Question

Find the sample size, *n*, needed to estimate the mean of
a normal population, *μ*, with σ = 9.25 and a maximum error
of *E* = 3.2 at a 99% level of confidence.

Answer #1

Solution :

Given that,

standard deviation = =9.25

Margin of error = E = 3.2

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58

sample size = n = [Z/2* / E] 2

n = ( 2.58* 9.25/ 3.2)2

n =55.61

Sample size = n =56

If σ = 12 then find sample size required to estimate a
population mean μ to within 3.6 with 99% confidence?

Assume that a sample is used to estimate a population mean μ μ .
Find the margin of error M.E. that corresponds to a sample of size
10 with a mean of 17.7 and a standard deviation of 17.6 at a
confidence level of 99.8%. Report ME accurate to one decimal place
because the sample statistics are presented with this accuracy.
M.E. =

Determine the sample size needed to obtain an estimate of µ if
the margin of error E = 0.06, σ = 0.75, and the confidence level is
99%
a) Find α for 99% confidence
level.
α =
(b) Find 1 − α/2.
1 − α/2 =
(c) Find zα/2 for a
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zα/2 =

The sample size needed to estimate a population mean to within
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a) 90%
b) 95%
c) 98%
d) 99%
e) None of the above.

Determine the sample size n needed to construct a 90?%
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? = 48 and the margin of error equals 8.
n =

A sample size of 26 is taken from a normal
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Using a 99% confidence level,
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E.

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a. 139 b. 98 c. 127 d. 239 e. 196

Use the given data to find the 95% confidence interval estimate
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______<μ<________

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x=54, n=14, σ=5, confidence level=99%
A. Use the one-mean z-interval procedure to find a confidence
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The confidence interval is from ___to___
B. Obtain the margin of error by taking half the length of the
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Given a (1- ∝)% confidence interval for population mean; μ is
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Find the value of: (a) the unbiased point estimator; ( sample
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(b) the margin of error, E.
(c) Hence, if population standard deviation; σ = 3.5 and the
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