Hypothesis Testingusing Z-Test, T-Test,

Linear – Correlation and Regression, and Chi-Square Distribution

Submitted by: ASPILI, Alleli A. ;AYERS, Anna V. ; LEYRAN, Kristine Joy F. ; EDLAGAN, Bianca Camille ; LOGAN, Cathrin V.

Submitted to: Mr. Marcus Alphonsus O. Lagazon

BSBAMM2A

Z test (one sample)• On the year 2007, Feel Fresh Inc., a manufacturer of

soap, claims that the average weight of a bath soap per pack is 200 grams. A retailer sampled 10 packs of this soap and got an average weight of 199.3 grams and a standard deviation of 5.

• Is it appropriate to conclude at the α = .05, that Feel Fresh Inc. fools customer?

)'.(200:.1 0 customersfooltdoesnIncFreshFeelgramsH ).(200:1 customersfoolsIncFreshFeelgramsH

05..2

...3 VCwithtesttailedOne

443.0442718872.0

516227766.3)7.0(5

10)2003.199(

)(

.4

nXz

knownissincetestZ

6451055

. is -value tabular or critical the, ., at α test- Ztailed-one a For.

10 Re6 ject H,. Accept H

-1.645 -0.443

acceptance

s.l customeres not foosh Inc. do. Feel Fre7

Z test (two samples)• On the next year, 2008, Feel Fresh Inc., decided to produce

two kinds of bath soap – the papaya and milk soap. The company directed experiments to both soaps to test its capability to whiten skin. Few scientists were asked and the results were as follows:

• Assuming that the standard deviation, σ , is known to be 8.6. Is it valid to conclude that Papaya soap whitens skin more than Milk soap?

n XPapaya Soap

21 61

Milk Soap 17 52.5

y)he same waens skin t soap whitp and MilkPapaya soaXXH mp (:.1 0

)(:1 soapthan milk skin more p whitens Papaya soaXXH mp 05..2

...3 VCwithtesttailedOne

065.3065083511.3805796308.2

5.8106442576.6.8

5.8171

2116.8

5.5261

11

.4

mp

mp

nn

XXz

knownissincetestZ

6451055.is lar value al or tabuthe critic

, . test, at α-tailed Z-.For a one

10Re6 ,Accept Hject H. 1.645 3.065

rejection

k soap.e than Mils skin moroap whiten. Papaya s7

T test (one sample)• This year of 2009, Feel Fresh Inc. claims that the

average weight of papaya essences from the Papaya soap per piece is 100 mg with a standard deviation of 3. An experiment was made from the 5 pieces of Papaya soap and got an average weight of papaya essences of 97.9 mg. Is it valid to conclude, at α = .05 that Feel Fresh Inc., still, doesn’t fool customers?

)'.(100:.1 0 customersfooltdoesnIncFreshFeelmgH

).(100:1 customersfoolsIncFreshFeelmgH

...3 VCwithtesttailedOne

05..2

5652.1565247584.1

32360679.2)1.2(

35)1009.97(

)(

30sin.4

nXz

ncetestT

132.2,405

41515

is -value tabular or critical the, df., at α test-t tailed-one a For

n.df

10 Re6 ject H,. Accept H

-2.132 -1.5652

acceptance

. customerss not foolstill, doesh Inc.,. Feel Fre7

Linear - Correlation• Five students in CEU were given 1 test in Statistics

and 1 test in Business Communication. Find the correlation between the scores in the two subjects, which are given as follows:

Student Scores in Statistics

(x)

Scores in Bus. Com.

(y)1 27 322 22 153 10 224 39 275 40 30

x=138 y=126

Computed values made from the table from the previous slide are as follows:

rx ry rx-ry (rx-ry)2 xy x2 y2

3 1 2 4 864 729 10244 5 -1 1 330 484 22515 4 1 1 220 100 4842 3 -1 1 1053 1521 7291 2 -1 1 1200 1600 900

(rx-ry)2=8 xy=3667 x2=4434 y2=3362

From the computed values in the table from the previous slide,compute for the SPEARMAN RANK-ORDER CORRELATIONCOEFFICIENT and conduct the hypothesis testing.

96.04.1120481

)15(5)8(61

1)(61

2

2

2

nn

ryrxrs

d)generalize be (CanrHd)generalize be (CannotrH

s

s

0: 0:.1

1

0

05..2 testtailedTwo .3

96..4 sr

001505 5

.5

. is , df .αat value Critical

ndf

10 Re6 ject H,. Accept H

-1.00 1.00.96

acceptance

d.generalize be cannot it therefore, nt, significanot is .96 r value, computed 7.The s

From the computed values from the previous table,compute for the PEARSON PRODUCT MOMENT CORRELATIONCOEFFICIENT and conduct the hypothesis testing.

5677.05676802817.0

708284.1708970

)934)(3126(970

)126()3362(5)138()4434(5)126)(138()3667(5

22

2222

yxnxxn

yxxynr

d)generalize be (CanrH

d)generalize be (CannotrH

0:

0:.1

1

0

05..2 testtailedTwo .3

5677.0.4 r

8783.505 3

252.5

is , df .αat value Critical

ndf

-.8783 .8783 .5677

rejection

10Re6 ,Accept Hject H.

d.generalize be can it

therefore, nt, significais .5677 r value, computed 7.The

Linear - Regressionx y xy x2 y2

3 5 15 9 252 7 17 4 4910 8 80 100 6412 9 108 144 816 2 12 36 4

x = 33 y = 31 xy = 232 x2 = 293 y2 = 223

6.6533

nxx

2.6531

nyy

3644.0

3643617021.0376137

1089146510231160

33)293(5)31)(33()232(5

2

22

xxn

yxxynm

3.794962.405042.6

)6.6)(3644(.2.6

xmyb

REGRESSION LINE

6169657949638221

794963536445

3.794963644

...

.)(. ?, then y if x

x . b mxd y predicte

STANDARD ERROR OF ESTIMATE

63.2634099467.26.93848

320.81544

384.5408-117.6437622325

)232)(3644(.)31(79496.32232

2

n

xymybySe

Chi Square Distribution• Fine Arts students from DLSU claim that among

the four most refreshing colors, students have these preference rates: 49% prefer green; 35% prefer white; 10% prefer blue; and 6% prefer yellow. A random of 250 students were chosen. Test the claim that the percentages given by the students are correct, using α = .05.

Color No. of studentsGreen 125White 84Blue 29Yellow 12

06.;10.;35.491 43210 PP;P.: P.H06.;10.;35.49 43211 PP;P.: PH

05..2 ...3 VCwithtesttailedOne

4310.1.42

2

E

EO

O E O-E (O-E)2

125 122.5 2.5 6.25 .05702040884 87.5 -3.5 12.25 .1429 25 4 16 .6412 15 -3 9 .6

= 1.431020408

E

2E)-O(

E

EO 2

815.7305 3

141.5

is , df .αat value Critical

kdf

10 Re6 ject H,. Accept H

1.431 7.815

acceptance

claimed.the values equal to rtions are.The propo7