ch7 statistics notes

INTRODUCTIONDERIVATION

NOTATIONS AND GENERAL FORM1 SAMPLE Z-INTERVAL FOR Μ1 SAMPLE T-INTERVAL FOR Μ

2 INDEPENDENT SAMPLE T-INTERVAL FOR Μ2 DEPENDENT SAMPLE T-INTERVAL FOR Μ

Z-INTERVAL FOR PSAMPLE SIZE

Chapter 7 Confidence Intervals

Introduction

In this chapter, we will estimate the population parameter using an interval.

This interval will ‘capture’ the true population parameter with a certain measure of precision.

Example: A 95% confidence interval for µ is (13, 18)

Notations and General Form

CI - confidence interval CV - critical value ME - margin of error SE - standard error SD - standard deviation pt.est. - point estimate Zα/2 - normal distribution critical value(use invnorm) t(n-1,α/2) - students t distribution critical value with n-1 degrees of freedom (use math solver or the invT)


General Formula:

SECV estPt .


ppX

ˆ

Parameter estimatePoint EstimatePoint

n)p̂-(1p̂ SE ,p̂for

nSD/ SE ,Xfor

(SE)Error Standard


Critical Value

Z-critical valueZ1-α/2 = InvNorm(1 - α/2)

example: 95% Confidence intervalα = 1 – 0.95 = 0.05

Z 1-α /2 = InvNorm(1- α/2) = InvNorm(1-(.05/2)) = 1.96


Critical Value

T-critical value t (1-α /2, df) = InvT(1-α/2, degrees of

freedom) or

t (1-α /2, df) = L on math solver (for TI-83)

Math -> solver -> tcdf(L,U,D) – AL = CVU = 9999D = degrees of freedomA = α/2


Example: 95% confidence interval based on a sample size of 20. (α = .05, df = n-1 = 19)

T-critical value t (1-α /2, df) = InvT(1-(0.05/2), 19)

Math->solver->tcdf(L,U,D) – AL = CV (highlight and then press alpha

enter)U = 9999D = 19A = .025

ANSWER: t (1-0.025, 19) = 2.0930

1 Sample Z-interval for µ

Error ofMargin )/(

Error Standard)/(

Value Critical ..

where)/(

Formula

2/1

2/1

2/

nZ

n

ZestptX

nZX

Population Standard Deviation (σ) is known


Suppose the time allotted for commercials on a primetime TV program is known to have a normal distribution with a standard deviation of 1.5 minutes. A study of 25 showings gave an average commercial time of 11 minutes. Find the 95% confidence interval for the true population mean, μ.

Given:

Confidence level = 0.95 Critical value = invNorm(1-.05/2) = 1.96

1125 n

deviation) standard populationknown theis (this 1.5

X


A 95% confidence interval for μ is

)588.11,412.10(588.011

3.0*96.111)25/5.1(*96.111

)/(*

*.

nZX

SECVestpt

We are 95% confident that the average time in commercials is between 10.412 and 11.588 minutes

Using 1-sampZInterval in TI-83/84Stat -> Tests -> ZInterval

Notes on intervals

Effect of confidence levelHigher confidence level results to a longer

confidence interval since CV increases as α decreases.

Example, Z = 1.645 when α = .10 while Z = 1.96 when α = .05

Effect of sample sizeIncreasing sample size (n) shortens the

confidence interval since SE = SD/sqrt(n).

Note: level of significance (α) confidence level (1- α)*100%

1 Sample t-interval for µ

1-n df Error ofMargin )/(

Error Standard/

Value Critical ..

where)/(

Formula

)1,2/1(

)1,2/1(

)1,2/(

nSDt

nSD

testptX

nSDtX

n

n

n

Population Standard Deviation (σ) is unknown

1 Sample t-interval for µ

A random sample of 12 graduates of a certain secretarial school typed an average of 79.3 words per minute with a standard deviation of 7.8 words per minute. Assuming normal distribution for the number of words typed per minute, find a 95% confidence interval for the average number of words typed by all graduates of this school.

Given:

Confidence level = 0.95Critical value = 2.201

8.73.79

12 n givennot

SDX

1 Sample t-interval for µA 95% confidence interval for μ is

)2559.84,3441.74(9559.43.79

2517.2*201.23.79)12/8.7(*201.23.79

)/(

nSDtX

We are 95% confident that the average number of words the graduates type per minute is between 74.3441 and 84.2559 words.

