AgentRed
New member
- Joined
- Oct 11, 2013
- Messages
- 1
Hi all,
I need help with my statistics assignment which is due in 2 days :/ I have been busy with lots of other assignments and work so any help would be greatly appreciated.
This is question 1 and refers to population probability and confidence levels.
A telephone survey asked 880 drivers to answer the following question:
“Recalling the last ten traffic lights you drove through, how many of them were red when you entered the intersections?”
Of the 880 respondents, 161 admitted that at least one light had been red. A random sample was completed to obtain the participants for this survey.
(a) (6 marks) If there is no reason to believe that one driver’s response influences another, estimate, with 95% confidence, the population proportion of drivers who ran one or more of the last ten red lights.
My answer so far:
x =161
n = 880
P^(hat) =161 / 880
=0.18295
≈0.183
q = 1-p
=0.817
Sd = sqrt (p*q / n)
=sqrt (0.183*0.817*880)
= 1.303%
(b) (4 marks) Check the procedure you used in part (a) is appropriate by checking all the necessary conditions, assumptions and/or ‘rules of thumb’.
I'm not too sure with this, but the condiitons and assumptions are they not where we :
use a multiplier of 1.0 for about 67 percent confidence,
a multiplier of 2.0 for about 95 percent confidence,
and a multiplier of 3.0 for about 99 percent confidence
or something?
(c) (2 marks) If the company wished for the confidence level to be increased to 99%, keeping the same sample size, what could happen to the width of the confidence interval (show detail)?
I am not too sure here
(d) (2 marks) Name two issues that you can see with this type of survey.
I need help with my statistics assignment which is due in 2 days :/ I have been busy with lots of other assignments and work so any help would be greatly appreciated.
This is question 1 and refers to population probability and confidence levels.
A telephone survey asked 880 drivers to answer the following question:
“Recalling the last ten traffic lights you drove through, how many of them were red when you entered the intersections?”
Of the 880 respondents, 161 admitted that at least one light had been red. A random sample was completed to obtain the participants for this survey.
(a) (6 marks) If there is no reason to believe that one driver’s response influences another, estimate, with 95% confidence, the population proportion of drivers who ran one or more of the last ten red lights.
My answer so far:
x =161
n = 880
P^(hat) =161 / 880
=0.18295
≈0.183
q = 1-p
=0.817
Sd = sqrt (p*q / n)
=sqrt (0.183*0.817*880)
= 1.303%
(b) (4 marks) Check the procedure you used in part (a) is appropriate by checking all the necessary conditions, assumptions and/or ‘rules of thumb’.
I'm not too sure with this, but the condiitons and assumptions are they not where we :
use a multiplier of 1.0 for about 67 percent confidence,
a multiplier of 2.0 for about 95 percent confidence,
and a multiplier of 3.0 for about 99 percent confidence
or something?
(c) (2 marks) If the company wished for the confidence level to be increased to 99%, keeping the same sample size, what could happen to the width of the confidence interval (show detail)?
I am not too sure here
(d) (2 marks) Name two issues that you can see with this type of survey.