ST 361 CI for Prob

ST 361 Estimation --- Interval Estimation for (§7.3)

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Topics:

I. Interval estimation: confidence interval

II. (Two-sided) Confidence interval for estimating population mean (§7.2, 7.4)

(a) When the population SD is known: use Z distribution (§7.2)

(b) When the population SD is NOT known: use t distribution (§7.4)

III. (Two-sided) confidence interval for estimating population proportion (§7.3)

IV. Two-sided confidence interval for estimating population mean difference (§7.5)

(a) when the population SD’s are known

(b) when the population SD’s are NOT unknown

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Confidence Interval for , the population proportion (or the success probability)

The natural statistic for estimating is the sample proportion of success p.
Recall that , where is total number of successes out of n trials.
Also recall that if n is large (i.e., ), the sampling distribution of can be approximated by

The CI for with confidence level () takes the following form:

Since in the above interval is unknown, replace it by the sample proportion p and actually use the following CI:

Note that here we still use a Z value from the Standard normal distribution instead of a t-distribution, but this time it requires a more stringent criterion for “large n”.

In conclusion,

Still use Z-distributionto find the critical valuez*

Need np > 10 and n(1-p)>10 for to be normally distributed.

Confidence interval for : (Assume n>30, np > 10 and n(1-p)>10)

The Confidence interval for is

Ex1. For a certain disease, there is a generic drug and a brand name. We want to estimate = P(the brand name is more preferred by a patient). So 100 patients were asked, and 20 of them preferred the brand name.

(a)What is your estimate of ?What is the estimated SE of your estimator?

Answer: A natural estimator of is the sample proportion p = 20/100=0.2. The estimated standard error (SE) of p is

(b)Find the 95% CI for

Answer: Here n = 100>30, np = 100*0.2=20>10, n(1-p)=100*0.8=80>10. So we can use the CI formula for given above.

So a 95% CI for is

=[0.2 - 1.96*0.04, 0.2+1.96*0.04]=[0.122,0.278].

(c)What is the width of your CI?

Answer: The width of the CI is 0.278 – 0.122=0.156.

(d)If we want to make the width of our 95%CI for be 0.1, how many patients should be sampled?

Answer: The width of our 95% CI is

If we use p=0.2, then

If we treat p unknown (it is unknown before the sampling), we can use p=0.5 in the formula and solve

Ex2. Suppose that the proportion of the left-handed students at a certain university is. A random sample of 200 students was collected and found that 40 out of the 200 students are left-handed.

(a)Use an unbiased estimator to compute a point estimate of .

Answer: an unbiased estimator of is the sample proportion p =40/200=0.2

(b)What is the approximate distribution of your estimator in (a)? Why?

Answer: It can be well approximated by a normal distribution since n=200>30, np=40>10 and n(1-p)=160>10.

(c)What is the estimated standard error of your estimator?

(Hint: if the population proportion is known, then )

Answer: The estimated standard error of p is

(d)Your estimator in (a) is unbiased because (circle one)

Its distribution is normal
Its mean is equal to
Its SE is equal to
It is based on a sample with size greater than 30

(e)What is a 95% confidence interval for ?

Answer: since n=200>30, np=40>10 and n(1-p)=160>10, we can use the formula given before:

[0.145, 0.255]