73626176 Cac Buoc Danh Gia Cldv Dung Servperf

7/29/2019 73626176 Cac Buoc Danh Gia Cldv Dung Servperf

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GVHD: Phm L Hng Nhung M Hnh SERPERF

O LNG CHT LNG DCH V

S DNG M HNH SERVPERF

I. GII THIU CHUNG V M HNH SERVPERF

Khi cnh tranh trong ngnh cng nghip dch v tng ln, cc khi nim v

cht lng dch v ngy cng quan trng, cht lng dch v c xc nh l

mt yu t quyt nh th phn, u t v gim chi ph. Xc nh v o lng

cht lng dch v c tm quan trng vi cc nh cung cp dch v. C rt nhiu

nh nghin cu c gng nh ngha v o lng cht lng dch v. Cht lng

dch v l khong cch gia s mong i v dch v ca khch hng trc khi sdng dch v v nhn thc ca h sau khi s dng dch v thc t. Cc nh nghin

cu u tin a ra thang o SERVQUAL o lng CLDV. Tuy nhin, c

mt s kh khn khi s dng thang o ny.

T nhng vn trn, mt bin th ca thang o SERVQUAL ra i, c tn l

thang o SERVPERF. Thang o SERVFERP c s dng o lng cm nhn

ca khch hng t xc nh cht lng dch v thay v o lng c cht lng

cm nhn ln k vng nh thang o SERVQUAL. Thang o SERVPERF c cc

tc gi Cronin & Taylor (1992) a ra da trn vic khc phc nhng kh khn khi

s dng thang o SERVQUAL cng vi 5 nhn t ca cht lng dch v: tin

cy, phn hi, s bo m, s cm thng v tnh hu hnh v 22 bin quan st

c s dng o lng 5 nhn t k trn.

II. CC BC THC HIN NGHIN CU CHT LNG DCH V

S DNG M HNH SERVPERFGm 2 bc:

1. Bc 1: Nghin cu nh tnh:

1.1. Nhm nghin cu t chc hp nhm, tho lun xc nh cc vn

sau:

Tng th: l nhng khch hng s dng dch v m chng ta ang

cn nh gi.

Mu: c th xc nh mu theo cng thc sau:

Bi Tp Nhm: QT Cht Lng Du Lch Nhm: 011


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Trong : n : c mu

N: tng th

e: sai s ti a

Cc bin quan st

=>Bng cu hi gm bao nhiu nhm v tiu ch (Da vo 5 nhm v 22 tiu

ch gc). Thang o no (Lirkert 5 hoc 7)

VD: Bng cu hi nh gi cht lng dch v ca cng ty XYZ , s dng

thang o Likert 5 mc . Bng cu hi c dng nh sau:

BNG CU HI

NH GI CHT LNG DCH V CA CNG TY XYZ

Xin cho, ti tn l ., thuc nhm nghin cu th

trng..Hin chng ti ang tin hnh kho st kin khch hng v cht

lng dch v ca cng ty XYZ. Cuc kho st ny s gip chng ti c c s

ci thin, nng cao cht lng phc v ca Cng ty XYZ nhm p ng nhu cungy cng cao ca qu khch. Do vy, ti rt mong nhn c s cng tc chn tnh

nht ca Anh/ Ch. Chng ti mong Anh/ch dnh cht thi gian khong 30 pht

tr li cc cu hi di y. Cc kin tr li ca Anh/ Ch s c gi b mt

tuyt i.


.n = N/(1+N *e2)

S th t mu: ..Ngy:.H tn phng vn vin:.


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I. PHN SNG LC (c th bao gm nhng cu hi sau)

Q1. C thnh vin no trong gia nh Anh/ Ch hoc ngi thn ca Anh/ Ch hin

ang lm vic ti Cng ty XYZ khng?

(1) C Dng (2) Khng Tip tcQ2. C thnh vin no trong gia nh Anh/ Ch hoc ngi thn ca Anh/ Ch hin

ang lm vic ti cc cng ty nghin cu th trng khng?

(1) C Dng (2) Khng Tip tc

Q3. Anh/ Ch tng s dng dch v ca Cng ty XYZ cha?

