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# T TEST - PowerPoint PPT Presentation

T TEST. DEFINITION OF TERMS. T – Test – the statistical test for comparing a mean with a norm or for comparing two means with small sample sizes (n≤30). Formula: __ X - µ

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### T TEST

• T – Test – the statistical test for comparing a

mean with a norm or for comparing

two means with small sample sizes

(n≤30).

Formula:

__

X - µ

t = ———

SD/√n

• Mean ( X ) – the most common measure of central

tendency, denoted by µ in the

population and by in the sample.

Formula: _

X = ΣX

n

Σ – summation or to add

µ - population mean

X – individual observations

n – number of observations

• Population – the entire collection of observations

or subjects that have something in

common and to which conclusions are

inferred.

• Standard Deviation – the most common measure of

σ in the population and SD or s

in the sample. It can be used with

mean to describe the distribution

of observations.

_

SD = √Σ (X – X)²

n – 1

Hypothesis Test – an approach to statistical inference

resulting in a decision to reject or

not to reject the null hypothesis.

STEP 1: State the research question in terms of

statistical hypotheses or formulate the hypothesis

Null Hypothesis (Ho) – a statement claiming that there is

no difference between the

assumed or hypothesized value

and the population mean.

Alternative Hypothesis (Hı)- a statement that disagrees

with the null hypothesis.

### Step 2: Decide on the appropriate test statistic.

__

X - µ

t = ———

SD/√n

Level of Significance – the probability of incorrectly

rejecting the null hypothesis in a

test of hypothesis, denoted by α

when it is actually true (and

concluding there is difference

when there is nor.

- traditional values used for α are

0.05, 0.01 and 0.001.

STEP 4: Determine the value the test statistic must attain to be declared significant. Determine the rejection region or the critical region.

Rejection Region – is a set of values of the test statistic for

which the null hypothesis is rejected

in a hypothesis test.

Acceptance Region – non-critical region

Critical Value – the value that a test statistic must exceed

for the null hypothesis to be rejected.

- refer to the critical values for t

distribution table.

deviation depending on a

complex concept related to

the sample size which is

related to the number of times

sample information is used.

- n -1 (df)

STEP 5: Perform the calculation.

STEP 6: Interpret and state the conclusion.

Oats is claiming that their product contains

8 grams of fiber per 35 gram-pack. VIN

Company, its direct competitor conducts an

independent test of the product. They test

15 samples of Yummy Oats and it found out

to have 8.00, 8.25, 8.16, 7.92, 7.81, 7.93,

8.01, 8.02, 8.50, 8.16, 7.71, 7.81, 7.62, 7.81

and 7.86 of fiber content. If 0.01 level of

significance is appropriate, would VIN

Company agree on the claim of YLO Co.?

STEP 1: H o: µ = 8 grams : Yummy Oats contains 8 grams of fiber/pack Hı: µ ≠ 8 grams : Yummy Oats does not contain 8 grams of fiber/pk

STEP 2: __

X X-X (X-X)²

8.00 0.03 0.0009

8.25 0.28 0.0784

8.16 0.19 0.0361

7.92 0.05 0.0025

7.81 0.16 0.0256

7.93 0.04 0.0016

8.01 0.04 0.0016

8.02 0.05 0.0025

8.50 0.53 0.2809

8.16 0.19 0.0361

7.71 0.26 0.0676 7.81 0.16 0.0256 7.62 0.35 0.1225 7.81 0.16 0.0256 7.86 0.11 0.0121

_

X = 119.57

15

= 7.97 Σ = 0.7196

7.97 – 8 -0.03 -0.03

T = ÷ = ÷ = ÷

0.2267/√15 0.2267/3.8729 0.0585

T = -0.5128

STEP 3: Level of Significance: 0.01

STEP 4:

Degrees of Freedom: 15 – 1 = 14

Critical Value : -2.624

RR AR

__________________________________

-2.624 -0.5128 0

the acceptance region.

Accepts Null Hypothesis

CONCLUSION:

VIN Company would agree to the claim

of YLO Company that Yummy Oats has

8 grams per 35 gram/pack of fiber

content.

that it contains 235mg of ascorbic

acid. An independent test was done

to prove whether the claim of the

manufacturer is true or not. Ten samples

were tested and the results are as follows:

230, 232, 228, 231, 218, 225, 221, 220, 223,

and 228. If 0.01 level of significance is

appropriate, determine whether to accept or

reject the claim of Vita C manufacturer.

STEP 1: H o: µ = 235 mg : Vita C contains 235 mg ascorbic acid Hı: µ ≠ 235 mg : Vita C does not contain 235 ascorbic a.

__

X X-X (X-X)²

230 4.4 19.36

232 6.4 40.96

228 2.4 5.76

231 5.4 29.16

218 7.6 57.76

225 0.6 0.36

221 4.6 21.16

220 5.6 31.36

223 2.6 6.76

228 2.4 5.76

_ X = 2256/10 = 225.6 Σ = 218.4

SD = √218.4/10-1 = √218.4/9 = √24.27 = 4.93

T = 225.6-235/4.93/√10 = -9.4/4.93/3.16 = -9.4/1.56

= -6.03

STEP 3: Level of Significance: 0.01

STEP 4: Degrees of Freedom: 10-1 = 9

RR AR _______________________________________ -6.03 -2.821 0

STEP 6: The computed value of t is -6.03 is at the

rejected region.

Rejects Null Hypothesis.

CONCLUSION: The independent test for Vita C

containing 235 mg of ascorbic

acid is rejected.