Find the interval [ μ−z σn√,μ+z σn√μ−z σn,μ+z σn ] within
which 95 percent of the sample means would be expected to fall,
assuming that each sample is from a normal population.
(a) μ = 179, σ = 10, n
= 33. (Round your answers to 2 decimal
places.)
The 95% range is from to .
(b) μ = 867, σ = 18, n
= 10. (Round your answers to 2 decimal
places.)
The 95% range is from to .
(c) μ...
Find the probability of each of the following, if Z~N(μ = 0,σ =
1).
a) P(Z < -1.88)
b) P(Z > 1.51) =
c) P(-0.61 < Z < 1.54) =
d) P(| Z | >1.78) =
e) P(Z < -1.27) =
f) P(Z > 1.02) =
g) P(-0.69 < Z < 1.78) =
h) P(| Z | >1.86) =
Consider the following hypotheses:
H0: μ = 38
HA: μ ≠ 38
Find the p-value for this test based on the following
sample information.
a. x¯x¯ = 33; s = 11.9;
n = 38
0.05 p-value < 0.10
0.02 p-value < 0.05
0.01 p-value < 0.02
p-value < 0.01
p-value 0.10
b. x¯x¯ = 43; s = 11.9;
n = 38
0.01 p-value < 0.02
p-value < 0.01
p-value 0.10
0.05 p-value < 0.10
0.02 p-value < 0.05
c. x¯x¯...
please be very specific on showing work done!!
If Z∼N(μ=0,σ2=1)Z∼N(μ=0,σ2=1), find the following
probabilities:
P(Z<1.58)=P(Z<1.58)=
P(Z=1.58)=P(Z=1.58)=
P(Z>−.27)=P(Z>−.27)=
P(−1.97<Z<2.46)=
Consider a two-sided confidence interval for the mean μ when σ
is known;
X̄-Zα1 σ/√n ≤ μ ≤ X̄+Zα2 σ/√n
where α1 + α2 = α. if α1= α2 = α/2, we have the usual 100(1-α)%
confidence interval for μ. In the above, when α1 ≠ α2 , the
interval is not symmetric about μ. Prove that the length of the
interval L is minimized when α1= α2= α/2. Hint remember that Φ Zα =
(1- α) , so Φ-1...
(a) Find P [Z > 1.26]. (b) Find P [Z > -1.37}. (c) find
P[-1.25
(a)
Find P [Z > 1.26]. (b) Find P [Z > -1.37}. (c) find
P[-1.25<Z<0.37). (d) find z such that P[Z>z]=0.05. (e)
find z such that P[-z<Z<z]=0.99. (f) find the value of k such
that P[k<Z<-0.18]=0.4197
Construct the confidence interval for the population mean μ.
c= 0.98, x̄=6.6, σ=0.7, n=41
A 98 % confidence interval for μ is ( _ , _ )
Round two decimal places as needed.