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.

answer:   expression= p+p*d%

price reduction + sale price = $625

sale price: $p+$p*d%

where d% is the percentage of discount

step-by-step explanation:

original price = p

price went down by $150 price reduction $p*d%

sale price= $ 475

therefore the price before sale is: $625

and the expression that shows the price before sale is: p+p*d%