The results that follow were obtained from an analysis of data obtained in a study to assess the relationship between percent increase in yield (Y) and base saturation (x1, pounds/acre), phosphate saturation (x2, BEC%), and soil pH (x3). Fifteen responses were analyzed in the study. The least-squares equation and other useful information follow. y^=38.83-0.0092x1-0.92x2-11.56x3, Syy=10965.46, SSE=1107.01

a. Is there sufficient evidence that, with all independent variables in the model, β2 < 0? Test at the α = .05 level of significance.
b. Give a 95% confidence interval for the mean percent increase in yield if x1 = 914, x2 = 65 and x3 = 6.

a. yes there is enough evidence

B. -22.83, 102.79

Step-by-step explanation:

H0: B2 >= 0

H1: B2 < 0

Alpha = 0.05

T critical is calculated to be -1.79

If t is less than -1.79 we reject the null hypothesis

Test statistic

S = √1107.01/15-3-1

= 10.032

T stat = -0.92 - 0/10.032√0.00081

= -3.222

-3.222 is less than -1.79

So we Reject the null hypothesis and conclude that enough evidence exists that B2<0

B. At 95% confidence

38.83 - 0.0092x1 - 0.92x2 + 11.56x3

= 38.83 - 0.0092x914 - 0.92x65 + 11.56x6

= 38.83 - 8.4088 - 59.8 + 69.36

= 39.98

39.98-2.2*28.55,39.85+2.2*28.55

= 39.98 - 62.81, 39.98 + 62.81

= -22.83, 102.79

$450.00 for 6 weeks

or

$300

step-by-step explanation:

$200.00 for first week $50.00 x5 equals $250 plus $200 equals $450.00 for six weeks.

$200 for first week

$200 x 0.10 = $20

$20 x 5 = $100

$200 + $100 = $300

hi there!

the answer will be d: 2