WARM UP

A statewide poll reveals the only 18% of Texas household participate in giving out Halloween Treats. To examine this you collect a Random Sample of 120 Texas households.

1. What is the Probability that less than 12% of Texas households give out Treats?

2. Comment on the validity of your results by checking the Rule of Thumb assumptions.

ˆ .12 ?P p ˆ

1

p pz

p p

n

.12 .18

?.18 1 .18

120

P z

1.7108 ?P z ( 99, 1.7108) 0.0436normalcdf E

# ?P z

RULE OF THUMB #2: 120 • .18 ≥ 10 AND RULE OF THUMB #2: 120 • .18 ≥ 10 AND 120 • (1 – .18) ≥ 10120 • (1 – .18) ≥ 10

RULE OF THUMB #1: Pop. of Texas Households ≥ 10 • 120RULE OF THUMB #1: Pop. of Texas Households ≥ 10 • 120

3. Draw the Sampling Distribution curve.

N(0.18, 0.035)

.075 .11 .145 .18 .215 .25 .285

.12

(1 ),p p

N pn

Sample Proportion, use Categorical DataData collected by COUNTING.

Sample Means, use Quantitative Data

Data collected by AVERAGING

Chapter 18 (continued)Chapter 18 (continued)

x

p̂

SAMPLE MEANS SAMPLE MEANS (Chapter 18 continued)(Chapter 18 continued)

The unknown Population Mean is a parameter with the symbol of: . The Population Standard Deviation has the symbol:

The Sample Mean is a statistic with the symbol of:

and the Sample Standard Deviation has the formula

.

x

I. THE BASICS

sn

The Test Statistic is: x

z

n

ASSUMPTIONS for Sample Means:

#1: The sample was collected using an SRS.

#2: Approximate Normal: 1. Large n = (n ≥ 30), OR 2. Stated that its Normal, OR 3. You construct an approximately Normal Histogram/Boxplot from the given data.

x ,Nn

It is known that each individual Halloween Treat given out has about 120 calories with standard deviation of 50. To examine this you collect a Random Sample of 40 Halloween treats.

1. What is the Probability that the sample average is less than 100 calories?

2. Comment on the validity of your results by checking the conditons/assumptions.

100 ?P x xz

n

100 120

?50

40

P z

2.5298 ?P z ( 99, z) 0.0057normalcdf E

Approximately Normal due to the Large n.Approximately Normal due to the Large n.Stated to be a Random sample Stated to be a Random sample

3. Draw the Sampling Distribution curve.

II.II.THE CALCULATIONSTHE CALCULATIONS

1.1. Find the Mean and Standard Deviation of the Find the Mean and Standard Deviation of the Sampling Distribution of .Sampling Distribution of .

2.2. Explain why you can assume normality in the Explain why you can assume normality in the sample.sample.

xMean = 3.5Mean = 3.5Std. Dev. = .1697Std. Dev. = .1697

1.2. .

50Std Dev

#1: The sample was collected by an SRS #1: The sample was collected by an SRS #2: Approximately Normal b/c Large n #2: Approximately Normal b/c Large n

EXAMPLE: EXAMPLE: A national political poll asked a SRS of 50 A national political poll asked a SRS of 50 adults to give an approval rating of President Obama’s adults to give an approval rating of President Obama’s Job in office using a scale from 1Job in office using a scale from 1(Low)(Low) to 10 to 10(High)(High). . Suppose in fact, nationally Obama’s rating has a Mean Suppose in fact, nationally Obama’s rating has a Mean of 3.5 and Standard Deviation of 1.2of 3.5 and Standard Deviation of 1.2

3. Find the probability Obama has a rating over 4.0.

II.THE CALCULATIONS of PROBABILITY

4.0 ?P x xz

n

4.0 3.51.2

50

P z

2.9463 ?P z

Probability (2.9463, 99) 0.0016normalcdf E

SAME EXAMPLE:SAME EXAMPLE: A national political poll asked a SRS of 50 adults to A national political poll asked a SRS of 50 adults to give an approval rating of President Obama’s Job in office using a scale give an approval rating of President Obama’s Job in office using a scale from 1(Low) to 10(High). Suppose in fact, nationally Obama’s rating has from 1(Low) to 10(High). Suppose in fact, nationally Obama’s rating has a Mean of 3.5 and Standard Deviation of 1.2a Mean of 3.5 and Standard Deviation of 1.2

# ?P z

4. Find the probability Obama’s approval rating is between 3.2 and 4.0.

3.2 4.0 ?P x

3.2 3.5 4.0 3.5?

1.2 1.250 50

P z

1.7678 2.9463 ?P z

Prob.= ( 1.7678, 2.9463) 0.9598normalcdf

SAME EXAMPLE:SAME EXAMPLE: A national political poll asked a SRS of 50 adults to A national political poll asked a SRS of 50 adults to give an approval rating of President Obama’s Job in office using a scale give an approval rating of President Obama’s Job in office using a scale from 1(Low) to 10(High). Suppose in fact, nationally Obama’s rating has from 1(Low) to 10(High). Suppose in fact, nationally Obama’s rating has a Mean of 3.5 and Standard Deviation of 1.2a Mean of 3.5 and Standard Deviation of 1.2

xz

n

Finding Sample Mean Observations, (x) from Probabilities

EXAMPLE: Assume that the duration of human pregnancies is Appr. Normal with mean: 266 days and Std Dev.: 16. At least how many days should the longest 25% of all pregnancies last?

218 234 250 266 282 298 314

x=?

z

1

x

z = invNorm(.75)= .6745

2660.6745

16

x

x = 276.79

25%25%

Finding Sample Mean Observations, (x) from Probabilities

EXAMPLE: Assume that the rainfall in Ithaca, NY is Appr. Normal with mean: 35.4 inches and Std Dev.: 4.2”. Less than how much rain falls in the driest 20% of all years

22.8 27 31.2 35.4 39.6 43.8 48

x=?

zx

z = invNorm(.20)= -.8416

35.4.8416

4.2

x

x = 31.865”

20%20%

1.The Test over Chapter ninteen has traditionally had a mean of 86.2% with a std.

deviation of 8.5%. Assume the data follows a Normal distribution.

a.) If a student is selected at random, what is the probability that he or she will score above 90%?

b.) If 16 students are selected at random, what is the probability that their average will be above 90%?

c.) Draw the Sampling Distribution Normal Curves for both a and b. On both graphs label all six standard deviations

(± 3s) for the .,Nn

With μ = 86.2% and σ = 8.5%.

a.) When n = 1 the

b.) When n = 16 the

c.).

90 ?P x 90 86.2?

8.5

1

P z

0.4471 ?P z

(0.4471, 99) 0.3274normalcdf E 1.Appr. Normal 2. .Stated Data Collected Randomly SRSand

90 ?P x 90 86.2?

8.5

16

P z

1.7882 ?P z

(1.7882, 99) 0.0369normalcdf E

60.7 69.2 77.7 86.2 94.7 103.2 111.760.7 69.2 77.7 86.2 94.7 103.2 111.7

86.2

86.2

88.3

88.3

92.6

92.6

90.5

90.5

79.8

79.8

82.0

82.0

84.1

84.1

8.586.2,

1N

8.586.2,

16N

86.2,8.5N 86.2,2.125N

1.Appr. Normal 2. .Stated Data Collected Randomly SRSand