Using 1-samptInterval in TI-83/84Stat -> Tests -> tInterval

2 Independent Sample t-interval for µ

Value Critical t2

)1()1(

where

11)(

Formula

)221,2/(

21

222

2112

21

2)221,2/(21

nn

p

pnn

nnSnSnS

nnStXX

.Confidence Interval for µ1 - µ2

Requirement 1: Both samples where taken from a normal distributionRequirement 2: The population standard deviations are equal (σ1=σ2=σ)


Do credit cards with no annual fee charge higher interest rates than cards that have annual fees? Among 29 cards surveyed, 17 had no annual fees while 12 charged an annual fee. Among the cards with no annual fee, the average interest rate was 19% (SD = 8%). Among cards with an annual fee, the average interest rate was 17% (SD = 3%).a. What assumptions do you need to get a confidence interval

for the difference in average interest rate?

b. Calculate the estimate of the common standard deviation.

c. Construct a 95% interval estimate for the difference in average interest rates.


a. What assumptions do you need to get a confidence interval for the difference in average interest rate?

The samples were taken from normally distributed populations and that they have a common standard deviation.


b. Calculate the estimate of the common standard deviation.

004159.021217

03.0)112(08.0)117(

2)1()1(

22

222

WWO

WWWOWOp

nnSnSnS

No Annual Fee With Annual FeeAverage ( ) 0.19 0.17 0.19 – 0.17 = 0.02SD (S) 0.08 0.03Sample Size (n) 17 12

WWO XX


c. Construct a 95% interval estimate for the difference in average interest rates.

Level of Confidence (1-α) = 0.95 α = 0.05Critical value (tα/2,df=n1+n2-2 = t0.05/2,df=27 = t0.025,27) = 2.052

A 95% confidence interval for is calculated as

)06989.0,0299.0(0498964063.002.0

)0243159875.0(052.202.0121

1710041592593.0052.202.0

11)( 21

2

nnStXX pWWO

Therefore, with 95% confidence, the difference in average interest rate will lie between -2.99% and 6.99%.


No Annual Fee With Annual FeeAverage ( ) 0.19 0.17 0.19 – 0.17 = 0.02SD (S) 0.08 0.03Sample Size (n) 17 12

Using 2-samptInterval in TI-83/84

Stat -> Tests -> 2samptInt


Suppose it is of interest to estimate the difference in the first exam scores of STAT 2160 male and female students. A random sample of 16 males and 17 females taking the course this semester was drawn and asked about their scores. Among male students, the average was found to be 15.28 with a standard deviation of 2.3, while among females, the average is 16.5 with a standard deviation of 1.6. Assuming the scores follow a normal distribution with equal variances, construct a 95% CI for the true difference in the first exam scores of male and female STAT 2160 students.


Step 1: Create a table of given values

Step 2: Compute for the common variance

MALE FEMALEn 16 17

Mean 15.28 16.5Standard Deviation

2.3 1.6

2)1()1(

21

222

2112

nn

SnSnS p


217166.1)117(3.2)116( 22

2

pS

880967742.32 pS


Step 3: Obtain the Critical Value (CV)

95% CI α = 0.05CV = invT(1-α/2, n1+n2-2) = invT(0.975, 31)

= 2.0395


Step 4: Plug-in the values in the CI formula

21

2)221,2/(21

11)( nn

StXX pnn

171

161880967742.30395.2)5.165.281(

)6861870765.0(0395.21.22

1795.0,6195.2

iClicker

2 Dependent Sample t-interval for µ

There is one sample but measured twice

sdifference theof SEn/SD

valueCritical sdifference theof averageX

where)/(

Formula

DIFF

)1,2/(

DIFF

DIFF)1,2/(

n

nDIFF

t

nSDtX


The table below shows the opening and closing prices of a sample of 10 active stocks on a certain day. Give an estimate of the difference in the average stock prices? What is the corresponding standard error of this estimate?

Opening Price Closing PriceIntel Corp 16.09 16.25Citigroup Inc 17.47 16.32Bank of America Corp 25.6 24.67JPMorgan Chase & Co 41.52 42.33General Electric Co 20.11 19.37Microsoft Corp 23.77 23.44Pfizer Inc 17.03 16.93Exxon Mobil Corp 67.69 62.92Coca-Cola Co 46.48 49.23Alcoa Inc 11.54 10.09


First, we need to take the difference between the two prices for each stock.

