With the information that you gather from the summary tables, test the following (you can use excel when appropriate): Determine if there is sufficient evidence to conclude the average amount of births is over 5000 in the United States and territories at the 0.05 level of significance. Determine if there is sufficient evidence to conclude the average amount of deaths is equal to 6000 in the United States and territories at the 0.10 level of significance. Determine if there is sufficient evidence to conclude the average amount of marriages is greater or equal to 2500 in the United States and territories at the .05 level of significance. Determine if there is sufficient evidence to conclude the average amount of divorces is less than or equal to 4000 in the United States and territories at the 0.10 level of significance.

Kindly check explanation

Step-by-step explanation:

A.)

H0 : μ = 5000

H0 : μ > 5000

xbar = 6671 ; s = 8185.21 ; n = 52 ; α = 0.05

The test statistic :

T = (xbar - μ) ÷ (s/√(n))

T = (6671 - 5000) ÷ (8185.21/√52)

T = 3.814

Pvalue from Test statistic : ; df = n - 1 = 52-1= 51

Pvalue at 3.814; 51 = 0.000185

Pvalue < α ; Reject H0 and conclude that average birth is greater than 5000

B)

H0 : μ = 6000

H0 : μ < 6000

xbar = 4187 ; s = 4386 ; n = 52 ; α = 0.01

The test statistic :

T = (xbar - μ) ÷ (s/√(n))

T = (4187 - 6000) ÷ (4386/√52)

T = - 2.981

Pvalue from Test statistic : ; df = n - 1 = 52-1= 51

Pvalue at - 2.981; 51 = 0.0022

Pvalue < α ; Reject H0 and conclude that average death is less than 6000

C.)

H0 : μ < 2500

H0 : μ ≥ 2500

xbar = 2744 ; s = 3134.41 ; n = 52 ; α = 0.05

The test statistic :

T = (xbar - μ) ÷ (s/√(n))

T = (2744 - 2500) ÷ (3134.41/√52)

T = 0.561

Pvalue from Test statistic : ; df = n - 1 = 52-1= 51

Pvalue at 0.561; 51 = 0.289

Pvalue > α ; Fail to Reject H0 and conclude that average marriage is not greater Tha or equal to 2500

D.)

H0 : μ = 4000

H0 : μ ≤ 4000

xbar = 1451 ; s = 1217 ; n = 52 ; α = 0.01

The test statistic :

T = (xbar - μ) ÷ (s/√(n))

T = (1451 - 4000) ÷ (1217/√52)

T = - 15.10

Pvalue from Test statistic : ; df = n - 1 = 52-1= 51

Pvalue at - 15.10; 51 = 0.000001

Pvalue < α ; Reject H0 and conclude that average divorce is less eqaul to 4000

expression #3 is not equivalent. the others are.

5x+40

5(x + 8)

7x - 49

7(x - 7)

18x + 9

9(2x + 1)

15x - 10

5(3x - 2)

step-by-step explanation:

for number one all you have to do is factor 5 out of 5x + 40

for number 2 factor 7 out of 7x - 49

for number three factor 9 out of 18x + 9

for number four factor 5 out of 15x - 10