Large sample inference: Let the test statistic, Z, have a normal distribution whenH0:μ=μ0istrue. Give as much information as you can about the p-value for each of the following situations:(a)Ha:μ > μ0, andZ= 2.0(b)Ha:μ < μ0, andZ=−2.4(c)Ha:μ6=μ0, andZ=−1.6(d)Ha:μ < μ0, andZ=−0.4.(e)Ha:μ < μ0andZ=−3.16(f)Ha:μ > μ0, andZ= 5.0(g)Ha:μ6=μ0, andZ= 2.8

The p-values are:

(a) p-value = 0.023        (e) p-value = 0.001

(b) p-value = 0.008        (f) p-value = 0

(c) p-value = 0.1096       (g) p-value = 0.005

(d) p-value = 0.3446

Step-by-step explanation:

The p-value is well-defined as per the probability, [under the null-hypothesis (H₀)], of attaining a result equivalent to or greater than what was truly observed.

(a)

The alternate hypothesis is:

Hₐ: μ > μ₀

The test statistic value is, Z = 2.0.

Compute the p-value as follows:

*Use a z-table.

The p-value of 0.023 implies that the probability of getting the mean value greater than or equal to μ₀ is 0.023.

(b)

The alternate hypothesis is:

Hₐ: μ < μ₀

The test statistic value is, Z = -2.4.

Compute the p-value as follows:

*Use a z-table.

The p-value of 0.008 implies that the probability of getting the mean value as extreme as μ₀ is 0.008.

(c)

The alternate hypothesis is:

Hₐ: μ ≠ μ₀

The test statistic value is, Z = -1.6.

Compute the p-value as follows:

*Use a z-table.

The p-value of 0.1096 implies that the probability of getting the mean value as extreme as μ₀ is 0.1096.

(d)

The alternate hypothesis is:

Hₐ: μ < μ₀

The test statistic value is, Z = -0.4.

Compute the p-value as follows:

*Use a z-table.

The p-value of 0.3446 implies that the probability of getting the mean value as extreme as μ₀ is 0.3446.

(e)

The alternate hypothesis is:

Hₐ: μ < μ₀

The test statistic value is, Z = -3.16.

Compute the p-value as follows:

*Use a z-table.

The p-value of 0.001 implies that the probability of getting the mean value as extreme as μ₀ is 0.001.

(f)

The alternate hypothesis is:

Hₐ: μ > μ₀

The test statistic value is, Z = 5.0.

Compute the p-value as follows:

*Use a z-table.

The p-value of 0 implies that the probability of getting the mean value as extreme as μ₀ is 0.

(g)

The alternate hypothesis is:

Hₐ: μ ≠ μ₀

The test statistic value is, Z = 2.8.

Compute the p-value as follows:

*Use a z-table.

The p-value of 0.005 implies that the probability of getting the mean value as extreme as μ₀ is 0.005.

q1 a person or thing situated away or detached from the main body or system. a cluster is a a mash of many things.

q2 do

step-by-step explanation:

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.