Inference about Mean(σ Unknown)

When σ is known, the sampling distribution for a sample mean is normal if conditions are satisfied.

For many years, it was thought that when σ was unknown, this was still the case. However, because of the increased variability introduced by not knowing σ, the sampling distribution for a sample mean with unknown σ is not normal.

This was discovered by W. S. Gosset, an Irish Brewery quality control inspector in 1908. He also discovered that when σ is unknown, we can still do inference using a sampling distribution model he discovered, the t-distribution.

1Section 9.1, Page 184

t-Distribution

•The t-Distribution is unimodal and symmetric•It has more area in that tails and is flatter in the middle than the normal distribution•It is a family of distributions, with a different curve for each μ, σ, and df (degrees of freedom) df=n-1, where n is the sample size.

Critical Values for 95% Confidence IntervalKnown σ: 1.960 from Normal DistributionUnknown σ with 3 df: 3.18 from t-DistributionUnknown σ with 10 df: 2.23 from t-DistributionUnknown σ with 1000 df: 1.962 from t-Distribution

2Section 9.1, Page 184

t-DistributionWhen to Use It

3Section 9.1, Page 184

Is the population parameter of interest a mean?

Yes

Is the value of the population standard deviation, σ, an

unknown value?

Yes

Use the t-Distribution

Confidence Interval(Unknown σ)

€ €

4Section 9.1, Page 186

Confidence Interval =

Sample Mean ± Margin of Error

± Critical Value × Standard Error of

€

x

€

x

€

x ± t(df ,α ) ×s

n

where s is the standard deviation of the sample and n is the sample size, and df is the degrees of freedom, n-1.

Conditions: The population must be normal or the sample is large (n ≥ 30)

Confidence Interval (Unknown σ)Illustrative Problem- TI 83 Add-in Programs

C.I. =Sample Mean ± Critical Value * Standard ErrorPRGM – CRITVAL – ENTER2: T-DISTCONF LEVEL = .95df = 19Answer: 2.0930PRGM – STDERROR-ENTER4: 1-MEANn=20Sx=1.76Answer: .3935

C.I. = 6.87 ± 2.0930*.3935 = 6.87 ± .8236=(6.87 - .8236, 6.87 + .8236) = (6.05, 7.69)

5Section 9.1, Page 186

Confidence Interval (Unknown σ)Illustrative Problem- TI 83 Black Box Program

STAT – TESTS – 8:TintervalInpt: Stats 6.87Sx: 1.76n: 20 C-Level: .95CalculateAnswer: (6.05, 7.69)

6Section 9.1, Page 186

€

x :

Problems

a. Find the 98% confidence interval.b. Find the critical valuec. Find the margin of error.d. Find the standard error.e. What assumption must we make about the the

population to have a t-sampling distribution.f. What are the proper words to describe the

confidence interval?g. If you wanted to have a margin of error of one

minute and the 98% confidence interval for this data, how large must the sample be?

7Problems, Page 205

Problems

a. Find the 93% confidence interval.b. Find the critical valuec. Find the margin of error.d. Find the standard error.e. What assumption must we make about the population

to have a t-sampling distribution.f. What are the proper words to describe the confidence

interval?g. If you wanted to have a margin of error of 25 lbs. and

a confidence level of 93% for this data, how many students would you have to monitor?

8Section 9.1, Page 205

Hypotheses Test (Unknown σ)Illustrative Problem- TI-83 Add-In

The EPA wanted to show that the mean carbon monoxide level is higher than 4.9 parts per million. Does a random sample of 22 readings (sample mean = 5.1, s=1.17) present sufficient evidence to support the EPA’s claim? Use α = .05. Assume the readings have approximately a normal distribution.

Ho: μ = 4.9 (no higher than)Ha: μ > 4.9 (higher than)

t-Distributiondf = 21

€

μx = 4.9

€

x = 5.1

P-value = .2158

PRGM – TDIST – ENTERLOWER BOUND = 5.1UPPER BOUND = 2ND EE99MEAN = 4.9 df = 21Answer: .2158 (Evidence is not sufficient to support the EPA’s claim.)

9Section 9.1, Page 187

€

1.17 / 22

€

SE(x ) =

€

P −Value = P(x > 5.1 given μ x = 4.9)

Hypotheses Test (Unknown σ)Illustrative Problem- TI-83 Black Box

The EPA wanted to show that the mean carbon monoxide level is higher than 4.9 parts per million. Does a random sample of 22 readings (sample mean = 5.1, s=1.17) present sufficient evidence to support the EPA’s claim? Use α = .05. Assume the readings have approximately a normal distribution.

