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Well these are the answers anyway....

Statistics 2 – June 2001 – Answers

1 5.

2 (i) Distribution of R is B(60, 0.05); since n > 50 and np < 5, a Poisson

approximation (R ~ Po(3)) would be appropriate.

(ii) Assign a number to each pupil in the school and use random number tables to

choose the sample of size 60.

3 0.887.

4 (i) ;

(ii) r  5; if r = 4, accept the null hypothesis – there is no evidence that

employee is late on more than 1 day in 10;

(iii) 0.630.

5 (i) Only a breaking strength less than that claimed is a cause for concern;

(ii) Reject null hypothesis (acceptance region is > 12953) – the company’s

claim that the breaking strength is 13000 N is not justified.

(iii) Central Limit Theorem needed because breaking strengths are not given to

be normally distributed.

6 (i) 3;

(ii) 0.125;

(iii) 2.75;

(iv) 2.79.

7 (i) The typing errors occur independently of each other;

(ii) 0.554;

(iii) 0.184;

(iv) 0.0923;

(v) 1.05 minutes.

Statistics 2 – June 2001 – Answers

1 5.

2 (i) Distribution of R is B(60, 0.05); since n > 50 and np < 5, a Poisson

approximation (R ~ Po(3)) would be appropriate.

(ii) Assign a number to each pupil in the school and use random number tables to

choose the sample of size 60.

3 0.887.

4 (i) ;

(ii) r  5; if r = 4, accept the null hypothesis – there is no evidence that

employee is late on more than 1 day in 10;

(iii) 0.630.

5 (i) Only a breaking strength less than that claimed is a cause for concern;

(ii) Reject null hypothesis (acceptance region is > 12953) – the company’s

claim that the breaking strength is 13000 N is not justified.

(iii) Central Limit Theorem needed because breaking strengths are not given to

be normally distributed.

6 (i) 3;

(ii) 0.125;

(iii) 2.75;

(iv) 2.79.

7 (i) The typing errors occur independently of each other;

(ii) 0.554;

(iii) 0.184;

(iv) 0.0923;

(v) 1.05 minutes.

Saiyan

Hey.... can anyone tell me how to do Question 4 on that paper (Jan 2001)?

The hypothesis question... I don't know what the question is asking for... Binomial hypothesis? Normal? Poission?

Thanks.

The hypothesis question... I don't know what the question is asking for... Binomial hypothesis? Normal? Poission?

Thanks.

i didnt get the answers correct for question 4 calculations.

but i reckon it will be poisson distribution for the hyposthesis, because he says he's late no more than 1 in 10 days, λ=1, therefore in 20 days, λ=2!

so for the hypothesis

H

H

is that what your'e asking?

Tazzie

i didnt get the answers correct for question 4 calculations.

but i reckon it will be poisson distribution for the hyposthesis, because he says he's late no more than 1 in 10 days, λ=1, therefore in 20 days, λ=2!

so for the hypothesis

H_{0} : λ=2

H_{1} : λ>2

is that what your'e asking?

but i reckon it will be poisson distribution for the hyposthesis, because he says he's late no more than 1 in 10 days, λ=1, therefore in 20 days, λ=2!

so for the hypothesis

H

H

is that what your'e asking?

haha... I've done it now..

it was

Ho: Mu = 1/10

H1: Mu > 1/10

los 5%

onetail test

Under Ho

X~B(20,1/10)

P(X<=3) = 0.8670

<---- 0.9

P(X<=4) = 0.9568

therefore,

P(X>3) = 0.133

<----0.1

P(X>4) = 0.0432

Decision,

Reject Ho if X>4 (X=>5)

r days = 4

not in rejection zone so... no evidence

etc etc (i think it was that binomial proportion thingy topic)

hmmm... that poisson way looks kind of dodgy? lol. Don't think it would work like that. I think you've thought of lamba value wrong.

Bye....

Tazzie

see i thought it was that too! i wrtoe it up, and then deleted the post!

imconfused!

can u guys please clarify why its one and the not the other!

imconfused!

can u guys please clarify why its one and the not the other!

hehe...I have the same problem... but in this case they didn't give us one average CONSTANT value for being away (poisson)... and ho and h1 were in fractions... so I guessed it would be binomial.... proportion testing..

Saiyan

hehe...I have the same problem... but in this case they didn't give us one average CONSTANT value for being away (poisson)... and ho and h1 were in fractions... so I guessed it would be binomial.... proportion testing..

and this binomial proportioney thing. i can work it in s3, but dont remember doing it in s2. would you just look up the tables to do it?

Tazzie

and this binomial proportioney thing. i can work it in s3, but dont remember doing it in s2. would you just look up the tables to do it?

oh wait sorry... I just realized that was just a normal binomial hypothesis question!!..

Hypothesis test of proportion is different.. here is an example question of it.

ho: p=1/4

h1: p>1/4

one tail test

los 5%

Carried out 100 times leading to 32 successes

under ho

X~B(100,0.25)

using normal approx

X~N(25,18.75)

to find C so that P(X>C-½) remember? you have to minus ½... can't remember why but i just know you have to

standardising

P(Z> (c-½-25)/√18.75) = 0.05

P(Z>1.645) = 0.05

Hence

(C-½-25)/(√18.75) = 1.645

therefore c = 32.6

decision

X>=33 Reject Ho

X<33 Accept Ho

Observed value x = 32

32<33 therefore accept Ho

No evidence that p is less than 1/4

nope, havent done the june 04 yet.

but i have a question anyway.

I have X, distributed by poisson, with mean 20.

PART 1: use the exact distribution to find P(X=22)

That's fine, just from the tables.

PART 2: use a suitable appoorximation to find P(X=22)

I need help with the seond part. Iaproximate it the normal, so thats

X-N(20,20)....But then?

but i have a question anyway.

I have X, distributed by poisson, with mean 20.

PART 1: use the exact distribution to find P(X=22)

That's fine, just from the tables.

PART 2: use a suitable appoorximation to find P(X=22)

I need help with the seond part. Iaproximate it the normal, so thats

X-N(20,20)....But then?

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can someone please explain what principle domain is and why the answer is a not c?Maths

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