OCR MEI Statistics (S1) - 25th January 2013

Apparently I lose one mark for not explicitly writing 'Two Tailed Test'

Is this true?

No not at all.

Aslong as you clearly state that the new H1 hypothesis is...

H1 ≠ 0.35 (NOT H1 > 0.35 or H1 < 0.35)

Then you are fine, as that pretty much insinuates a two tail test. And as long as you carry out the two tail, then you're home and dry...

Reply 141

likkle

Original post by Magenta96

You know for the outer region of the venn diagram, did anyone get 0.892?

i did

Reply 142

username1107833

2

Original post by likkle

i did

ah you got it wrong mate, sorry, it was 0.92

Reply 143

username1107833

2

Original post by Sam_94

Does anyone remember ALL of the questions??? it would be helpful if you knew the marks for it too! , hopefully someone has photographic memory haha!!

Ask away, I'm pretty sure I can recall most if not all of the questions and might know the marks for them aswell.

Reply 144

Dr.Monstaa

I remember most of it.

1)Stem and leaf diagram. Find mean and standard deviation.

2)I think it was Alan's probability question. (Venn diagrams) etc.

3)Dog shows. Combinations question, and probability.

4)Malik has a fair dice. Find the probability.

5)Expectation and variance.

6)Histograms, mean and median.

7)Two tailed hypothesis testing. Involving binomials before the hypothesis questions

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Reply 145

Attz_09

sum 1 make a unofficial mark scheme
1) had 2 parts it asked about skewness part i
ii) find mean and varience using calculator

2i) i think it was find p(LnW) where p(L|W) is 0.4 P(luB) =0.08 P(A) = 0.07
ii) draw ven diagram for the data
iii) state whethere they are independant or not from the ven diagram

3i) 11 dogs want to have 3 in competition how many different selections
3ii) 5 terriers in the 11 find probability that out of the 3 dogs chosen 2 are terriers (5 marks)

4) cant remember malik one that well i thin part i was throw it 3 times what is probability gets it right 3rd time (fair dice)
4ii) again camt remember aswell i think it was at most/atleast 10 throws get 1 6 find probability (fair dice)

5i) summut like this draw a probability distribution table for r = 2,3,4,5 where formula is x= k(xsquared - 1)
5ii) calulate var(x) and E(x)

6) had 5 parts cant remember all i)i think it was draw histogram
ii) calculate median
iii) calculate mean
iv) find amount of girls exceeding 160 cm and amount of boy exceeding 160cm
v) ??

7i) = 5 parts cant remember both i) was just a standard binormial easy (3 marks)
iA) at most
ib) at least
ii) the ******* huuge 9 marker cant remember so well as time finished
iii) never even got to read

After all my hard work putting all this ****e in sum1 clever plz do a rough unofficial mark scheme. i dont know if my answers are right thats y. Add questions i missed if ya want and add marks you can remember :P

i dun my bit

(edited 11 years ago)

Reply 146

Attz_09

Original post by theCreator

Ask away, I'm pretty sure I can recall most if not all of the questions and might know the marks for them aswell.

do it do it nooow!!!! unofficial mark scheme

Reply 147

Attz_09

fir 9 marker was your null and alternative hypothesis
H0 = 0.5
h1 not equal 0.5
i was gonna do 0.35 but it said differing so ehhh
i got mostly everthing else (method right for two tailed test but never got time to finish it
i think i found critical region and probability of x less than or equal to 0.5 and hos half way through that upper value
never got time to finish they shud giv me extra time lol

Reply 148

Sam_94

Original post by theCreator

Ask away, I'm pretty sure I can recall most if not all of the questions and might know the marks for them aswell.

could you please list them?, much appreciated

Reply 149

username1107833

2

UNOFFICIAL TEST PAPER - S1 January 2013

1)i) What is the skew of the data?
ii) Find the means and standard deviation using a calculator

2i) Find P(LnW)
ii) Draw a Venn diagram to represent the data
iii) Are the events L and W independant

3i) How many different dogs selections can be made (11C3)
ii) Find the probability that atleast 2 of the 3 dogs are terriers

4)i) Whats the probability that Malik gets a 6 on his third throw
ii) Probability of it requiring at most 10 throws to get a 6

5)i) Draw a probability distribution table for r= k(r^2 - 1) for when r=2,3,4,5
5ii) Calculate the Var(X) and E(X)

6) Forgot the order but these were the questions:
i) Estimate the number of boys taller than 155cm
ii) Draw the histogram
iii) Calculate the mean height
iv) How many MORE girls than boys exceed 160cm
v) Calculate the median of the boys (I think)

7i)A) probability of 5 using internet
B) probability of at least 5
C) Calculate the expectation of X~B(20, 0.35)
ii) Test the hypothesis of whether 10 people using internet access out of a sample of 20. Something like that, where X~B(20, 0.35)
iii) Given a sample of 90 people using the internet out of 200, test the hypothesis at the same significance level

(edited 11 years ago)

Reply 150

username1107833

2

UNOFFICIAL MARK SCHEME - S1 January 2013

1)i) Positive skew
ii) mean = 5.064, sd = 1.32

2i) 0.028
ii) Neither L nor W: 0.92
P(L) on its own = 0.01
P(LnW) = 0.028
P(W) on its own = 0.042
iii) Not independant as P(L) * P(W) does not equal 0.028

