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STAT 400 UIUC

8.1 - Hypothesis Tests for One Mean Stepanov

Dalpiaz Nguyen

Hypotheses Testing for the population mean µ Null Alternative H 0 : µ ³ µ 0 vs. H 1 : µ < µ 0 Left - tailed. H 0 : µ £ µ 0 vs. H 1 : µ > µ 0 Right - tailed. H 0 : µ = µ 0 vs. H 1 : µ ¹ µ 0 Two - tailed.

Test Statistic: OR OR

Rejection Region:

H 0 : µ ³ µ 0

H 1 : µ < µ 0

Left - tailed.

H 0 : µ £ µ 0

H 1 : µ > µ 0

Right - tailed.

H 0 : µ = µ 0

H 1 : µ ¹ µ 0

Two - tailed.

Reject H 0 if

Z < – z a

Reject H 0 if

Z > z a

Reject H 0 if

Z < – z a/2

or Z > z a/2

Reject H 0 if

< µ 0 – z a

Reject H 0 if

> µ 0 + z a

Reject H 0 if

< µ 0 – z a/2

or > µ 0 + z a/2

n σμ0X

Z-

=

ns

0μXT

-= X

xn s x

n s x

n s

xn s

→assume

pop. isnormal

some

f numberor sample size is

large enough( >25)

0

pop. std sample Std

I - pie⇒ T - Za

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Example 1: The overall standard deviation of the diameters of the ball bearings is s = 0.005 mm. The overall mean diameter of the ball bearings must be 4.300 mm. A sample of 81 ball bearings had a sample mean diameter of 4.299 mm. Is there a reason to believe that the actual overall mean diameter of the ball bearings is not 4.300 mm? a) Perform the appropriate test using a 10% level of significance. Claim:

H 0 : vs. H 1 : Test Statistic: Rejection Region: Decision:

P-value: Decision:

Confidence Interval: Decision: b) State your decision (Accept H 0 or Reject H 0 ) for the significance level a = 0.05.

2=-0. I

6 is known → E- test

The overall mean diameter of the ball bearing = 4.300 mm

µ=4 µ f- 4.300 mmMo ↳ two - tailed test

z ⇐I-M°_

=4.299-2.3004 =

X

¥¥±→ % 0.00%87 prob . of observing theobsessed

42 I(Z f = O- 95 I test statistic as extreme or→% / more extreme in the direction

Reject to if ✓ oft, assuming Ho is true.

Z f - zag or Z >Z p- value = 2x I (Z L - 1.8 )

⇐ Z f - I . 645 or Z> 1.645= 2×0 .

0359 = 0.0718= I (Z C - 1.8) t I ( Z> 1.8)

Reject to at 2=0 . I Reject to since

p-value = 0.0718 ( 2=0. I

6 is known Reject Ho at 2--0. IA 90% CI for µ :

I I Zz = 4.299 I 1.645 . Q°g ⇐ (4.2981.4.299#does not include 4.300

mm

since p- value = 0.0718 > • = 0.05, fail to reject Ho .

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Example 2: A trucking firm believes that its mean weekly loss due to damaged shipments is at most $1800. Half a year (26 weeks) of operation shows a sample mean weekly loss of $1921.54 with a sample standard deviation of $249.39. a) Perform the appropriate test. Use the significance level a = 0.10. Claim:

H 0 : vs. H 1 : Test Statistic: Rejection Region: Decision:

P-value: Decision:

b) State your decision (Accept H 0 or Reject H 0 ) for the significance level a = 0.05. The t Distribution r t 0.40 t 0.25 t 0.20 t 0.15 t 0.10 t 0.05 t 0.025 t 0.02 t 0.01 t 0.005

25 0.256 0.684 0.856 1.058 1.316 1.708 2.060 2.167 2.485 2.787

t - test 1€ ¥#EReject Ts - ta ta - take +42

Reject if T > to Reject if TC - theORT>

+42.

s = $249.39

Mean weekly loss f $1800 pright - tailed

Mµ.

µ > 1800

6 is unknown ⇒ t- test

1- =I - Mo 1921.54 - 1800-

=-I 2.485

4pm 249.391kg€¥¥ Ni,2=0.1 -1T

2.485Reject Ho tf T > ta

, df - 25 P - value = Rft 72.485 )off -- 25

tf T > 1.316 = 0.01

Reject tho at- 2--0.1 Since p- value = 0.0122=0.1

,

since -1=2.4857 1.318 . reject Ho .

Since p- value = 0.0122=0.05

, reject Ho .

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Example 3: Metaltech Industries manufactures carbide drill tips used in drilling oil wells. The life of a carbide drill tip is measured by how many feet can be drilled before the tip wears out. Metaltech claims that under typical drilling conditions, the life of a carbide tip follows a normal distribution with mean of at least 32 feet. Suppose some customers disagree with Metaltech’s claims and argue that Metaltech is overstating the mean (i.e. the mean is actually less than 32). Metaltech agrees to examine a random sample of 25 carbide tips to test its claim against the customers’ claim. If the Metaltech’s claim is rejected, Metaltech has agreed to give customers a price rebate on past purchases. Suppose Metaltech decided to use a 5% level of significance and the observed sample mean is 30.5 feet with the sample variance 16 feet

2. Perform the appropriate test. Claim:

H 0 : vs. H 1 : Test Statistic: Rejection Region: Decision:

P-value: Decision:

The t Distribution r t 0.40 t 0.25 t 0.20 t 0.15 t 0.10 t 0.05 t 0.025 t 0.02 t 0.01 t 0.005

24 0.256 0.685 0.857 1.059 1.318 1.711 2.064 2.172 2.492 2.797

yes, 32p left

- tailed

µ > 32 µ ( 32

6 is unknown ⇒ t - test

f- =I - Mo 30.5 - 32- =

⇒= - 1.875

Trn p- value

xx = 0.05

NReject Ho of p - value -- IC s)

%, Tt - ta,df=2y- = I ( tdf=2y > 1.875)- ta TL - 1.711

⇒ 0.025 ( p - value ( 0.05

Since T= - 1.875C- 1.711,we reject since p

- value ( 0.05,

Ho at 2=0.05 reject to at 2=0.05

p1. 875