Confidence intervals

Question 1

Why do we even need the sample proportion?

Answer 1

We need it because we might not know the data for the population proportion. So, we must have an unbiased estimator sample proportion to best estimate the population proportion.

Question 2

What is a confidence interval? (in easy terms)

Answer 2

It tells us how close the sample proportion is to the population proportion when we do not know the population proportion.

Question 3

Confidence interval definition

Answer 3

Simply, a range of values over which a population statistic occurs with a certain probability

Question 4

What is the first thing you need to construct a confidence interval?

Answer 4

A confidence level

Question 5

What is a confidence level?

Answer 5

The percentage by which we are confident that the confidence interval will contain the population proportion.

Question 6

what is another word for population?

Answer 6

true

Question 7

Confidence level example

Answer 7

If i choose a 90% confidence level, then 90% of the time our confidence interval will contain the true proportion

Question 8

Confidence interval steps

Answer 8

1) Choose confidence level
2) Know associated critical z-score
3) calculate the margin of error (E)
4) Confidence interval is then the interval ((p hat) minus (E), (p hat) plus (E))

Question 9

Finding critical z-score steps

Answer 9

1) Take the complement of the confidence level aka the alpha (a)
1. 1) complement of 90% is 10%, a=10%
2) Divide alpha in half (a/2 = 10%/2 = 5%)
3) Take complement of a/2 (Complement of 5% is 95%)
4) Use calc invnorm and calculate (invnorm(0.95))=1.645

Question 10

What is alpha?

Answer 10

The complement percentage of the confidence level, such as how the complement of 90% is 10%, so 10% = alpha

Question 11

Invnorm on calc

Answer 11

“2nd” “vars” “3” aka invnorm

Question 12

Margin of error formula

Answer 12

E = (associated critical z-score) times (square root ((p hat times q hat)/(n))

Question 13

(A1) Example question about students emailing teacher

Answer 13

A teacher wants to know the # of his students that send him at least 1 email. He picks 40 of his students at random and checks to see how many of the 40 sent him an email. Suppose 14 have sent him an email.

Question 14

(A2) What is the sample proportion?

Answer 14

Sample proportion or (p hat) equals 14/40 which equals 0.35 or 35%, since 14 students out of the 40 students sent him an email

Question 15

(A3) What confidence level does the teacher choose?

Answer 15

He chooses 90% confidence level

Question 16

(A4) What is the associated critical z-score if the confidence level is 90%?

Answer 16

The complement of 90% is 10%. Now divide 10% in half = 5%. The complement of 5% is 95%. So then use calc invnorm(0.95) and the answer is 1.645

Question 17

(A5) What is q hat?

Answer 17

It is the complement of the sample proportion. So the complement of 35% is 65%

Question 18

(A6) Now what is the margin of error aka E?

Answer 18

I multiply (1.645) and the square root of ((0.35 times 0.65) divided by (40)) = 0.124 = E

Question 19

(A7) What is the confidence interval?

Answer 19

[{(p hat) minus (E)} comma {(p hat) plus (E)}] which is (0.35-0.124, 0.35+0.124) = (0.226, 0.474) or 0.226 < p < 0.474

Question 20

(A8) What is the answer?

Answer 20

The teacher can be 90% confident that the percentage of his students (the population) sending him an email is on the interval from 22.6% to 47.4%

Question 21

(A9) Can the teacher conclude that 20% of his students are sending him an email?

Answer 21

Yes because the interval for the sample proportion is entirely above 20%.

Question 22

(B1) Example question about the proportion of trees that have a disease.

Answer 22

A biologist takes a sample of 250 trees and discovers 42 have a disease. What is the 99% confidence interval for the proportions of trees having a disease?

Question 23

(B2) Identify variables, (p hat), (critical z score), (q hat), (n)

Answer 23

Sample proportion aka (p hat)=42/250=0.168
Critical z score=2.575
Complement of (p hat)=(q hat)=0.832
n=250

Question 24

(B3) Finding critical z score aka z_(a/2)

Answer 24

1) 99% confidence interval
2) Complement of 99% is 1%
3) 1% divided by 2 = 0.5%
4) Complement of 0.5% = 95.5%
5) Use chart or invnorm to find the number for 0.955
6) z_(a/2)=2.575

Question 25

(B4) Find the margin of error E

Answer 25

E = 2.575 (square root ((0.168 times 0.832) divided by 250)) = 0.609

Question 26

(B5) Finding the confidence interval (p hat minus E, p hat plus E)