3i) 11C3
ii) P( exactly 2 + exactly 3) = 4/11 + 2/33 = 14/33 (0.42 recurring)

4)i) 5/6 * 5/6 * 1/6
ii) Very long but its (1/6) + (5/6 * 1/6) + (5/6^2 * 1/6) + (5/6^3 * 1/6) and then up until 5/6 gets to the power of 9 and you should get like 0.83849

5)i) k = 0.02, then forgot the actual probabilities but they went 3k, 8k, 15k, 24k, so just work it out.
5ii) E(X) = 0.42 Var(X) = 0.84 i think

6)i) sorry can't remember
ii) Draw the histogram
iii) sorry can't remember
iv) There were 5 MORE girls than boys with a height of 160cm
v) Median height of boys: 144.4 (this is someone else's result but i believe it is wrong as the median was the 50th value not 50.5 as since it is grouped data you would do n/2 and not n+1/2)

7i)A) 0.1536
B) P(X>=5) = 1 - P(X<=4) so = 0.2485 (for X~B(10,0.35))
C) 10*0.35 = 3.5 expected
ii) Insufficient evidence to accept H1, so accept H0 and reject H1
iii) Significant evidence to accept H1

Reply 151

Rubyturner94

Original post by theCreator

UNOFFICIAL MARK SCHEME - S1 January 2013

1)i) Positive skew
ii) mean = 5.064, sd = 1.32

2i) 0.028
ii) Neither L nor W: 0.92
P(L) on its own = 0.01
P(LnW) = 0.028
P(W) on its own = 0.042
iii) Not independant as P(L) * P(W) does not equal 0.028

3i) 11C3
ii) P( exactly 2 + exactly 3) = 4/11 + 2/33 = 14/33 (0.42 recurring)

4)i) 5/6 * 5/6 * 1/6
ii) Very long but its (1/6) + (5/6 * 1/6) + (5/6^2 * 1/6) + (5/6^3 * 1/6) and then up until 5/6 gets to the power of 9 and you should get like 0.83849

5)i) k = 0.02, then forgot the actual probabilities but they went 3k, 8k, 15k, 24k, so just work it out.
5ii) E(X) = 0.42 Var(X) = 0.84 i think

6)i) sorry can't remember
ii) Draw the histogram
iii) sorry can't remember
iv) There were 5 MORE girls than boys with a height of 160cm
v) Median height of boys: 144.4 (this is someone else's result but i believe it is wrong as the median was the 50th value not 50.5 as since it is grouped data you would do n/2 and not n+1/2)

7i)A) 0.1536
B) P(X>=5) = 1 - P(X<=4) so = 0.2485 (for X~B(10,0.35))
C) 10*0.35 = 3.5 expected
ii) Insufficient evidence to accept H1, so accept H0 and reject H1
iii) Significant evidence to accept H1

im sure there was a question on outliers

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Reply 152

Rapidzx

Original post by Rubyturner94

im sure there was a question on outliers

Posted from TSR Mobile

There was.
The only outlier was 8.1 I believe.

Reply 153

username1107833

2

Original post by Rubyturner94

im sure there was a question on outliers

Posted from TSR Mobile

Sorry guys I do apologize, question 1)iii) was on outliers and yeah 8.1 was the outlier

Reply 154

lapuntru

Not seen the paper but think Q5ii s/be E(X)=4.2.
What was the significance level in Q7, please?

Reply 155

Enraged Papaya

5

It was 5% but a two tailed test, so 2.5 at either end.

Reply 156

Rapidzx

Original post by lapuntru

Not seen the paper but think Q5ii s/be E(X)=4.2.
What was the significance level in Q7, please?

5% i think

Reply 157

Rubyturner94

guys do you remembed what you got for the mean for girls? :/

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Reply 158

Enraged Papaya

5

Second last Q:
The probability that someone in a cafe is using the internet is 0.35. However now they believe it is no longer 0.35 because blah blah blah... A sample size of 20 customers is carried out, and 10 are found to be using the internet. Test at 5% significance

p represents probability of a randomly selected customer using the internet

H0: p = 0.35
H1: p is not 0.35
Significance level = 5%
X~B(20, 0.35)

P(X less than or equal to 3) = 0.0444 = 4.44% > 2.5%
P(X less than or equal to 2) = 0.0121 = 1.21% < 2.5%

P(X greater than or equal to 10) = 1 - P(X less than or equal to 9) = 1 - 0.8782 = 0.1218 = 12.18% > 2.5%
P(X greater than or equal to 11) = 1 - P(X less than or equal to 10) = 1 - 0.9468 = 0.0532 = 5.32% > 2.5%
P(X greater than or equal to 12) = 1 - P(X less than or equal to 11) = 1 - 0.9804 = 0.0196 = 1.96% < 2.5%

Critical region = X is less than or equal to 2 AND X is greater than or equal to 12
Hence X = 10 does not lie in the critical region
=> There is not enough evidence to reject H0 in favour of H1
There is not enough evidence to suggest the probability of a randomly selected customer using the internet is different to 0.35

Last Q: Another test is conducted except sample size is now 200, and 90 are found using the internet. P(X greater than or equal to 90) = 0.0022. Carry out the test

P(X greater than or equal to 90) = 0.0022 = 0.22% < 2.5%

Hence X = 90 lies in the critical region
=> There is enough evidence to reject H0 in favour of H1
There is enough evidence to suggest the probability of a randomly selected customer using the internet is different to 0.35