Answer:
Step-by-step explanation:
Hello!
You need to estimate the population mean using a 95% confidence interval using the information of several samples:
a.
n=75
X[bar]= 28
S²= 12
X[bar] ± [tex]Z_{1-\alpha /2}[/tex] * [tex]\frac{S}{\sqrt{n} }[/tex]
[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]
28±1.965*[tex]\frac{3.46}{75}[/tex]
[27.91;28.09]
b.
n= 225
X[bar]= 115
S²= 23
X[bar] ± [tex]Z_{1-\alpha /2}[/tex] * [tex]\frac{S}{\sqrt{n} }[/tex]
[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]
115±1.965*[tex]\frac{4.80}{255}[/tex]
[114.96;115.04]
c.
n= 105
X[bar]= 14
S²= 0.30
X[bar] ± [tex]Z_{1-\alpha /2}[/tex] * [tex]\frac{S}{\sqrt{n} }[/tex]
[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]
14±1.965*[tex]\frac{0.55}{105}[/tex]
[13.98;14.02]
d.
n= 105
X[bar]= 4.14
S²= 0.83
X[bar] ± [tex]Z_{1-\alpha /2}[/tex] * [tex]\frac{S}{\sqrt{n} }[/tex]
[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]
4.14±1.965*[tex]\frac{0.91}{105}[/tex]
[4.12;4.16]
e.
There is no need to assume that any of the population measurements are normally distributed. All samples are large enough to be valid to apply the Central Limit Theorem. This theorem says that if a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
X[bar]≈N(μ;σ2/n)
I hope it helps!