Ho: μ = 4.9 (no higher than)Ha: μ > 4.9 (higher than)

€

STAT − TESTS − TTest

Inpt :Stats

μ0 :4.9

x : 5.1

Sx :1.17

n = 22

μ :> μ0

Calculate

Answer : p − value = .2158

10Section 9.1, Page 187

There is not sufficient evidence to support

the EPA claim.

Problems

a. State the correct hypothesisb. Find the p-value.c. State your conclusion.d. What is the name of the probability model

used for the sampling distributione. What is the mean of the sampling

distribution?f. What is the value of the standard error?g. If your conclusion is in error, what type of

error is it?

11Problems, Page 209

In a study of computer use, 95 randomly selected Internet Canadian users were asked how much time they spend using the Internet in a typical week. The mean of the sample was 12.7 hours and the standard deviation is 4.1 hours. Is this convincing evidence that Canadians use the internet more than the USA average of 12.3 hours a week?

Problems

a. State the correct hypothesis.b. Find the p-value.c. State your conclusion.d. What is the name of the probability model used

for the sampling distributione. What is the mean of the sampling distribution?f. What is the value of the standard error?g. If your conclusion is in error, what type of error is

it?

12Problems, Page 205

The recommended number of hours of sleep per night is 8 hours, but everybody “knows” that the average college student sleeps less than 7 hours. The number of hours slept last night by 10 randomly selected college students is listed here.

5.2 6.8 5.5 7.8 5.8 7.1 8.1 6.9 5.7 7.2

Is there convincing evidence that the students sleep less than 7 hours? Assume the population is normal.

Problems

a. State the appropriate hypotheses.b. Find the p-value.c. State your conclusion.d. What is the name of the probability model used

for the sampling distributione. What is the mean of the sampling distribution?f. What is the value of the standard error?g. If your conclusion is in error, what type of error is

it?

13Problems, Page 206

Inference about Proportions

A binomial experiment is a experiment that has only two outcomes. Binomial experiments relate to categorical variables. Consider the variable, Supports Obama. The variable has two categories, yes and no. We will consider the yes category as a “success” and the no category a “failure”.

Suppose we have a population of 1000 voters, and 550 support Obama. We define proportion of success for the population, p = 550/1000 = .55. The proportion of failures is q = 1-p = .45.

Suppose we take a sample to estimate p, the true proportion of voters that support Obama. We take a random sample of 100 voters and 53 support Obama. Our sample proportion is p’ = 53/100=.53. Our sample q’ = 1-p’ = .47.

14Section 9.2, Page 192

Sampling Distribution for a Proportion

15Section 9.2, Page 192

In practice, the following conditions will insure the sampling distribution for a proportion is normal.1.The sample size is greater than 20

2.The product np and nq are both greater than 5. Where we do not know p, we substitute p’, np’ = # success in sample must be > 5 and nq’ =n(1-p’) = # failures in sample must be > 5.

3.The sample consists of less than 10% of the population.

Confidence Interval for a Proportion, p

16Section 9.2, Page 193

Given a sample proportion p’, that has a normal sampling distribution, the confidence interval for the true population proportion, p, is:

sample proportion ± margin of error

sample proportion ± critical value × standard error of p’

€

p'±z(α ) ×p'q'

n

where q’ = proportion of failures = 1-p’ and n is the sample size.

Confidence Interval for ProportionIllustrative Example TI-83 Add-in Programs

In a discussion about the cars that fellow students drive, several statements were made about types, ages, makes colors, and so on. Dana decided he wanted to estimate the proportion of convertibles students drive, so he randomly identified 200 cars in the student parking lot and found 17 to be convertibles. Find the 90% confidence Interval for p, the true proportion of convertibles.

