Statistics Question

Edward01 · 26 Aug 2009 at 19:50

A friend wanted some help with her stats questions and it's been a while since I did statistics but I managed the first 3 happily enough, but I'm a bit confused with this one.

4. A survey of the car park at a Farmers Market indicates 40 out of 100 vehicles are four-wheel drive. Construct a 95% confidence interval for percentage of vehicles in the UK which are four-wheel drive. Do you have any reservations about this survey?

Does it want her to go and research figures for 4 x 4 usage in the UK from external sources? or does it want us to provide a hypothetical answer as we don't have enough variables to provide an accurate one? e.g. 40 – (1.96(σ/10)) ≤ µ ≤ 40 + (1.96(σ/10)) Or am I just being thick and missing something?

Any ideas?

Tombstone · 26 Aug 2009 at 19:54

Ironically enough, I did A-Level statistics many decades ago. To be honest though I dont remember a damn thing about it so I'm afraid I'm not of much help on this.

In fact, come to think of it...it was pretty useless of me to even post to say that...er...I'll just get my coat

monkentage · 26 Aug 2009 at 20:01

Edward01 said:
A friend wanted some help with her stats questions and it's been a while since I did statistics but I managed the first 3 happily enough, but I'm a bit confused with this one.

4. A survey of the car park at a Farmers Market indicates 40 out of 100 vehicles are four-wheel drive. Construct a 95% confidence interval for percentage of vehicles in the UK which are four-wheel drive. Do you have any reservations about this survey?

Does it want her to go and research figures for 4 x 4 usage in the UK from external sources? or does it want us to provide a hypothetical answer as we don't have enough variables to provide an accurate one? e.g. 40 – (1.96(σ/10)) ≤ µ ≤ 40 + (1.96(σ/10)) Or am I just being thick and missing something?

Any ideas?

Isn't there a formula for estimating σ, and when its estimated a different distribution table is used?

mattheman · 26 Aug 2009 at 22:42

this is cofidence interval with an unknown σ

x bar +/_ T n-1 * s/( sq root of n)

you need to work out the Degree of freedom.

Estimation for proportion

n=100
p= 40/100= AP
1-AP= bp

P+/_ z * the square root of this AP*BP/N

mattheman · 26 Aug 2009 at 22:49

yes EDIT

Alex UK · 26 Aug 2009 at 22:51

mattheman said:
this is cofidence interval with an unknown σ

x bar +/_ T n-1 * s/( sq root of n)

you need to work out the Degree of freedom.
estimation for proportion

n=100
p= 40/100= AP
1-AP= bp

P+/_ z * the square root of this AP*BP/N

That's what I was going to say.

mattheman · 26 Aug 2009 at 23:15

Alex UK said:
That's what I was going to say.

would you like to say it then.

Edward01 · 27 Aug 2009 at 00:01

Does this look right then??

N = 100
AP = .40
α = .05
1.96 = Critical Value

Therefore:

.4 - 1.96 SqRt ((.4 * .6) / 100) less or equal p less or equal .4 + 1.96 SqRt ((.4 * .6) / 100)

0.304 ls or eq P ls or eq 0.496

National 4 x 4 usage is between 30% and 49%

mattheman · 27 Aug 2009 at 02:02

Edward01 said:
Does this look right then??

N = 100
AP = .40
α = .05
1.96 = Critical Value

Therefore:

.4 - 1.96 ((.4 * .6) / 100) less or equal p less or equal .4 + 1.96 SqRt ((.4 * .6) / 100)

0.304 ls or eq P ls or eq 0.496

National 4 x 4 usage is between 30% and 49%

.4 -/+ 1.96 * square root this ------> (.4 * .6 / 100)
P _/+ z * square root this ------> p(1-P)/N

if thats what you got then its correct.