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can any plz do question 3 of june 2001 and explain. i can write the question if any wants the script

http://www.ajimal.com/download/alevel/stats/s2sol/jun01.pdf

this helped me, although i havent been taught to use the critical regions unless im told to.

this helped me, although i havent been taught to use the critical regions unless im told to.

In a sack containing a large number of beads 1/4 are coloured gold and the remainder are of different colours. A group of children use some of the beads in a craft lesson and do not replace them. Afterwards the teacher wishes to know whether or not the proportion of gold beads left in the sack has changed. He selects a random sample of 20 beads and finds that 2 of them are coloured gold.

Stating your hypotheses clearly test, at the 10% level of significance, whether or not there is evidence that the proportion of gold beads has changed.

H0: The proportion of gold beads is 1/4.

H1: The proportion of gold beads is not 1/4.

Let X be the number of gold beads in the sample of 20. Then under H0 X~Bin(20, 1/4), so X~Normal(5, 15/4) approximately. Since Phi(1.65) = 0.95, we should reject H0 if X is more than 1.65 standard deviations from 5, ie, if X is more than 1.65*sqrt(15/4) = 3.19521 from 5, ie, if X is less than 5 - 3.19521 or more than 5 + 3.19521.

The observed value of X is 2, so we do not reject H0.

There is insufficient evidence at the 10% to conclude that the proportion of gold beads has changed.

Stating your hypotheses clearly test, at the 10% level of significance, whether or not there is evidence that the proportion of gold beads has changed.

H0: The proportion of gold beads is 1/4.

H1: The proportion of gold beads is not 1/4.

Let X be the number of gold beads in the sample of 20. Then under H0 X~Bin(20, 1/4), so X~Normal(5, 15/4) approximately. Since Phi(1.65) = 0.95, we should reject H0 if X is more than 1.65 standard deviations from 5, ie, if X is more than 1.65*sqrt(15/4) = 3.19521 from 5, ie, if X is less than 5 - 3.19521 or more than 5 + 3.19521.

The observed value of X is 2, so we do not reject H0.

There is insufficient evidence at the 10% to conclude that the proportion of gold beads has changed.

It would probably help if you mentioned which board you're on. I doubt it's OCR, because the question is "A fair sided dice is thrown 540 times. Use a suitable approximation to calculate the probability that at least 80 sixes are obtained"

The answer to that question would be:

X~B(540, 1/6);

using the normal to approximate the binomial (np > 5, nq > 5)

Y~N(90, 75);

P(X >= 80) = 1-P(X<80)

= 1-P(Y<79.5)

= 1-P(Z<(79.5-90)/sqrt(75))

= P(Z<1.21) = 0.887

The chances are that's not the question you were talking about, but it doesn't hurt for me to get practice. lol.

Yeh, you're probably on Edexcel, that question seems more confusing than this one, although I'm happy to post my solution to this question here for practice, as I said. And hey, If I'm wrong, someone correct me

The answer to that question would be:

X~B(540, 1/6);

using the normal to approximate the binomial (np > 5, nq > 5)

Y~N(90, 75);

P(X >= 80) = 1-P(X<80)

= 1-P(Y<79.5)

= 1-P(Z<(79.5-90)/sqrt(75))

= P(Z<1.21) = 0.887

The chances are that's not the question you were talking about, but it doesn't hurt for me to get practice. lol.

Yeh, you're probably on Edexcel, that question seems more confusing than this one, although I'm happy to post my solution to this question here for practice, as I said. And hey, If I'm wrong, someone correct me

zxcvbnm

can any plz do question 3 of june 2001 and explain. i can write the question if any wants the script

The distribution is binomial X-Bin (20.0.25) X-RV "The number of gold beads out of 20

Ho:=0.25

Hi:=0.25(The equal sign should have a line dashed through it because the probability has changed therefore it could have increased or decreased.

Next you have to find the tails of the critical region. As the significence level is 10% each tail should be close to 5%.0.005

p(X<2)=0.0913

p(X<1)=0.0243 The critical region lies between 1 and 2

Next you find the tails on the other side

P(X>8)=1-(P<7)

1-0.898=0.1018

p(x>7)=1-(PX<6)

=1-0.7858=0.2142

P(X>6)=1-(Px<5)

1-0.6712=0.3828

Critical regionlies between 6 and 7.

Therefore the Test value of 2 does not lie it the critical region. Do not reject H0 As there is insignificent evidence that the number of beads has changed.

zxcvbnm

thanks every one for the help .. but one more very little question

X-B(20,0.4) find P(5<x<15) ??

X-B(20,0.4) find P(5<x<15) ??

Use a normal aproximation so find np and npq.

Then use the continuity correction so it becomes (5.5<x<15.5)

Then standardise the data by taking away the mean (np), and dividing by the variance square rooted (root npq)

i'll try and explain why...although it mite turn into waffle...

Basically you want the probability x is greater than 5 but less than 15. This means the values which need to be included are 6,7,8... all the way upto 14.

So the way to do this is either manually i.e. stick in all the values into a calculator and take about 20mins to complete, or just use the table.

Basically you want the probability x is greater than 5 but less than 15. This means the values which need to be included are 6,7,8... all the way upto 14.

So the way to do this is either manually i.e. stick in all the values into a calculator and take about 20mins to complete, or just use the table.

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