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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  1. 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)

  1. Its distribution is normal
  2. Its mean is equal to
  3. Its SE is equal to
  4. 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]

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