Introduction to Probability and Statistics - Questions with Solutions | STAT 20, Assignments of Statistics

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Statistics 20: Extra Problems V Solutions

Estimating Proportions

1. In a random sample of 200 eligible voters interviewed in a large city, 126 objected to the use of public funds for the construction of a new professional football stadium.

(a) Construct a 95% confidence interval for the proportion of city residents who approve the use of public funds.

Let p be the true proportion of eligible voters who approve of the use of public funds. We wish to estimate p with the sample’s empirical proportion ˆp = 126/200 = 0.63 of approving voters. Proportions are inherently means of n independent, identically distributed random variables, and hence ˆp is approximately Normal under the Central Limit Theorem. Let’s figure out its parameters:

E[ˆp] = p ≈ pˆ, the expected approval of any one voter, and

SD(ˆp) =

√ p(1−p) n ≈

√ pˆ(1−ˆp) n.

The above approximations fill in ˆp as an estimate for p. We then have enough information to construct the 95% confidence interval:

pˆ ± z 0. 95

√ ˆp(1−pˆ) n = 0.^63 ±^1.^96

√ 0 .63(1− 0 .63) 200 = (0.^563 ,^0 .697).

(b) Construct a 95% confidence interval for the proportion of city residents who object to the use of public funds.

Now let q represent the true proportion who object to the use of public funds. In this case, the sample proportion is ˆq = 200200 −^126 = 0.37. Then, like the previous problem, the 95% confidence interval for q is given by:

qˆ ± z 0. 95

√ qˆ(1−ˆq) n = 0.^37 ±^1.^96

√ 0 .37(1− 0 .37) 200 = (0.^303 ,^0 .437).

(c)Construct a 99% confidence interval for the proportion of city residents who approve the use of public funds.

Once again our quantity of interest is p, the proportion who approve. We can use the form of the confidence interval in (a) by substituting z 0. 99 in place of z 0. 95.

pˆ ± z 0. 99

√ ˆp(1−pˆ) n = 0.^63 ±^2.^58

√ 0 .63(1− 0 .63) 200 = (0.^542 ,^0 .718).

2. In a random sample of 120 cheerleaders, 54 had suffered moderate to severe damage to their voices. With 90% confidence, what can we assert about the maximum error if the sample proportion 54/120 = 0.45 is used as an estimate of the true proportion of cheerleaders who are afflicted in this way?

If the true proportion p were 0.45, then for any representative sample of 120 cheerleaders, the absolute difference between the value of ˆp obtained in the sample and the true value of p will be less than the margin of error with probability 0.9. Then we just need to calculate the margin of error:

M OE = z 0. 9

√ pˆ(1−ˆp) n = 1.^65

√ 0 .45(1− 0 .45) 120 = 0.0749.

3. A random sample of 300 shoppers at a large supermarket includes 234 who regularly use dis- count coupons. Construct a 98% confidence interval for the probability that any one shopper at that supermarket, randomly chosen for an interview, will confirm that he or she regularly uses discount coupons.

We are interested in p, the probability that a randomly selected shopper uses coupons. With pˆ = 234/300 = 0.78, a 98% confidence interval for p is:

pˆ ± z 0. 98

√ ˆp(1−pˆ) n = 0.^78 ±^2.^33

√ 0 .78(1− 0 .78) 300 = (0.^724 ,^0 .836).