In: Statistics and Probability

For a normal population with known variance (sigma)^2, what is the confidence level for the CI (mean)-2.14(sigma)/√n≤μ≤(mean)+2.14(sigma)/√n ?

For a normal population with known variance and the confidence level for the CI

Here,

is the standard error value and 2.14 is the Standard score for the confidence level hence using Z score table shown below we can find that

The confidence level would be around 96.76 % or 97 %.

The Table shown below

Q1.
Assume a normal distribution with a known variance. Calculate the
Lower Confidence Level (LCL) and Upper Confidence Level (UCL) for
each of the following:
a.
X-Bar = 50; n = 64; σ = 40; α = 0.05
LCL ==>
UCL ==>
b.
X-Bar = 85; n = 225; σ2 = 400; α = 0.01
LCL ==>
UCL ==>
c.
X-Bar = 510; n = 485; σ = 50; α = 0.10
LCL ==>
UCL ==>

(Topic: Confidence Interval Estimation for the Mean of a Normal
Distribution: Population Variance Known)
The number of bolts produced each hour
from a particular machine is normally distributed with a standard
deviation of 7.4. For a random sample of 15 hours, the average
number of bolts produced was 587.3.
Find the upper and lower confidence
limits of a 98% confidence interval for the population mean number
of bolts produced per hour.
Find the answer by hand calculation.
Find the answer...

Use technology to construct the confidence intervals for the
population variance sigma squared and the population standard
deviation sigma. Assume the sample is taken from a normally
distributed population. cequals0.98, ssquaredequals7.84,
nequals28 The confidence interval for the population variance
is

Use technology to construct the confidence intervals for the
population variance sigma squared and the population standard
deviation sigma. Assume the sample is taken from a normally
distributed population, C= 0.90, S^2=10.24, N= 30
The confidence interval for the population variance is:
The confidence interval for the population standard deviation
is:

For a normal population with known variance
σ2, what value of
zα/2 in the Equation below gives a 98%
CI?
x-zα/2σn≤μ≤x+zα/2σn

For a normal population with known variance s2, answer the
following questions:
(a) What is the confidence level for
the interval: ?− 1.85?/√?≤ ?≤ ?+ 1.85?/√??
(b) What is the confidence level for
the interval: ?≤ ?+ 1.4?/√?

Consider a normal population distribution with the value of \(\sigma\) known.(a) What is the confidence level for the interval \(\bar{x} \pm 2.88 \sigma / \sqrt{n} ?\) (Round your answer to one decimal place.)\(\%\)(b) What is the confidence level for the interval \(\bar{x} \pm 1.47 \sigma / \sqrt{n} ?\) (Round your answer to one decimal place.) \(\%\)(c) What value of \(z_{\alpha / 2}\) in the CI formula below results in a confidence level of \(99.7 \% ?\) (Round your answer to...

For a normal population with known variance σ2, answer the
following questions:
d. What is the conﬁdence level for the interval μ ≤ x+ 2.00σ∕ √
n?
e. What is the conﬁdence level for the interval x−1.96σ∕ √ n ≤
μ?

Sigma known
Ratings
Population Standard Deviation
2.64
Mean
42.95384615
95% CI Margin of Error
Standard Error
0.327747828
0.641805331
Median
43
95% Confidence Interval
Mode
44
42.31204082
43.59565148
Standard Deviation
2.642387469
99% CI Margin of Error
Sample Variance
6.982211538
0.843188126
Kurtosis
-0.392154755
99% Confidence Interval
Skewness
-0.446608112
42.11065803
Range
12
43.79703428
Minimum
36
Maximum
48
Sum
2792
Sigma Unknown
Count
65
95% Confidence Interval
Confidence Level(95.0%)
0.654751556
42.2990946
43.60859771
State the 95% confidence interval for sigma
unknown.
What is the multiplier...

For normal population with known standard deviation, the x%
confidence interval for the population mean µ has a lower end point
(z times the standard error) below the sample mean and a upper end
point (z times the standard error) above the sample mean. That is,
the x% CI is given as
Sample mean ± z *standard error
For 95% CI, the value for z is
Answer 1Choose...2.581.6451.280.9751.96
For 80% CI, the value for z is
Answer 2Choose...2.581.6451.280.9751.96
For 90%...

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