An agricultural company is trying to decide which type of fertilizer to use on its crops. The company concerned with the average yield on its crops as well as the variance of their yields. In a sample of 41 different fields with using fertilizer A, the average crop yield per acre is 185 tons with a standard deviation of 35 tons. In a separate sample of 61 different fields using fertilizer B, the average crop yield is 175 tons with a standard deviation of 28 tons. When testing the hypothesis that there is no difference between the average yields using the different fertilizers, if the test statistic is 1.53 and the critical value is 2.57 then what is your conclusion concerning the null hypothesis?

We do not reject the null

Step-by-step explanation:

Given that :

Sample size, n = 41

Test statistic ; Zstatistic ;= 1.53

Critical value ; Zcritical ; = 2.57

The null hypothesis H0 ;

H0: μ1 - μ2 = 0

Alternative hypothesis, H1 ;

H1: μ1 ≠ μ2

Decision region :

Reject Null ; if |Zstatistic ≥ Zcritical|

Zstatistic > Zcritical

Result :

1.53 < 2.57

Zstatistic < Zcritical

Hence, we do not reject the null ; there is no significant evidence that there is difference in the average yield of the two samples.

at first, in 1635, the evening meetings had been just for women . .

what relationship between ideas is shown in the text above?

question 1 options:

location of events

similarity of events

timing of events

examples of an idea

step-by-step explanation:

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18