Question
In: Statistics and Probability
Calculate the lower confidence limit (LCL) of the mean for the following: x¯= 600, n =...
Calculate the lower confidence limit (LCL) of the mean for the following: x¯= 600, n = 94, σ = 8 (sigma), α = 0.10 (alpha).
Solutions
Expert Solution
Solution :
Given that,
= 600
= 8.8
n = 94
A ) At 95% confidence level the z is ,
=
0.10
/ 2 = 0.10 / 2 = 0.05
Z/2
= Z0.05 = 1.645
Margin of error = E = Z/2*
(
= 1.645 * (8 /
= 1.36
At 80% confidence interval estimate of the population mean is,
- E < 
<
+ E
600 - 1.36 <
< 600 + 1.36
598.64 <
< 601.36
The lower confidence limit = 598.64
orchestra answered 1 year agoRelated Solutions
Calculate the lower confidence limit (LCL) of the mean for the following: x¯= 600, n =...
Calculate the lower confidence limit (LCL) of the mean for the
following: x¯= 600, n = 94, σ = 8 (sigma), α = 0.10 (alpha).
Q1. Assume a normal distribution with a known variance. Calculate the Lower Confidence Level (LCL) and...
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 ==>
CALCULATE THE MEAN value : Mean(x) = x0 = 1/N * ∑in xi and the Stardandar...
CALCULATE THE MEAN value : Mean(x) = x0 = 1/N * ∑in xi and the
Stardandar deviacion : σ = √ (∑in (xi-x0)2
/ (N-1) BOTH USING MATLAB EMERGENCY PLEASE . THE DATA IS BELOW
....SHOW THE STEPS AND THE GRAPHS ...THERE ARE R1 AND R5 for the
data
R1
1.3004560111742451e+01
1.2989489582476999e+01
1.2992918145494006e+01
1.3015926219975810e+01
1.3003113745531683e+01
1.3009060588116746e+01
1.3006255988139626e+01
1.3006525855308004e+01
1.3015986719424999e+01
1.3015607658994400e+01
R5
1.2896465697037076e+01
1.3078048708539365e+01
1.2947635768569747e+01
1.3036655293457565e+01
1.2970848809276067e+01
1.3044710696683399e+01
1.3036858101292060e+01
1.3033168310496990e+01
1.2872334019855321e+01
1.2853995527962450e+01
CALCULATE THE MEAN value : Mean(x) = x0 = 1/N * ∑in xi and the Stardandar...
CALCULATE THE MEAN value : Mean(x) = x0 = 1/N *
∑in xi and the Stardandar deviacion :
σ = √ (∑in
(xi-x0)2 / (N-1) BOTH USING MATLAB EMERGENCY PLEASE .
THE DATA IS BELOW ....SHOW THE STEPS AND THE GRAPHS ...THERE ARE R1
AND R5 for the data
R1
1.3004560111742451e+01
1.2989489582476999e+01
1.2992918145494006e+01
1.3015926219975810e+01
1.3003113745531683e+01
1.3009060588116746e+01
1.3006255988139626e+01
1.3006525855308004e+01
1.3015986719424999e+01
1.3015607658994400e+01
R5
1.2896465697037076e+01
1.3078048708539365e+01
1.2947635768569747e+01
1.3036655293457565e+01
1.2970848809276067e+01
1.3044710696683399e+01
1.3036858101292060e+01
1.3033168310496990e+01
1.2872334019855321e+01
1.2853995527962450e+01
11. The lower limit of a 95% confidence interval for the population proportion P given a...
11. The lower limit of a 95% confidence interval for the
population proportion P given a sample size n = 100 and sample
proportion P̂ = 0.62 is equal to: Show your calculations.
Show your calculations.
A. 0.699
B. 0.525
C. 0.440
D. 0.715
12. The 95% confidence interval for the population proportion P,
given a sample size n = 2200 and sample proportion P̂ = 0.214 is
computed as: Show your calculations. Show your
calculations.
A. 0.214 ± 0.017...
Find the pointwise limit f(x) of the sequence of functions fn(x) = x^n/(n+x^n) on [0, ∞)....
Find the pointwise limit f(x) of the sequence of functions fn(x)
= x^n/(n+x^n) on [0, ∞). Explain why this sequence does not
converge to f uniformly on [0,∞). Given a > 1, show that this
sequence converges uniformly on the intervals [0, 1] and [a,∞) for
any a > 1.
Use the sample information x=44, a=6, n=17 to calculate the following confidence intervals for u assuming...
Use the sample information x=44, a=6, n=17 to calculate the
following confidence intervals for u assuming the sample is from a
normal population
(a) 90% confidence interval is ? to ?
(b) 95% confidence interval is ? to ?
(c) 99% confidence interval is ? to ?
2.A college wants to know the lower limit of a two-sided 80% confidence interval for SAT...
2.A college wants to know the lower limit of a two-sided 80%
confidence interval for SAT scores on the Math section (range
equals 200 to 800). They sampled 1000 students and found that the
sample mean was 580 and the sample standard deviation was 40.
3.A Rollercoaster's auditors estimate that the average daily
loss from those illegally riding without tickets is at least
(greater or equal) $250, but wants to determine the accuracy of
this statistic. The company researcher takes...
For the given confidence level and values of x and n, find the following.x=48, n=97, confidence...
For the given confidence level and values of x and n, find the
following.x=48, n=97, confidence level 99%
The point estimate for the given data is .4948.
The standard error for the given data is .0502.
(c) Find the margin of error. Round the answers to at
least four decimal places, if necessary.
How do I find the lower limit of a confidence interval using a z-table. Having trouble...
How do I find the lower limit of a confidence interval using a
z-table. Having trouble understanding where the 1.645 number came
from in this solution
Sample proportion {p} = 367 / 561 = 0.654 Lower limit for 95%
confidence interval is {p} - Z\alpha * sqrt( {p} ( 1 - {p} ) / n) =
0.654 - 1.645 * sqrt( 0.654 * 0.346 / 561) = 0.617
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