Lecture 7: Confidence Intervals in Statistics, Lecture notes of Statistics

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Lecture 7: Confidence Intervals

Outline

Lecture 7: Confidence Intervals Example – Birth Weight Theory – Confidence Interval Confidence Interval for μ Example – Paint Primer Thickness Example – Concrete Tensile Strength Example – Clinton vs Trump Polls Approximation CI for proportion Conservative CI for proportion Justifying the conservative CI for proportion Formulae for test statistic & confidence intervals

Example – Birth Weight

mean(x)

[1] 3.

(i) An estimate for μ is x¯ = 3. 179 (ii) X¯ ∼ N (μ, 0. 5252 /8) → Z =

X¯ − μ

1. 525 /

∼ N (0, 1).

0. 95 = P (− 1. 96 < Z < 1 .96)

= P

X¯ − μ

1. 525 /

= P

X¯ − 1. 96 · 0. 525 /√ 8 < μ < X¯ + 1. 96 · 0. 525 /√ 8

Now, 3. 179 ± 1. 96 · 0. 525 /

8 , giving a 95% confidence interval of (2. 82 , 3 .54).

Revision – Probability

What is the green area?

qnorm(0.975)

[1] 1.

Distribution calculator

−1.96 0.00 1.96 5/

Confidence Interval for μ

I (^) If the sampling population is N (μ, σ^2 ) with known σ, 100(1 − α)% confidence interval for μ is ¯x ± z 1 −α/ 2 · √σn

where z 1 −α/ 2 = z∗^ such that P (Z < z∗) = 1 − α 2. I (^) In practice σ is unknown so we use x ¯ ± t 1 −α/ 2 ,n− 1 · √sn

where t 1 −α/ 2 ,n− 1 = t∗^ such that P (tn− 1 < t∗) = 1 − α 2. I (^) z 1 −α/ 2 and t 1 −α/ 2 ,n− 1 are called critical values. I √σn and √sn are standard errors of the estimate of μ. I (^) Remember tn− 1 → N (0, 1) as n → ∞!

Confidence Interval for μ

I (^) If the sampling population is N (μ, σ^2 ) with known σ, 100(1 − α)% confidence interval for μ is ¯x ± z 1 −α/ 2 · √σn

where z 1 −α/ 2 = z∗^ such that P (Z < z∗) = 1 − α 2. I (^) In practice σ is unknown so we use x ¯ ± t 1 −α/ 2 ,n− 1 · √sn

where t 1 −α/ 2 ,n− 1 = t∗^ such that P (tn− 1 < t∗) = 1 − α 2. I (^) z 1 −α/ 2 and t 1 −α/ 2 ,n− 1 are called critical values. I √σn and √sn are standard errors of the estimate of μ. I (^) Remember tn− 1 → N (0, 1) as n → ∞! I (^) What if the distribution of the sampling population is unknown?

Example – Paint Primer Thickness

Assume that paint primer thickness can be modelled by X ∼ N (μ, σ^2 ). In an ongoing process of quality control in an industrial system, the following first sample of values was obtained:

x = c(1.3, 1.1, 1.2, 1.25, 1.05, 0.95, 1.1, 1.16, 1.37, 0.98)

(i) What is a 95% confidence interval for the primer thickness? (ii) What is a 99% confidence interval for the primer thickness? (iii) The company advertises that the primer thickness is 1.25. What would you conclude?

Theory – Size of confidence interval

I (^) What happens if we increase the sample size n?

As we increase n, the confidence interval gets narrower, and so x¯ is a better estimate for the long term estimate of μ. I (^) What happens if we increase the confidence level 1 − α?

As we increase the confidence level, the confidence interval gets wider.

Example – Paint Primer Thickness

(i) A 95% CI for μ is: ¯x ± t 0. 975 , 9 · s √ n which is 1. 146 ± 2. 262 · 0 √.^13610 , giving (1. 05 , 1 .24). qt(0.975, 9)

[1] 2.

(ii) A 99% CI for μ is: ¯x ± t 0. 995 , 9 · s √ n which is 1. 146 ± 3. 250 · 0 √.^13610 , giving (1. 01 , 1 .29). qt(0.995, 9)

[1] 3.

Theory – Relationship with hypothesis testing

I (^) For a false positive rate of α, a set of possible hypothesis {H 0 : μ = μ 0 } which will be retained if μ 0 is within the 100(1 − α)% confidence interval. I (^) Note: this is a two-sided hypothesis test. I (^) Although we don’t explicitly mention, we implicitly assume confidence intervals are two-sided. Example – Paint Primer Thickness (iii) The CI does not contain H 0 : μ = 1. 25 hence the data provide evidence against the company’s advertising if we use false positive rate of 0.05.

Example – Concrete Tensile Strength

(i) Find a 95% CI for the mean tensile strength of small specimens, assuming that the strengths can be modelled by N (μ, 10002 ). (ii) Find a 90% CI for the mean difference in tensile strengths, assuming that the differences can be modelled by N (μ, σ^2 ).

Example – Concrete Tensile Strength

small = c(4404, 4326, 3788, 3475, 3418, 2262, 7415, 6993) mean(small)

[1] 4510.

sd(small)

[1] 1792.

(i) Assuming that the strengths can be modelled by N (μ, 10002 ), a 95% CI for the mean tensile strength of small specimens is: x ¯ ± 1. 96 · σ √ n which is

4510. 125 ± 1. 96 · 1000 /

which gives (3817, 5203).

Example – Clinton vs Trump Polls

From a recent report on the USA Election 2016^ , we find the following quotes: “The tighter race in Florida showed Clinton edging Trump 45% to 42% among likely voters, with Johnson at 5% and Stein at 3%. That three-point lead was within the poll’s margin of error.”

Image: Wikimedia Commons CC BY-SA 4.

“The NBC/WSJ/Marist poll Florida poll surveyed 700 likely voters between October 3-5 with a margin of error of plus or minus 3.7 percentage points.”

“The Pennsylvania poll surveyed 709 likely voters between October 3-6 with a margin of error of plus or minus 3.7 percentage points.”

A random survey of 2000 voters found that 1165 were going to vote for Hilary Clinton.

(i) Find a 95 % CI for the proportion of voters p that will vote for Hillary. (ii) What is the margin of error?

(iii) What is the smallest sample size needed to give a 95 % CI for p with width at most ± 0. 03?