Check conditions for normal sampling condition:n = 200 > 20# successes = 17 > 5# failures = 200 – 17 = 183 > 5Conditions are satisfied.PRGM: CRITVAL 1 – CONF LEVEL = .90Answer: 1.6449PRGM: STDERROR – 1:1 PROP – p’ = 17/200; n = 200Answer: .0197

Confidence Interval = 17/200 ± 1.6449*.0197 =0.0850 ± 0.0324 = (.0850 - .0324, .0850 + .0324)=(.0526, .1174)

17Section 9.2, Page 193

Confidence Interval for ProportionIllustrative Example TI-83 Black Box Program

In a discussion about the cars that fellow students drive, several statements were made about types, ages, makes colors, and so on. Dana decided he wanted to estimate the proportion of convertibles students drive, so he randomly identified 200 cars in the student parking lot and found 17 to be convertibles. Find the 90% confidence Interval for p, the true proportion of convertibles.

STAT – TESTS – A:1 – PROPZIntx: 17 (The number of success in p’: must be integer)n: 200C-Level: .90CalculateAnswer: (.0526, .1174)

We are 90% confident that the true proportion of convertibles is in the interval.

18Section 9.2, Page 193

Problems

19Problems, Page 206

Problems

20Problems, Page 206

A bank randomly selected 250 checking account customers and found that 110 of them also had savings accounts at the same bank.

a.Construct a 90% confidence interval for the true proportion of checking account customers who also have savings accounts.

b.What is the name of the sampling distribution used for the confidence interval?

c.Show that the necessary conditions for a sampling distribution are satisfied.

d.What is the standard error of the sampling distribution?.

Problems

21Problems, Page 206

In a sample of 60 randomly selected students, only 22 favored the amount being budgeted for next year’s intramural and interscholastic sports.

a.Construct the 99% confidence for the proportion of students who support the proposed budget amount.

b.What is the name of the sampling distribution?

c.Show that the necessary conditions for a sampling distribution are satisfied.

d.What is the standard error of the sampling distribution?.

Confidence Interval for ProportionRequired Sample Size

€

ME = z(α ) ×pq

n= z(α ) ×

p(1− p)

nSolving for n :

n =z(α )

ME

⎛

⎝ ⎜

⎞

⎠ ⎟2

× p(1− p)

22Section 9.2, Page 195

ME = Critical Value × Standard Error

If we have a good estimate of p, we use it. If we have no good estimate of p, we estimate p = .5. This will produce the largest sample for the given conditions. If p’ turns out to be different that p, our ME will be less than we initially required.

Required Sample SizeIllustrative Problem TI-83

Consider a manufacturer that purchases bolts from a supplier who claims the bolts are approximately 5% defective. How large a sample do we need to estimate the true proportion to be within ± .02 with 90% confidence?

PRGM – SAMPLSIZ – 1:PROPORTIONCONF LEVEL = .90ME = .02p EST = .05 (Problem estimate)Answer: 322

23Section 9.2, Page 195

Problems

24Problems, Page 208

Hypotheses Test - One ProportionIllustrative Problem – TI-83 Add-In

Ho: p = .61Ha: p > .61

p=.61 p’ = 235/350 = .6714

PRGM – NORMDIST – 1LOWER BOUND = .6714 – UPPERBOUND = 2ND EE99MEAN = .61 ; Answer: p-value = .0093 (Reject Ho, We proved that more than 61% sleep more that 7 or more hours)

p-value

25Section 9.2, Page 195€

σp ' = .61(1− .61) /350)

Sampling Distribution for p’

€

P −Value = P( p' > .6714 given p = .61)

p’

Hypotheses Test - One ProportionIllustrative Problem – TI-83 Black

Box

Ho: p = .61Ha: p > .61

STAT – TESTS – 5:1-PropZTestpo: .61x: 235 (# of successes in p’; must be integer)n: 350prop > po

CalculateAnswer: p-value = .0092 (Ho is rejected)

26Section 9.2, Page 196

Problems

a. Check the conditions for a normal sampling distribution.

b. State the hypotheses.

c. Find the p-value.

d. State your conclusion.

e. If you make an error in your conclusion, what type is it?

f. Find the mean of the sampling distribution.

g. Find the standard error of the sampling distribution.

27Problems, Page 207

A politician claims that she will receive at least 60% of the vote in an upcoming election. The results of a properly designed random sample of 100 voters showed that 57 of them will vote for her. Is the sample evidence sufficient to prove that her claim is false?

Problems

a. Check the conditions for a normal sampling distribution.

b. State the hypotheses.

c. Find the p-value.

d. State your conclusion

e. If you make an error in your conclusion, what type is it?

f. Find the mean of the sampling distribution.

g. Find the standard error of the sampling distribution.

28Problems, Page 207