A sample of size n = 16 is made from a normal distribution with
mean μ....
A sample of size n = 16 is made from a normal distribution with
mean μ. It
turns out that the sample mean is x = 23 and the sample standard
deviation is s = 6.
Construct a 90% confidence interval for μ.
Solutions
Expert Solution
Given that,
= 23
s =6
n = 16
Degrees of freedom = df = n - 1 =16 - 1 = 15
At 90% confidence level the t is ,
= 1 - 90% = 1 - 0.90 = 0.1
/ 2 = 0.1 / 2 = 0.05
t
/2,df = t0.05,15 =1.753 ( using
student t table)
Margin of error = E = t/2,df
* (s /n)
= 1.753 * (6 /
16) = 2.6295
The 90% confidence interval estimate of the population mean
is,
Let X be the mean of a random sample of size n from a N(μ,9)
distribution.
a. Find n so that X −1< μ < X +1 is a confidence interval
estimate of μ with a confidence level of at least 90%.
b.Find n so that X−e < μ < X+e is a confidence interval
estimate of μ withaconfidence levelofatleast (1−α)⋅100%.
A random sample of size n = 55 is taken from a
population with mean μ = −10.5 and standard deviation
σ = 2. [You may find it useful to reference
the z table.]
a. Calculate the expected value and the
standard error for the sampling distribution of the sample mean.
(Negative values should be indicated by a minus sign. Round
"expected value" to 1 decimal place and "standard error" to 4
decimal places.)
b. What is the probability that...
A random sample of size n = 50 is taken from a
population with mean μ = −9.5 and standard deviation
σ = 2. [You may find it useful to reference
the z table.]
a. Calculate the expected value and the
standard error for the sampling distribution of the sample mean.
(Negative values should be indicated by a minus sign.
Round "expected value" to 1 decimal place and
"standard deviation" to 4 decimal places.)
Expected Value=
Standard Error=
b. What...
A random sample of n =100 is selected from a normal
population with mean μ = 24 and standard deviation
σ = 1.25.
Find the probability that is less than 24.3
A random sample of n = 25 is selected from a normal
population with mean
μ = 101
and standard deviation
σ = 13.
(a) Find the probability that x exceeds 107. (Round
your answer to four decimal places.)
(b) Find the probability that the sample mean deviates from the
population mean μ = 101 by no more than 2. (Round your
answer to four decimal places.)
A random sample of size 16 from a normal distribution with known
population standard deviation � = 3.1 yields sample average � =
23.2.
What probability distribution should we use for our sampling
distributions of the means?
a) Normal Distribution
b) T-distribution
c) Binomial Distribution
d) Poisson Distribution
What is the error bound (error) for this sample average for a
90% confidence interval?
What is the 90% confidence interval for the population mean?
Random sample of size n=19 is taken from a Normal population,
sample mean is 11.5895, standard deviation is 1.0883.
1. At the 2% level, test whether it is reasonable to believe
that the true population variance is larger than 1.
Using the scenario from above, do the following
2. Derive the power function. Show your work.
3. Using R, graph the power function for 0.5 < sigma^2 <
3.5.
4. Pretend that the sample size was actually 56. Plot this...
In R, get a sample of size n=5 from the Cauchy distribution and
calculate its mean.
(mean(rcauchy(5)). Repeat this 1000 times and plot the histogram
of the resulting 1000 means. Please show your R codes.
n <- 5
listofmeans <- c()
for(i in 1:1000){
}listofmeans <- c(listofmeans, mean(rcauchy(n)))
hist(listofmeans)
Create other histograms, changing the value of n to 10, 30, 100,
and 500. Comment on what these graphs show in light of the Central
Limit Theorem.
Assuming the population has an approximate normal distribution,
if a sample size n=10 has a sample mean ¯x=36 with a sample
standard deviation s=9, find the margin of error at a 90%
confidence level. THEN YOU MUST ROUND ANSWER TO TWO DECIMALS
PLACES.
A random sample of size n1 = 16 is selected
from a normal population with a mean of 74 and a standard deviation
of 9. A second random sample of size n2 = 7 is
taken from another normal population with mean 68 and standard
deviation 11. Let X1and X2 be the two sample means. Find:
(a) The probability that X1-X2 exceeds 4.
(b) The probability that 4.8 ≤X1-X2≤ 5.6.
Round your answers to two decimal places (e.g. 98.76).