[r213] Suppose we are testing
HI] : cfw_:I3 = 0':
versus
H1133 :- 0E?
The sample size is #2 =15.
Pvalue:
The are to the right ofthe test statistic: and is, using the f2 table.
The degrees of freedom is
arm1
=151
af=14
We have to look for the area to the r

[ejThe 98% eonfidenee interval for ,61 is
From the above multiple regression output, we note that
s, = o. isssis, = o.ossss,:,_, = asssoos.
The lower 98% eonfidenee interval is 61 r.m93i8'bl
= 0.19681 2365002 [002566)
= 0.19681 0.060686
= 0.136124
95-.
Th

Instructions to obtain Multiple regression output using .MiinitablT:
* Store the Amount of Tip), Amount ofE-illliXI, and the Number of Dinners (X3)
in the columns of the Minitab worksheet.
* IChoose Stat Regression Regression 4; Fit Regression Model
* Sel

[all The multiple regression equati an is
.3: +41+zfa+gs+e444+fs
EI=BEBT+U.332X1U332XE+ UHTXE +0.?34J 43.275 X5
(in) The sample size is as = 56.
The number of independent variables is E: = 5.

[l) The regression equation is
A
l = 2533+914X1+ETEX2
[2) The predicted balanee for a eustorner with X1213 ATM transaetions, anng 2 4 ATM transaetions is
The regression equation is
A
I" = 2533+914X1+ETEXE
= 258.3+91.4 [13)+sv.s[4j
= 2583+11882+2Tlo
=1?13.

[:13] SSE 21486.3and SSE = 2230.3
SST: SSE +333
= l4863+22303
= 3316.6
SST=3316.6
The formula for calculating the r2 value is
3 Regression surn ofsquares
r =
Total sum of squares
_ ssa
_
_ 1486.3
_ stuns
= ossssos
r3 = ossss
Interpretation:
R2 = 39.99%

(a) From the given information,
Sample 1: n1: 3", sample variance 321:17090236
Sample 2: as: = 3, sample variance 3% 2 5300.500
Step1:
The null and alternative hvpothes es:
H0: 0'21: J22
H1 : 021 # a:
Step2: The level of signficance is ct = 0.10.
Step3:
T

(at) The leasts quares predieti on equation for the interstate highway model is
f = a + arose
= l.82305+ 0.19681X1+U.0001?X2
[e] Interpretati on:
We estimate the mean number of erashes per 3 years will increased by ,8 for eaeh additional mile ofroad away,

(a) The degrees offreedom is
121+:2 2 =1C|+1C| 2
= 20 2
calf =18
The given information is
m=m,=4aa=ns
%=40,g=21,%=t6
The value ofthe test statistic: for xl x: is computed as follows.
[331113211'1392 1)th
921+322 2
[101)U.64+[1U1:I2.56
10+1U2
_J556163
_ 20

[33:] Area: The formula for calculating the mean is f = l
a!
1
3 1335
43': :1
2: 28.28531
2%-
Thc fcrmula fur calculating the standard dcvi ati an is s =
2s lLUsssg)
431
2s liming)
48
3 3133913
31331-413
ss-
EX:-
i-l

The owner ofan automobile repair shop knows that the probabilities of 0, 1, 2, 3 and-4 requests for towing services are
0.10, 0.25, 0.35, 0.20, and 0.10 per dav.
The number of requests for towing services can the owner expect on a given dav is
Eb?) =XPX=

The sample is given by 120, 130, 100, 295, 135, and 220.
_120+130+190+205+185+229
6
_ 95E]
_ a
The femiula fer calculating the value of standard desiati an is
1 Etta2:12
ssl 1-_1
Lassa)
51
12 50
5K:
W
l
= 2490
= 4969909

Main Effects:
The variance tahle enables us that the main effect Sample is having the Fvalue is 0.52 and the corresponding
pvalue is 0.59. It means that it is not significant.
The variance tahle enables us that the main effect Columns is having the Fvalue

[as] The sample variance for each of the three samples is
as = 4 for each of the three samples.
The sample variance for the first sample is
33
921
varl=
| as
l
mlmh
D
66666?
varl = 0.66666?
The sample variance for the seccuncl sample is
33
921
var
2:
_ 4

[raj Multiple Regression Analysis: Report
Instruetions to obtain a Multiple Regression output using Erroel:
3 Store the data in the erreel worksheet.
3 Choose Data A; Data Analysis Regression.
* I131i el: ole.
3 Seleet the Input I" range and then seleet t