(1) C Tip tc (2) Khng Dng

II. PHN CT LI

Q4. Anh/ Ch hy cho bit mc ng ca Anh/ Ch v cc pht biu di y

(bng cch nh du x vo cc thch hp) ?

1. Rt khng ng 2. Khng ng 3. Trung bnh 4. ng 5. Rt ng

CC PHT BIU NH GI CA

KH1 2 3 4

S hu hnh (Tangible) Cng ty XYZ c trang thit b hin i Cng ty XYZ c c s vt cht trng rt hp dn

Nhn vin ca cng ty XYZ c trang phc gn gng, cn thn C s vt cht ca cng ty XYZ c hnh thc, kiu dng hp dn v

lin quan n dch v cung cp

S tin cy (Reliability) 5. Khi ha s thc hin 1 iu g vo khong thi gian c th,cng tyXYZ s thc hin

6.Khi bn c vn , cng ty XYZ th hin s quan tm chn thnh khigii quyt vn

7. Cng ty XYZ thc hin dch v ngay t ln u tin

8. Cng ty XYZ cung cp dch v ng vo thi im cng ty ha sthc hin



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9. Cng ty XYZ lun lu tm n vic gi gn ting tm ca mnh.

Tinh thn trch nhim (Responsiveness) 10. Cng ty XYZ cho khch hng bit khi no h cung cp dch v cho

11.Nhn vin ca cng ty XYZ lun nhanh chng phc v dch v chokhch hng.

12.Nhn vin ca cng ty XYZ lun sn lng gip khch hng

13.Nhn vin cng ty XYZ khng bao gi qu bn rn khng png yu cu ca khch hng.

S m bo (Assurance)

14.Khch hng c th tin tng mi hnh vi ca nhn vin trongcng tyXYZ.

15.Khch hng cm thy an ton khi thc hin giao dch vi cng tyXYZ.

16.Nhn vin XYZ lun nim n, lch s v nh nhn vi khch hng

17.Nhn vin lun c cng ty XYZ cp nht kin thc, nng cao tayngh thc hin tt cng vic ca h

S cm thong (Empathy): 18.Cng ty XYZ lun dnh s ch n tng c nhn khch hng

19.Nhn vin cng ty XYZ lun quan tm n khch hng

20.Nhn vin cng ty XYZ hiu r nhu cu ca khch hng.

21.Nhn vin cng ty XYZ lun t li ch ca khch hng ln hng

u.

22.Cng ty XYZ lm vic vo nhng gi thun tin cho khch hng cah

Q5. Anh/ Ch hy cho bit mc hi long ca Anh/ Ch v dch v ca Cng ty

XYZ ?

(1) Rt khng hi lng (2) Khng hi lng (3) Trung bnh(4) Hi lng (5) Rt hi lng



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III. PHN PHN LOI ( c th gm nhng cu hi sau:)

Xin Anh/ Ch vui lng cho bit mt s thng tin c nhn sau:

Q6. Anh/ Ch cn c than hay lp gia nh?

(1) c thn (2) C gia nhQ7. Vui lng cho bit trnh hc vn ca Anh/ Ch ?

(1) M ch (2) Tiu hc ( 3) Trung hc

(4) Cao ng (5) i hc (6) Sau i hc

Q8. Xin vui lng cho bit Anh/ Ch thuc nhm tui no sau y:

(1)18-24 (2) 25-31 (3) 32-38

(4) 39-45 (5) >45

Q9. Xin cho bit thu nhp hng thng ca Anh/ Ch

(1) < 2 triu (2) 2-5 triu (3) 5-10 triu (4) >10 triu

Q10. Xin vui lng cho bit gii tnh ca Anh/ Ch:

(1) Nam (2) N

CHNG TI XIN CHN THNH CM N S GIP CA ANH/ CH!

1.2. M hnh nh th no



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VD:

1.3. Xc nh gi thit

VD: Cc gi thit cn kim nh nh sau: H1: Phng tin hu hnh c quan h dng vi s hi long, ngha l

phng tin hu hnh ca dch v c khch hang nh gi cng cao, th mc hi

lng ca khch hang i vi dch v cng ln v ngc li.

