The Student Room Group

Advanced statistics help :)

Hi, please could someone help me on this question, thanks :smile:

Suppose we are interested in whether the Conservatives have a majority with
a null hypothesis that the voters are equally split between Conservative and
the rest.

(a) Write down the null and alternative hypotheses in terms of Pi

(b) Using the sample data (2 = Conservative, 1 = Other)
use a classical hypothesis test procedure to test this null
hypothesis against the alternative.

n=765 (total number of votes)
votes for 2 = 407
votes for 1 = 358

2

Suppose that you wish to estimate Pi using P to within 3% with a probability exceeding 95%; i.e., you require
Pr (P-Pi< 0.03) > 0.95.
What is the smallest sample size, n; you should take in order to be sure that
this requirement is met?
(edited 12 years ago)
Reply 1
(a) have you got the hypotheses? I guess you're using Pi instead of p for the proportion of voters in the population who vote Tory?
Reply 2

Original post by vc94
(a) have you got the hypotheses? I guess you're using Pi instead of p for the proportion of voters in the population who vote Tory?


yes thats right, i think p=pi
but i am not too sure :/ stats is not my forte
Reply 3
H0: p=0.5, H1: p>0.5 (since you're looking for a majority)

p(hat)= 407/765 is your sample statistic.
Since n is large the central limit theorem says that the distribution of proportions is approx normal.

You need to calculate z=((phat) - p)/sqrt(pq/n)

where p=0.5 under H0, q=1-p and n=765. It's a one tail proportion hypothesis test.
Compare your z value with a critical value...have you done this before?
Reply 4

Original post by vc94
H0: p=0.5, H1: p&gt;0.5 (since you're looking for a majority)

p(hat)= 407/765 is your sample statistic.
Since n is large the central limit theorem says that the distribution of proportions is approx normal.

You need to calculate z=((phat) - p)/sqrt(pq/n)

where p=0.5 under H0, q=1-p and n=765. It's a one tail proportion hypothesis test.
Compare your z value with a critical value...have you done this before?


hey, we usually use the following formula:

z=( mean bar x-null probability)/SE[Xbar]

is that the same thing? and i havent done this before...just done the basics :/
Reply 5
You use the sampling distribution of Xbar if your test statistic is a sample mean.
Here you have a sample proportion, so you want the distribution of sample proportions, which because n is large, is approx normal, with mean=p and stdDev= pq/n.

So you haven't done this yet?
Reply 6
hey thanks :smile: and no i havent really done a question like this one before :/
Reply 7

Original post by vc94
You use the sampling distribution of Xbar if your test statistic is a sample mean.
Here you have a sample proportion, so you want the distribution of sample proportions, which because n is large, is approx normal, with mean=p and stdDev= pq/n.

So you haven't done this yet?



i get z=6.56... is that correct? :smile:
Reply 8
z= ((407/765) - 0.5) / sqrt(0.5*(1-0.5)/765)
Reply 9

Original post by vc94
z= ((407/765) - 0.5) / sqrt(0.5*(1-0.5)/765)


thank you :smile: so do i just find the probabilty of z>1.7716 from the normal tables now?
Reply 10
Yes, then compare with your sig level, 5% I guess?
Reply 11

Original post by vc94
Yes, then compare with your sig level, 5% I guess?


Aah yes...i get 0.9616>0.95 so accept null hypothesis :smile:
thank you :smile:
Reply 12
P(Z>1.77) = 1 - P(Z<1.77)
=1-0.9616 =0.0384 using tables
< 0.05, so reject H0!
Reply 13
aah, i see :smile: thanks

Quick Reply