(a) svstolie hp:
The level of signifieanee is l :12 = [195
The sample size is 22 = 61
The sample mean is E 2122
The standard deviation is s :1 l.
The formula for ealeulating the 95% eonfi denee interval is
S
5 ,2 222+:
321 :1: 22192
11 11
=122 _ :I,u:112

(e) Suppose we are testing
HI] : cfw_:I3 = 0':
versus
H1133 :- 0E?
The sample size is so =15.
Pvalue:
The are to the right ofthe test statistic: and is, using the :2 tah1e.
The degrees of freedom is
arm1
=151
af=14
We have to look for the area to the righ

The mean finish time far a yearly amateur autd raee is ,3! 2135.35.
The standard deyiatinn is J: [3.352 minutes.
Chii s:
The finish time for auto raee is 21 2134.35 minutes.
_135.35134.33
[1.352
[1.99
:5.
[.1352
252.3125
3: 2.3125
The mean finish time far

(at) Suppose we are testing
H.139j = 0':
versus
H1133 :- 0E?
The sample size is #2 =15.
Pvalue:
The are to the right ofthe test statistic: and is, using the f2 table.
The degrees of freedom is
arm1
=151
af=14
We have to look for the area to the right of 2

(at) Suppose we are testing
H.139j = 0':
versus
H1133 :- 0E?
The sample size is u 2 2'2].
Pvalue:
The are to the right ofthe test statistic: and is, using the f2 table.
The degrees of freedom is
arm1
=2oi
af=is
We have to look for the area to the right of

[45) Suppose we are testing
HI] : cfw_:I3 = 0':
versus
H1133 :- 0E?
The sample size is #2 =15.
Pvalue:
The are to the right ofthe test statistic: and is, using the f2 table.
The degrees of freedom is
arm1
=151
af=14
We have to look for the area to the rig

[raj Stepl:We setup the null hvpc-thesis the tvpe cf ceremonial ranking and pcuttenv
tvpe are dependent at the [1.05 level cf significance.
He : The carvings and cocaine are relapse dependent.
H1 :The carvings and cocaine are relapse independent.
Step?
Th

Insti'uetions to obtain Chis quare test for Asseoieati on output using Mnitah:
* Store the parti eipation in the eoluinn Cl and Sea in the rest of the eolunins ofthe h-iTinitah
worksheet.
* Choose Stat Tahles 4; Chi-s quare test for Assoieaition.
* Choose

(a) LetX be the random variable and the time between two sueeessive arrivals [assumed units is hours).
X ~EXp [.11 =1)
The expected time between two sueeessive arrivals is:
Elixj=n
[15:] Let X be the random variable and the time between two successive arr

(a) The pmf ofthe random variable x is
2 52
CE I j-,x=o,1,2.
The value of C is calculated is as follows.
2
Z p|:x) :1, we have
:Ir-IJ
:>p|:0)+p[:l)+p|:2:|=l
250 9C C
:>+=
7" 7" 7"
1
The value of C is calculated is as follows.
2
z p|:x) :1, w

(as) The sample size is as: 25.
The sample mean isf = 1221:
as
1-1
= 2;[26+2?+.+21)
_ ass
as
E: 22.4
The known standard desiati an is J: 3.
Te eenstruet a 93% eenfidenee interval estimate efthe mean breaking strength efthe wire.
_ J C"
XEsngx+3
a at?
2241

The simple rand-am sample pf size as = 125 SAT see-res.
The sample mean isp.r = 1522.
The standard deviatien is J: 333.
rCensti'net a 95% eenfidenee interval estimate of the mean SAT seere is
20EE+ET
=15221.96[ 333 gig in _1522+1. 96[ 333]
.1125
=15221.

(g) Inee-me:
The level of signicance is 1 cfw_1: = 5.99
The sample size is as = 91
The sample mean is I? = 43
The standard desiatinn is 3 =123
The formula fer ealeulating the 99% ennfi denee interval is
_ .S'
1: 53+
4T .55 EraINF;
243%? 12s 12s
_1 E in E

Instructions to obtain Chi-s quare test for Asscoicati on output using Minitab:
* Store the Sea: in the column Cl and Polytical party in the rest ofthe columns of the Minitab
worksheet.
* Choose Stat Tables 4,: Chis quare test for Assoicaition.
* Choose S