16 - Confidence Intervals

Question 1

What is a confidence interval?

Answer 1

1. range of feasible values for an unknown population parameter
   
   • µ (pop mean), p (pop proportion)
2. statement conveying the confidence that the range of feasible values really does include the unknown population value

Question 2

Because proportions are averages, the CLT implies…

Answer 2

a normal model for the sampling distribution of p^ if the sample size n is large enough:

p^ ~ N[p, p(1-p)/n]

Question 3

SE (p^)

Answer 3

sqr[p(1-p)/n]

Question 4

If we use the percentile of the normal distribution, z_0.025, then

Answer 4

z_0.025 = 1.9, and

P(-1.96 SE(p^) <= p^ - p <= 1.96 SE (p^)) = 0.95

p^ lies within 1.96 standard errors of p in 95% of samples

Question 5

se(p^) =

Answer 5

sqr[p^(1-p^)/n]

Question 6

CI for p

The 100(1-a)% z-interval for p is the interval from…

Answer 6

p^-z_a/2sqr[p^(1-p^)/2] to p^+z_a/2sqr[p^(1-p^)/2]

Question 7

Xbar =

xbar =

Answer 7

mean of a randomly chosen sample

mean of the observed sample

Question 8

SE(Xbar) =

se(Xbar) =

Answer 8

σ/sqr(n)

s/sqr(n)

Question 9

Student’s t-distribution

Answer 9

very similar to the normal distribution, but the t has fatter tails

incorporates excess variability

as sample size n gets larger, the t-distribution convergs to the standard normal distribution

Question 10

defining the t-distribution

Answer 10

any normal random variable, its Z-score:

(Xbar - µ)/(σ/sqr(n)) = Z ~ N(0, 1)

replace σ with s (sample SD), its Z-score:

(Xbar - µ)/(s/sqr(n)) = T ~ T_n-1

• T_n-1 → a random variable with n-1 degrees of freedom

Question 11

Student’s t-distribution compensates for…

Exact sampling distribution of random variable T_n-1…

Answer 11

substituting s for σ in the standard error.

The exact sampling distribution of the random variable T_n-1 = (Xbar - µ)/(S/sqr(n))

Question 12

degrees of freedom

Answer 12

n-1; larger n = better estimate of a standard normal distribution

Question 13

Degrees of freedom is necessary because…

Answer 13

mimics sample size,

there will be more variability in s for small sample sizes than for large sample sizes

Question 14

Confidence interval for µ

The 100(1-a)% confidence t-interval for µ is

Answer 14

xbar - t_a/2,n-1 s/sqr(n) to xbar + t_a/2,n-1 s/sqr(n)

Question 15

Interpreting CI’s

Answer 15

95% of intervals created according to this procedure are expected to contain μ.

Question 16

From the CLT, we know:

E(Xbar) =

Var(Xbar) =

SD(Xbar) = SE(Xbar) =

Answer 16

sample mean:

μ

σ²/n

σ/sqr(n)

Question 17

Manipulating CI’s

Answer 17

you can transform the ends of the CI to obtain a new CI for the transformed parameter

Ex. MPG → L/100km

Ex. probabilities → odds ratios

Question 18

_________ provides a 95% confidence interval for µ

Answer 18

Xbar +/- 1.96 σ/sqr(n)

or

Xbar +/- 2 σ/sqr(n)

Question 19

sample mean of 0/1 variables

Answer 19

sample proportion of 1’s

this means we can still apply CLT to sample proportion

Question 20

p =

p^ =

Answer 20

population proportion

sample proportion

Question 21

Sampling distribution of p^

Answer 21

p^ ~ N(p, p(1-p)/n)

Question 22

key assumptions of sampling distribution of p^

Answer 22

1. we have an independent sample from the pop
2. sample size is large enough for CLT to be applied
3. sample size rule: np^ and n(1-p^) > 10

Question 23

Rule for the sample size

Answer 23

both np^ and n(1-p^) > 10

*#of successes and failures each have to be greater than 10*

Question 24

_______ provides a 95% confidence interval for the population proportion

Answer 24

p^ +/- 2se(p^) = p^ +/- 2sqr[p(1-p)/n]

but since we don’t know p, we replace it with p^:

p^ +/- 2sqr[p^(1-p^)/n]

Question 25

What is MoE?

Answer 25

Margin of Error is the distance from the center to the edge of the interval MoE = 2 s/sqr(n) ⇒ **critical value \* standard error** Ex. if you want a 95% confidence level, MoE = 1.96 σ/sqr(n)

Question 26

Sample size formula for a population mean (µ) with 95% z-interval

Answer 26

n= (critical value \* σ/MoE)²

Question 27

sample size for estimating a population proportion

Answer 27

n = (1/MoE)² **use only if:** * **you want a 95% CI** * ***p* (pop proportion) lies between 0.25 & 0.75**

Question 28

Ex. a company has estimated the proportion of doctors who say they'll prescribe a new drug as 30%. They used a sample size of 100, but didn't provide a CI. What was the MoE for a 95% CI?

Answer 28

n= (1/MoE)² ⇔ MoE = 1/sqr(n) MoE = 1/10 = 0.1 CI is approx (30% +/- 10% ) = (20%, 40%)