http://www.xtremepapers.com/Edexcel/...%202006-01.pdf
Look at question 7 on the last page
so I know that for (ai)
H0: p = 0.2
H1: p <> 0.2 (does not equal)
Now when I carry out the test, if it was p>0.2 I would say
P(X>=9).... and then work out the answer
equall if it was p<0.2 I would say P(X<=9)
But given that it is p <>0.2 how do I carry out the test. Please can you explain as well!
For (b) I am again faced with the same issue
The distribution is
X-N(20,16)
H1 again is p<>0.2
so for my normal approx to I use
P(x>=18) or p(x<=18)
I will continuty correct once I understand the principle
thanks for the help!
http://www.edexcel.com/migrationdocu...anuary2006.doc
here is another website
The null hypothesis is that , and the alternative is that .
We reject the null hypothesis if the probability of the observed result or more extreme conditioned on the null-hypothesis being true is "small".
That is if:
Now since under the null hypothesis the distribution of the number of Deano readers in a sample of size has a binomial distribution we can work out the probability on the left (and it is ), which is less than so we reject the null hypothesis.
However exact binomial calculations of the size required by this problem are beyond what is reasonable to expect in an exam, so you will either have to use an approximation or tables of critical values provided for this purpose.
You will note that the alternative hypothesis does not figure in this test.
Part (b) works in essentially the same manner except the sample size is now 100, and large(ish) sample approximations are probably appropriate.
CB
well what I was asking was how do I decide if I should use P(X<=9) or P(X>=9)
Is it becasue the E(X) - mean - is 4.
9 is above the mean so I calcuate P(X>=9) NOT P(X<=9)
for normal, mean is 20, value is 18 SO here is say P(X<=18) NOT P(X>=18)
thanks