test 2 probability and statistics

8/3/2019 TEST 2 Probability and Statistics

1/5

Ariel Alvarez

EM 561 Probability and Statistics

TEST 2

Problem 1

a) One Sample Z

The assumed standard deviation = 0.0412

N Mean SE Mean 95% CI

75 2.01800 0.00476 (2.00868, 2.02732)

b) Using the confidence interval it can be concluded that the mean length ofits nails is NOT 2.00 inches.

c) Test and CI for one Standard Deviation: CI

Method

Null Hypothesis Sigma = 0.047

Alternative hypothesis Sigma = < 0.047

The Standard Method is only for the normal distribution.

The adjusted Method is for any continuous distribution.

Statistics

N StDev Variance

75 0.0412 0.00170

99% One-sided Confidence Intervals

Upper Bound Upper Bound

Method for StDev for Variance

Standard 0.0508 0.00258

Test

Method Chi-Square DF P-value

Standard 56.86 74 0.070

H0: 202H1:

2


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Problem 2

a) Estadsticas descriptivas: C1

Variable N Mean SE Mean Desv.Est. Mediana

C1 16 7.9750 0.0289 0.1156 7.9700

Test and CI for one Standard Deviation: CI

Method

The Standard Method is only for the normal distribution.

The adjusted Method is for any continuous distribution.

Statistics

N St.Dev Variance

16 0.116 0.0134

90% One-sided Confidence Intervals

Upper Bound Upper Bound

Method for StDev for Variance

Standard 0.153 0.0235

0.153

--------------------------(0.15)

------------------------------- (0.153)

The Cardinal Coffee Company conclude the standard deviation of the amount ofcoffee in its cans is NOT less than 0.15 ounce.

b) NO

c) H0: 0H0: >0

One-Sample T

Test of mu = 7.95 vs. > 7.95

95% Lower

Variable N Mean St.Dev. SE Mean Bound T P

C1 16 7.9750 0.1156 0.0289 7.9243 0.86 0.200

0.200>0.05 so I accept H0Its not believable that the mean amount of coffee in Cardinale Coffee Company

cans is at least 7.95.

d) NO


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Problem 3

Intervals # of observations Expected # obs X02

[25.0,30.0) 30 28.5714 0.0714315

[30.0,34.0) 28 22.8571 1.15716431

[34.0,39.0) 38 28.5714 3.11145054

[39.0,46.0) 30 40 2.5

[46.0,50.0) 22 22.8571 0.0321397

[50.0,60.0) 52 57.1429 0.46286451

7.33505057

Critical value = X20.05, 5 = 11.07

7.3351


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Problem 4

a) A straight line would be a good fit to this data.

706050403020100

30

25

20

15

10

Cavity Depth (in mm)

Strength(inMPa)

Grfica de dispersin de Strength (in MPa) vs. Cavity Depth (in mm)

b) Strength (in MPa) vs. Cavity Depth (in mm)

The regression equation is

Strength (in MPa) = 27.2 - 0.298 Cavity Depth (in mm)

Predictor Coef SE Coef T P

Constant 27.183 1.651 16.46 0.000

Cavity Depth (in mm) -0.29756 0.04116 -7.23 0.000

S = 2.86403 R-Sq = 76.6% R-Sq(adj) = 75.1%

Analysis of Variance

Source DF SS MS F P

Regression 1 428.62 428.62 52.25 0.000

Residual error 16 131.24 8.20

Total 17 559.86

With R-Sq = 76.6%, a simple linear regression is a good fit.

c) Yes, Cavity Depth does have a significant effect on Tooth strength.

We know this because

If p


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test 2 probability and statistics

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