Lecture 21- CI proportion

Question 1

How to get a proportion?

Answer 1

number who have positive test/ total number in sample

Question 2

What is the broad term for how we find out if our estimate of the proportion from our sample truly reflects the population proportion?

Answer 2

Calculate a confidence interval

Question 3

What properties must a data set have for it to be described by a binomial distribution?

Answer 3

• 2 outcomes
• N samples
• Independent
• Same probability of success (pi)

Question 4

What does central limit theory state?

Answer 4

The Central Limit Theorem says that as long as the samples
are ‘large enough’, the sampling distribution for any parameter (means, proportions) will
follow a normal distribution.

Question 5

In what cases does the shape of a binomial appear normal?

Answer 5

• large n

- π close to 0.5

Question 6

What do we use in order to determine if a normal distribution is a good enough approximation of a binomial?

Answer 6

npi +/- 3 x standard deviations

standard deviation is given by the square root of npi(1-pi)

gives two values between 0 and n. If one of values is less than 0 the distribution can not be represented by a normal distribution (will get cut off).

Question 7

Which binomial curve could be approximated by a normal?

1) n = 20 , π = 0.5
2) n = 20 , π = 0.1

Answer 7

1) (3.29, 16.71) yes
2) (−2.02, 6.02) no

look to slides for working

Question 8

How are the mean and standard deviation for the proportion worked out?

Answer 8

mean= pi

sd= square root of (pi(1-pi)/n)

Question 9

what is the general formular for calculating a confidence interval?

Answer 9

mean +/- multiplier x standard error

Question 10

What are the multipliers for calculating confidence intervals of proportions?

Answer 10

normal (Z) multipliers 1.96 (95%) and 2.58 (99%) are used for
calculating confidence intervals for proportions because we are using
the normal sampling distribution. If the CLT can’t be assumed to
hold (so we can’t assume a normal sampling distribution) we need a
different method that is not covered in this course.

Question 11

In a random sample of Aucklanders taken in 1996, 173 of the 500
respondents supported aerial spraying to eradicate tussock moth.

Estimate the proportion (π) of Aucklanders who support aerial spraying to eradicate tussock moth…

Include the margin of error way of interpreting the confidence interval….

Answer 11

0.305 < π < 0.38

‘The survey indicates that 34.6% of the population support the spraying, with a margin of error of 4.1%’.

working on slide 411

Question 12

In the moth example what else should be considered?

Answer 12

-Bias due to non-response should also be considered in our

interpretation of an estimate. i.e. maybe only people who are really care about the issue are responding.