Opening Price Closing Price Difference Intel Corp 16.09 16.25 -0.16Citigroup Inc 17.47 16.32 1.15Bank of America Corp 25.6 24.67 0.93JPMorgan Chase & Co 41.52 42.33 -0.81General Electric Co 20.11 19.37 0.74Microsoft Corp 23.77 23.44 0.33Pfizer Inc 17.03 16.93 0.1Exxon Mobil Corp 67.69 62.92 4.77Coca-Cola Co 46.48 49.23 -2.75Alcoa Inc 11.54 10.09 1.45


Then calculate the mean of the differences and the corresponding standard deviation then calculate the standard error.

6022850008.010

904592403.1)(

904592403.1575.0

nSXSE

SX

DIFFDIFF

DIFF

DIFF


Construct a 95% confidence interval for the difference in opening and closing stock prices.

Use 1-samptInterval in TI-83/84 since we converted the two samples into a sample of differences.

Answer: (-0.7875, 1.9375)

Z-interval for P

Error Standard/)1(

Value Critical where

/)1(

Formula

2/

2/

npp

Z

nppZp

Z-interval for P

In a random sample of 500 families owning television sets in the city of Hamilton, Canada, it was found that 340 subscribed to HBO. Find a 95% confidence interval for the actual proportion of families in this city who subscribe to HBO.

Given:

Find: A 95% CI for the population proportion, p. Level of Confidence (1-α) = 0.95 α = 0.05Critical value (z1-α/2 = z1-0.05/2 = z0.975) = InvNorm(0.975) = 1.96

68.0500340

size samplesample in the successes #

p

Z-interval for P

The confidence interval is calculated as follows:

7209.0,6391.00408884375.068.0

)0208614477.0(96.168.0500

)68.01(*68.096.168.0

/)1(

nppzp

Therefore, we are 95% confident that the actual proportion of families in this city who subscribe to HBO is between 64% and 72%.

A function in TI-83/84 is 1propZInt. Stat -> Tests -> 1propZInt

iClicker

Sample Size

In order to be (1-α) x 100% confident that the sample mean is within a distance ME of the mean μ, choose a sample size equal to

n = z2σ2/M2

For computing sample size for estimating population proportion, the formula is

2

2 )1(M

ppzn

Sample Size

A consumer group wishes to estimate the average electric bills for the month of July for single-family homes in a large city. Based on studies conducted in other cities, the standard deviation is assumed to be $25. The group wants to estimate the average bill for July to be within $5 of the true average with 95% confidence. a. How many single-family homes should be selected? b. If the group wants to be correct to within $10, what sample size is necessary? c. If 99% confidence and a sampling error of $5 are desired, how many single-family homes are necessary? Given: ME = 5

Level of Confidence (1-α) = 0.95 α = 0.05Critical value (zα/2 = z0.05/2 = z0.025) = InvNorm(0.025) = -1.96σ = 25

Sample Size

homesfamily -single 9704.96

)5()25()96.1(

2

22

2

22

M

zn

a. How many single-family homes should be selected?

Given: ME = 5Level of Confidence (1-α) = 0.95 α = 0.05Critical value (zα/2 = z0.05/2 = z0.025) = InvNorm(0.025) = -1.96σ = 25

Sample Size

b. If the group wants to be correct to within $10, what sample size is necessary?


)10()25()96.1(

2

22

2

22

M

zn

Sample Size

c. If 99% confidence and a sampling error of $5 are desired, how many single-family homes are necessary?

M = 5Level of Confidence (1-α) = 0.99 α = 0.01Critical value (zα/2 = z0.01/2 = z0.005) = InvNorm(0.01/2) = -

2.576σ = 25


)5()25()576.2(

2

22

2

22

M

zn

iClicker

Sample Size

In a random sample of 500 families owning television sets in the city of Hamilton, Canada, it was found that 340 subscribed to HBO. Suppose that a CI for the proportion of families who subscribe to HBO is computed at 90% confidence level and within ± 0.05, what will the new sample size be?

23653.235

05.0)68.1(*68.*1.645

)1(

0.05 ME 1.6450/2)invNorm(.1 Z

68.0 :Given

2

2

2

2

/2

MEppzn

p

iClicker

ch7 statistics notes

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