Given the following values: μ = 6.0, m = 7.6, n = 36, σ = 6, conduct a one-sample z test at a 0.05 level of significance. for a one-tailed test, upper-tail critical, what is the decision?
a. to reject the null hypothesis
b. to retain the null hypothesis
c. there is not enough information since the sample size is not given.

b) We accept H₀

Step-by-step explanation:

We assume Normal Distribution

Population mean   μ₀   =  6

Sample size   =  n        n  = 36           n  >  30

Sample mean  =  7.6

Sample standard deviation  s  = 6

1.- Hypothesis test:

We are ask for a one-tail test upper-tail critical then

H₀          null hipothesis          μ₀   =  6

Hₐ  Alternative hypothesis    μ₀   >  6  

2.-    

Level of significance   α =  0.05   from z table we find  value of  z(c)

z(c)  =  1.64

3.-Compute  z(s)

z(s)  =  (  μ  -   μ₀ ) / s/√n      ⇒  z(s)  =  ( 7.6  - 6 ) / 6/√36

z(s)  =  1.6

4.-Compare

z(s)   and  z(c)

z(s)  = 1.6       and  z(c)   = 1.64  then

z(s)  <  z(c)

z(s)  is inside the acceptance region. We accept H₀

the f

last graph i believe

step-by-step explanation:

worked it out i believe

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15