Answer 26

(0. 168 - 0.061, 0.168 + 0.061) = (0.107, 0.229) or

0. 107 < p < 0.229

Question 27

Researchers need an appropriate amount of data (n) or sample size in order for the margin of error to be smaller

Answer 27

Since a smaller amount of data means that E (standard error) will be bigger and the confidence interval bigger

Question 28

Formula that solves for the required sample size

Answer 28

n = ([critical z-score]^2 (p hat) (q hat)) / E^2

Question 29

(C1) Example problem solving for sample size, publisher, and foreign online orders

Answer 29

A publisher wants to know what proportion on average his online orders are from overseas. He wants a margin of error of less than 5%. How big of a sample do they have to look at?

Question 30

(C2) Define variables E and z_a/2

Answer 30

E=0.05

z_a/2=1.96

Question 31

(C3) Finding z_a/2

Answer 31

1) 95% complement
2) 5% / 2
3) 2.5% complement
4) 97.5%
5) 0.975
6) z_a/2= 1.9+0.06 = 1.96

Question 32

(C4) For p hat and q hat

Answer 32

Use 0.5 since it “assumes the worst” so (p hat=0.5) and (q hat = 0.5)

Question 33

(C5) Using sample size formula

Answer 33

n = [(1.96^2)(0.5)(0.5)]/(0.05^2)=384.16=385

Question 34

(C6) Answer in sentence form

Answer 34

The publisher needs to examine at least 384.16 orders or 385 orders to be certain that E will be no more than 0.05 or 5%

Question 35

When the population standard deviation is not known, the sample standard deviation is used to create a confidence interval

Answer 35

And the distribution is not a normal distribution, but instead, an alternate distribution called the “Student’s t” or “t distribution”

Question 36

What is the difference between the normal distribution and “t distribution”?

Answer 36

Both look similar in terms of shape since it is a hill, BUT are not. The tails of the “t distribution” is wider aka the tails are a little higher from the x line as the sample size gets smaller

Question 37

What is t_a/2?

Answer 37

t-distribution critical values

Question 38

What is a t-table?

Answer 38

The table has critical values for 90%, 95%, 98%, and 99% confidence intervals for selected values of n

Question 39

How to find critical values on a t-table?

Answer 39

Find the degrees of freedom aka n-1

Question 40

Whats the point of degrees of freedom?

Answer 40

a good way to think about degrees of freedom is that one of the data values gets “used up” in a mathematical
sense when the standard deviation is calculated so that effectively there are only n – 1 data values left.

Question 41

(D1) Example problem to find t_a/2

Answer 41

1) A confidence level of 90% is chosen
2) a=10% and a/2=5%
3) On a graph, the distribution will look like a hill
4) 90% is closest to the mean, so there will be 5% on both tails
5) The t_a/2 is for the 95th percentile since the 5% on the left side + 90% middle = 95% or 95th percentile

Question 42

(D2) Margin of error formula

Answer 42

E = (t_a/2) times (s/square root(n))

Question 43

(D3) Confidence intervals for proportions can be expressed in either two ways

Answer 43

1) An inequality: (x bar minus E < Myu < x bar plus E)

2) An interval: (x bar-E, x bar+E)

Question 44

(E1) Example problem, worm length and population mean

Answer 44

A researcher has a random sample of 41 worms and calculates mean length as 62.4 mm, SD is 14.7. Construct 90% confidence interval for the population mean

Question 45

(E2) What is critical value t_a/2

Answer 45

On t-table for 90% confidence interval, the area in each tail of distribution is 5% or in total 10%.
So critical value is 1.684 ofc with degrees of freedom in mind

Question 46

(E3) What is the degrees of freedom?

Answer 46

(n - 1) is (41 - 1 = 40)

DOF is 40

Question 47

(E4) What is margin of error E

Answer 47

E = t_(a/2) times ([standard deviation] divided by [square root of n])

E=1.684 (14.7/square root of 41) = 3.866 or 3.9

E=3.9

Question 48

(E5) Put into interval expression

Answer 48

(x bar minus E, x bar plus E)

62. 4-3.9=58.5, 62.4+3.9=66.3
(58. 5, 66.3)

Question 49

(E5.5) Put into inequality form

Answer 49

58.5 < myu < 66.3

aka myu is mean

Question 50

(E6) Answer in sentence form

Answer 50

The mean length for population of worms is 58.5 and 66.3 with 90% confidence