H2: Tin cy quan h dng vi s hi lng.

H3: p ng quan h dng vi s hi lng.

H4: Kh nng phc v quan h dng vi s hi lng

H5: Cm thng quan h dng vi s hi lng

1.4. Cng thc chung cho m hnh SEVRPERF

Trong : 1

(k

SQ I P =

=

1.4.1. SQ: Cht lng DV

1.4.2. Pij: Cm nhn ca KH i i vi DV j no 1.4.3. Ij: trng s ca DV j



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1.4.4. k: s lng DV p dng trong m hnh

2. Bc 2: Nghin cu nh lng:

Sau khi c s liu, dng phn mm thng k (SPSS) kim nh cctham s v gi thit sau y:

2.1. Thng k m t cc bin quan st.

Tm h s Alpha Cronbach kim nh tin cy ca m hnh.



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VD:



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Tnh im trung bnh ca cc bin quan st:VD:

=> Thng qua gi tr trung bnh (mean), gi tr ln nht (max), nh nht

(min) v lch chun chng ta c th nhn bit nh gi chung ca khch hng

i vi cht lng dch v m cng ty cung cp.

ngha ca gi tr trung bnh

Gi tr khong cch = (Max min) / n = (5 -1)/5 = 0.8



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Gi tr trung bnh ngha1.00 1.80 Rt khng tt1.81 2.60 Khng tt2.61 3.40 Trung bnh3.41 4.20 Tt4.21 5.00 Rt tt

Nh vy, theo v d trn ta thy cht lng dch v gio dc ca STU c

sinh vin nh gi tng i tt. Cc gi tr trung bnh dao ng t 3.29 3.78.

2.2. t gi thit

VD:

H1: Thnh phn tin cy c tng quan dng vi s hi lng ca KH

H2: Thnh phn p ng c tng quan dng vi s hi lng ca KH

H3: Thnh phn nng lc phc v c tng quan dng vi s hi lng ca

KH

H4: Thnh phn ng cm c tng quan dng vi s hi long ca KHH5: Thnh phn phng tin hu hnh c tng quan dng vi s hi lng

ca KH

2.3. Xc nh hm hi quy a bin. Phn tch hi quy kim nh m hnh.

VD:



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2.4. T h s Bta suy ra mc nh hng ca 5 nhn t n s hi lng

ca khch hang. H s ny cng cao th nhn t cng c nh hng.

VD:

2.5. Da vo kt qu thu c, rt ra kt lun: nhn t no c nh hng ln

nht n cht lng DV, nhn t no cn khc phc, t a ra cc gp v kin

ngh cho doanh nghip.

III. U, NHC IM CA M HNH SERVPERF

1. u im

- n gin, d thc hin , t tn thi gian v chi ph kho st.

- Cho kt qu chnh xc, tin cy cao hn so vi SERVQUAL v:

+ Khi c hi v mc cm nhn, ngi tr li c xu hng so snh

gia mong mun v cm nhn trong u tr li bn cu hi, lm gim sai s v

sai lch trong kt qu.

+ M hnh SERVPERF b qua phn hi v k vng ca khch hng khi

nim kh m h i vi ngi tr li, gip hn ch c sai st trong cht lng dliu thu thp.


M hnh SERVPERF1. Phng tin hu hnh (0,284)2. Nng lc phc v ( 0,201)3. p ng (0,182)4. ng cm (0,165)5. Tin cy (0,140)

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- Bng cu hi ca SERVPERF ngn gn hn mt na so vi SERQUAL,

khng gy nhm chn v mt thi gian cho ngi tr.

2. Nhc im:

- M hnh SERVPERF khng th hin c thng tin then cht nh hngn mc cm nhn cht lng dch v ca khch hng, l s k vng ca

khch hng.

- Mt s ngi cho rng SERVPERF cha kim tra c ton din v trit

cc nhu cu dch v v cht lng dch v cung cp.


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73626176 Cac Buoc Danh Gia Cldv Dung Servperf