An article reports the following data on yield (y), mean temperature over the period between date of coming into hops and date of picking (x1), and mean percentage of sunshine during the same period (x2) for the Fuggle variety of hop:
x1 x2 y
16.7 30 206
17.4 42 108
18.4 47 101
16.8 47 104
18.9 43 96
17.1 41 76
17.3 48 71
18.2 44 65
21.3 43 70
21.2 50 54
20.7 56 40
18.5 60 31
Here is partial Minitab output from fitting the first-order model Y = ß0 + ß1x1 + ß2x2 + e used in the article:
Predictor Coef StDev t-ratio p
Constant 415.11 82.52 5.03 0
Temp -6.593 4.859 -1.36 0.208
Sunshine -4.504 1.071 -4.20 0.002
s = 24.45 R-sq = 76.8% R-sq(adj) = 71.6%
(a) What is Y · 17.3, 48, and what is the corresponding residual, e? (Round your answers to four decimal places.)
Y · 17.3, 48 = 84.8591
e = -11.8591
b) Test H0: ß1 = ß2 = 0 versus Ha: either ß1 or ß2 ? 0 at level 0.05.
State the rejection region and compute the test statistic value. Round your answers to two decimal places.
rejection region f = 4.26
test statistic f = 14.90
(c) The estimated standard deviation of 0 + 1x1 + 2x2 when x1 = 17.3 and x2 = 48 is 10.13. Use this to obtain a 95% CI for µY · 17.3, 48. (Round your answers to two decimal places.)
(d) Use the information in part (c) to obtain a 95% PI for yield in a future experiment when x1 = 17.3 and x2 = 48. (Round your answers to two decimal places.)

x1=c(16.7,17.4,18.4,16.8,18.9,17.1,17.3,18.2,21.3,21.2,20.7,18.5)

x2=c(30,42,47,47,43,41,48,44,43,50,56,60)

y=c(210,110,103,103,91,76,73,70,68,53,45,31)

mod=lm(y~x1+x2)

summary(mod)

R output: Call:

lm(formula = y ~ x1 + x2)

Residuals:  

   Min      1Q Median      3Q     Max

-41.730 -12.174   0.791 12.374 40.093

Coefficients:

        Estimate Std. Error t value Pr(>|t|)    

(Intercept) 415.113     82.517   5.031 0.000709 ***  

x1            -6.593      4.859 -1.357 0.207913    

x2            -4.504      1.071 -4.204 0.002292 **  

---  

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1  

Residual standard error: 24.45 on 9 degrees of freedom  

Multiple R-squared: 0.768,     Adjusted R-squared: 0.7164  

F-statistic: 14.9 on 2 and 9 DF, p-value: 0.001395

a).  y=415.113 +(-6.593)x1 +(-4.504)x2

b). s=24.45

c).  y =415.113 +(-6.593)*21.3 +(-4.504)*43 =81.0101

residual =68-81.0101 = -13.0101

d). F=14.9

P=0.0014

There is convincing evidence at least one of the explanatory variables is significant predictor of the response.

e).  newdata=data.frame(x1=21.3, x2=43)

# confidence interval

predict(mod, newdata, interval="confidence")

#prediction interval

predict(mod, newdata, interval="predict")

confidence interval

> predict(mod, newdata, interval="confidence",level=.95)

      fit      lwr      upr

1 81.03364 43.52379 118.5435

95% CI = (43.52, 118.54)

f).  #prediction interval

> predict(mod, newdata, interval="predict",level=.95)

      fit      lwr      upr

1 81.03364 14.19586 147.8714

95% PI=(14.20, 147.87)

g).  No, there is not evidence this factor is significant. It should be dropped from the model.

answer:

the position of equilibrium will shift to the left.

step-by-step explanation:

the general equation for the dissociation of a weak base is

b + h₂o   ⇌   bh⁺ + oh⁻

according to le châtelier's principle, when a stress is applied to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

if you apply a stress by adding more bh⁺ or oh⁻,   the system will respond by moving in a direction that gets rid of the added ions.

the position of equilibrium will shift to the left, forming more of the undissociated base.

bull

step-by-step explanation: