A random sample of n measurements was selected from a population with unknown mean muμ and standard deviation sigmaσ. Calculate a95% confidence interval for muμ for each of the following situations. Complete parts a through e. a. n=75, x overbar=28, s^2=12 (Round to the nearest thousandth as needed.)b. n=225, x overbar=115, s^2=23 (Round to the nearest thousandth as needed.)c. n=105, x overbar=14, s=0.30 (Round to the nearest thousandth as needed.)d. n=105, x overbar =4.14, s=0.83 (Round to the nearest thousandth as needed.)e. Is the assumption that the underlying population of measurements is normally distributed necessary to ensure the validity of the confidence intervals in parts aminus−d? Yes No

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] ± *

28±1.965*

[27.91;28.09]

b.

n= 225

X[bar]= 115

S²= 23

X[bar] ± *

115±1.965*

[114.96;115.04]

c.

n= 105

X[bar]= 14

S²= 0.30

X[bar] ± *

14±1.965*

[13.98;14.02]

d.

n= 105

X[bar]= 4.14

S²= 0.83

X[bar] ± *

4.14±1.965*

[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!

m∠5 + m∠3 = m∠1   is true.

m∠5 + m∠6 = 180 is true

m∠2 +m∠3 = m∠6   is true

m∠2 +m∠3 +m∠5 = 180   is true