Test Of Hypothesis | EssayIvy.com

Surname 1

Student’s Name

Instructor’s Name

Course Name

Date

Test of Hypothesis

T-test

The ages of actresses when they won an acting award is summarized by the statistics n=78. x

with a line over it =35.2 years, and s=11.6 years. Use a 0.01 significance level to test the claim

that the mean age of actresses when they win an acting award is 34 years old

Hypothesis;

H0:  = 34

H1:  ≠ 34

T-statistics

Standard score, z=









Z= 





Z=0.9136

P(X=34) = P(Z=0.9136) = 1.36

α/2

=1.96

Surname 2

Since p(x=34) < Z

α/2

, do not reject H0 and conclude that the mean age of actresses when they

win an acting award is 34 years old

ANOVA

Suppose the National Transportation Safety Board (NTSB) wants to examine the safety of

compact cars, midsize cars, and full-size cars. It collects a sample of three for each of the

treatments (cars types). Using the hypothetical data provided below, test whether the mean

pressure applied to the driver’s head during a crash test is equal for each types of car. Use α =

5%.

A table of hypothetical data

Compact cars

Midsize cars

Full-size cars

643

469

484

655

427

456

702

525

402



666.67

473.67

447.33

31.18

49.17

41.68

Hypothesis

H0: µ

= µ

H1: At least one mean pressure is not statistically equal

T-statistics

Surname 3

Total sum of squares (SST)=

 

  









 where r is the number of rows in the table

and c is the number of columns in the table,  is the grand mean and x

is the i

observation in

the j

column.

Using the above data;

=











=529.22

SST= (643-529.66)

+(655-529.66)

+(702-529.66)

+(469-529.66)

+…+(402-529.66)

=96303.55

Treatment sum of squares (SSTR)=



  )^2

= [3*(666.67-529.22)

] + [3*(473.67-529.22)

] + [3*(447.33-529.22)

]

=86049.55

Error sum of squares(SSE)=

 

  ^2

= [(643-666.67)

+(655-666.67)

+(702-666.67)

] + [(469-473.67)

+(427-473.67)

+(525-

473.67)

] + [(484-447.33)

+(456-447.33)

+(402-447.33)

]

=10254

SST=SSTR+SSE (96303.55 = 86049.55 + 10254)

Total Mean Square (MST)=





where N is the total number of observations.

MST=





=12037.94

Mean Square Treatment (MSTR)=





Surname 4

MSTR=





=43024.78

Mean Square Error (MSE)=





MSE=





=1709









=25.17

Critical value;

The degrees of freedom are given as df1 = 3 - 1 = 2 and df2 = 9 - 3 = 6. Given that α = 5%, from

the F tables, FCV 2,6 = 5.14

Decision Rule

Reject the null hypothesis if: F (observed value) > FCV (critical value). Since 25.17

> 5.14, we reject the null hypothesis.

Chi-square test

The table below shows the age distributions of women in the sample and all women in Toronto

between the ages of 20 and 44.

Age

Number in sample

Per cent in Census

20-24

103

25-34

216

35-44

171

Total

490

100

Surname 5

Using the data in the above table, conduct a chi square goodness of fit test to determine whether

the sample does provide a good match to the known age distribution of Toronto women. Use the

0.05 level of significance.

Hypothesis:

H0: The age distribution of respondents in the sample is the same as the age distribution of

Toronto women, based on the Census.

H1: The age distribution of respondents in the sample differs from the age distribution of women

in the Census.

The frequencies of occurrence of women in each age group are the observed values Oi presented

in the middle column of table above. There are k = 3 categories into which the data has been

grouped so that i = 1; 2; 3

Suppose the age distribution of women in the sample conforms exactly with the age distribution

of all Toronto women as determined from the Census. Then the expected values for each

category, Ei, can be determined. With these observed and expected numbers of cases, the

hypotheses can be rewritten as

H0 : Oi = Ei

H1 : Oi ≠E

The test

For these hypotheses, the test statistic is









, with the degrees of freedom given as d=k-1

Surname 6

Let the 20-24 age group be category i=1. The 20-24 age group contains 18% of all the women,

and if the null hypothesis were to be exactly correct, there would be 18% of the 490 cases in

category 1. This is

=490x





=490x0.18=88.2

For the second category 25-34, there would be 50% of the total number of cases, so that

=490x





=490x0.5=245

=490x





=490x0.32=156.8

The table below shows the goodness of Fit Calculations for Sample of Toronto Women

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