MAT 2375, Assignment 4 1

Solution to assignment 4,MAT 2375

Total= 100 points

[7.6-13]

Solution. We have

n∑i=1

[Yi − α− β(xi − x)]2 =n∑i=1

[Yi − α + α− α + (β − β − β)(xi − x)]2

=n∑i=1

[Yi − α− β(xi − x)]2 + n(α− α)2 + (β − β)2n∑i=1

(xi − x)2.

It is important to notice that all the cross products are zero. For example, since∑ni=1(xi − x) = 0,

we get

(α− α)n∑i=1

(xi − x) = 0

Also

(α− α)n∑i=1

((Yi − α)− β(xi − x)) = nY − α− βn∑i=1

(xi − x) = 0.

[7.6-20]

Solution

x=c(32,23,23,23,26,30,17,20,17,18,26,16,21,24,30)

y=c(28,25,24,32,31,27,23,30,18,18,32,22,28,31,26)

> mean(x)

[1] 23.06667

> mean(y)

[1] 26.33333

a=lm(y~x)

summary(a)

Call:

lm(formula = y ~ x)

Residuals:

Min 1Q Median 3Q Max

-5.7688 -2.8489 -0.7564 3.6819 5.7004

Coefficients:

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

(Intercept) 14.6578 5.0382 2.909 0.0122 *

x 0.5062 0.2137 2.369 0.0340 *

MAT 2375, Assignment 4 2

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

Residual standard error: 4.037 on 13 degrees of freedom

Multiple R-squared: 0.3015,Adjusted R-squared: 0.2477

F-statistic: 5.611 on 1 and 13 DF, p-value: 0.03402

>hsigma=sum(model$resid**2)/(length(x)-2)

[1] 211.8864

We find confidence interval for β0 and β1 and σ2 in the model

Yi = β0 + β1xi + εi.

C.I. for β0 and β1 (this is similar for α and β) are

14.6578± (2.160369)(5.0382) = 14.6578± 10.88437

and0.5062± (2.160369)(0.2137) = 0.5062± 0.4616709.

For the modelYi = α + β(xi − x) + εi

the 95% c.i. for α is

α± t0.025(13)

√√√√ n∑i=1

e2i /(n(n− 2)).

Using R we get

>sigma=sqrt(sum(e**2)/((length(x)-2)*length(x)))

> mean(y)

[1] 26.33333

> mean(y)+qt(0.975,13)*sigma

[1] 28.5853

> mean(y)-qt(0.975,13)*sigma

[1] 24.08137

The 95% confidence interval for α is [24.08137, 28.5853]. The confidence interval for β1 and β arethe same.

Sincen∑i=1

e2i /σ2 ∼ χ2(13).

The confidence interval for σ2 is

(n∑i=1

e2i /χ20.975(13),

n∑i=1

e2i /χ20.025(13)).

We have∑ni=1 e

2i = 211.8658.

MAT 2375, Assignment 4 3

> qchisq(0.975,13)

[1] 24.7356

> qchisq(0.025,13)

[1] 5.008751

The confidence interval for σ2 is

(211.8658/24.7356, 211.8658/5.008751) = (8.565218, 42.29913).

[8.1-6].

Solution (a, b)H0 : µ = 3.4 H1 : µ > 3.4

(c)X ≥ 3.4 + tα(8)s/

√9⇒ RH0.

The test statistics is

T =X − 3.4

s/√n.

(d)X ≥ 3.4 + s/

√9⇒ RH0.

(e)

> qt(0.95,8)

[1] 1.859548

> x=c(3.4,3.6,3.8,3.3,3.4,3.5,3.7,3.6,3.7)

> length(x)

[1] 9

> mean(x)

[1] 3.555556

> sd(x)

[1] 0.1666667

We have3.4 + 1.859548s/

√9 = 3.4 + (1.859548)(0.1666667)/3 = 3.503308.

Since3.555556 > 3.503308

we reject H0.

Tobserved =X − 3.4

s/√n

=3.555556− 3.4

0.1666667/3= 2.800007.

Thereforep− value = P (t(8) > 2.800007).

MAT 2375, Assignment 4 4

> pt(2.800007,8)

[1] 0.9884012

> 1-0.9884012

[1] 0.0115988

p− value = 0.0115988

This confirms the we should reject H0 with α = 0.05.

[8.1-12].

Solution. We get

di = xi − yi : 13,−4, 3, 14,−1, 17, 11, 13, 17, 14, 7, 2,−7,−10, 0,−1,−7.

Calculate

d =1

n

17∑i=1

di = 4.764706

and

sd =

√√√√ 1

n− 1

n∑i=1

(di − d)2 = 9.086593.

Therefore

tobserved =d

sd/√n

=4.764706

9.086593/sqrt17= 2.162019.

Since t16(0.05) = 1.746, we reject H0.

We can also do this in R. Notice that a paired t-test is a one sample test on D = X − Y .

> x

[1] 265 272 246 260 274 263 255 258 276 274 274 269 244 212 235 254 224

> y

[1] 252 276 243 246 275 246 244 245 259 260 267 267 251 222 235 255 231

>d=x-y

> t.test(d,alternative="greater")

One Sample t-test

data: d

t = 2.162, df = 16, p-value = 0.02306

alternative hypothesis: true mean is greater than 0

95 percent confidence interval:

0.917088 Inf

sample estimates:

mean of x

4.764706

MAT 2375, Assignment 4 5

[8.1-14]

Solution. (a) We reject H0 if

(n− 1)S2

100< χ2

0.975(22) = 10.98.

Since(n− 1)S2

100= 7.8144

Therefore, we reject H0.

(b) If(n− 1)S2

100< 10.98 = χ2

0.975(22) or(n− 1)S2

100> χ2

0.025(22) = 36.78

then we reject H0. Based on the data provided, we should reject H0.

Here are commands and output in R for comparison. (a) From R we get

> (22)*(35.52)/100

[1] 7.8144

> qchisq(0.025,22)

[1] 10.98232

We should reject H0.

(b) This is a two-sided test

> qchisq(0.025,22)

[1] 10.98232

> qchisq(0.975,22)

[1] 36.78071

Since10.98232 > 7.8144

we reject H0.

[8.2-10]

Solution. We can use From the text seems the assumption of equality of variances can be assumeddue to similarity of conditions. Therefore

tobserved =x− y

sp√

1/n+ 1/m= −1.24621.

For simplicity we did our calculations in R as follows:

MAT 2375, Assignment 4 6

> x=c(4.67,4.21,2.18,3.91,4.09,5.24,2.94,4.71,4.04,5.79,3.8,4.38)

> y=c(5.04,4.52,6.18,7.01,4.36,1.81,6.22,5.7)

> mean(x)

[1] 4.163333

> mean(y)

[1] 5.105

> var(x)

[1] 0.9142606

> var(y)

[1] 2.591486

> length(x)

[1] 12

> length(y)

[1] 8

> sp=(11*0.9142606+7*2.591486)/(12+8-2)

> T=(mean(x)-mean(y))/(sp*sqrt(1/12+1/8))

> T

[1] -1.316992

H0 = µ1 = µ2 H1 : µ1 < µ2.

p− value = P (t(18) < −1.316992) = 0.1021812.

With α = 0.05 or 0.1 we accept H0.