and a
standard deviation of 13.3 at a confidence level of 99%.
Report ME accurate to one decimal place because the sample
statistics are presented with...

Use the standard normal distribution or the t-distribution to
construct a 99% confidence interval for the population mean.
Justify your decision. If neither distribution can be used,
explain why. Interpret the results. In a random sample of 13
mortgage institutions, the mean interest rate was 3.59% and the
standard deviation was 0.42%. Assume the interest rates are
normally distributed.
Select the correct choice below and, if necessary, fill in any
answer boxes to complete your choice.
A. The 99% confidence...

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