Chapter 20Confidence Intervals and Hypothesis Tests for a

Population Mean ; t distributions• t distributions• confidence intervals for a

population mean using t distributions

• hypothesis tests for a population mean using t distributions

The Importance of the Central Limit Theorem

When we select simple random samples of size n, the sample means we find will vary from sample to sample. We can model the distribution of these sample means with a probability model that is

,Nn

Since the sampling model for x is the normal model, when we standardize x we get the

standard normal z

n

xz

nxSD

)( that Note

If is unknown, we probably don’t know either.

The sample standard deviation s provides an estimate of

the population standard deviation

For a sample of size n,

the sample standard deviation s is:

n − 1 is the “degrees of freedom.”

The value s/√n is called the standard error of x , denoted

SE(x).

nxSD

)(

2)(1

1xx

ns i

n

sxSE )(

Standardize using s for

Substitute s (sample standard deviation) for

n

xz

ssss sss s

n

xz

Note quite correct

Not knowing means using z is no longer correct

t-distributions

Suppose that a Simple Random Sample of size n is drawn from a

population whose distribution can be approximated by a N(µ, σ) model.

When is known, the sampling model for the mean x is N(/√n).

When is estimated from the sample standard deviation s, the sampling model for the mean x follows a t distribution t(, s/√n) with degrees of freedom n − 1.

is the 1-sample t statistic

t x s n

Confidence Interval Estimates CONFIDENCE CONFIDENCE

INTERVAL for INTERVAL for

where: t = Critical value from t-

distribution with n-1 degrees of freedom

= Sample mean s = Sample standard

deviation n = Sample size

For very small samples (n < 15), the data should follow a Normal model very closely.

For moderate sample sizes (n between 15 and 40), t methods will work well as long as the data are unimodal and reasonably symmetric.

For sample sizes larger than 40, t methods are safe to use unless the data are extremely skewed. If outliers are present, analyses can be performed twice, with the outliers and without.

n

stx

x

t distributions Very similar to z~N(0, 1) Sometimes called Student’s t

distribution; Gossett, brewery employee Properties:

i) symmetric around 0 (like z)

ii)degrees of freedom if > 1, E( ) = 0

if > 2, = - 2, which is always

bigger than 1.

t

-3 -2 -1 0 1 2 3

Z

0 1 2 3-1-2-3

z = x - x

x

t =

x -

s, s =

s

nx

x

x

Student’s t Distribution

-3 -2 -1 0 1 2 3

Z

t

0 1 2 3-1-2-3

n

x-x

= z

n

s - x

=t x

Student’s t Distribution

Figure 11.3, Page 372

-3 -2 -1 0 1 2 3

Z

t1

0 1 2 3-1-2-3

n

sx - x

=t

Student’s t Distribution

Figure 11.3, Page 372

Degrees of Freedom

s = s2

s =

(X X)

n -12

i2

i=1

n

-3 -2 -1 0 1 2 3

Z

t1

0 1 2 3-1-2-3

t7

Student’s t Distribution

Figure 11.3, Page 372

n

sx - x

=t Degrees of Freedom

s = s2

s =

(X X)

n -12

i2

i=1

n

Degrees of Freedom1 3.0777 6.314 12.706 31.821 63.6572 1.8856 2.9200 4.3027 6.9645 9.9250. . . . . .. . . . . .

10 1.3722 1.8125 2.2281 2.7638 3.1693. . . . . .. . . . . .

100 1.2901 1.6604 1.9840 2.3642 2.62591.282 1.6449 1.9600 2.3263 2.5758

0.80 0.90 0.95 0.98 0.99

t-Table: text- inside back cover

90% confidence interval; df = n-1 = 10

118125.1 :interval confidence%90

sx

0 1.8125

Student’s t Distribution

P(t > 1.8125) = .05

-1.8125.05.05

.90

t10

P(t < -1.8125) = .05

Comparing t and z Critical Values

Conf.

level n = 30

z = 1.645 90% t = 1.6991

z = 1.96 95% t = 2.0452

z = 2.33 98% t = 2.4620

z = 2.58 99% t = 2.7564

Example – An investor is trying to estimate the return

on investment in companies that won quality awards last year.

– A random sample of 41 such companies is selected, and the return on investment is recorded for each company. The data for the 41 companies have

– Construct a 95% confidence interval for the

mean return.

18.875.14 sx

40141freedom of degrees

18.875.14

sx

awards.quality

that wincompaniesfor investmenton return

mean population thecontains 17.36) (12.14,

interval that theconfident 95% are We

36.17,14.1261.275.1441

18.80211.275.14

2.0211 t table,- tfrom

n

stx

1.. nfdn

stx 1.. nfd

n

stx

Example Because cardiac deaths increase after heavy

snowfalls, a study was conducted to measure the cardiac demands of shoveling snow by hand

The maximum heart rates for 10 adult males were recorded while shoveling snow. The sample mean and sample standard deviation were

Find a 90% CI for the population mean max. heart rate for those who shovel snow.

15,175 sx

Solution 1.. nfdn

stx 1.. nfd

n

stx

shovelers snowfor rateheart maximum

mean thecontains 183.70) (166.30,

interval that theconfident 90% are We

)70.183,30.166(

70.817510

158331.1175

1.8331 ttable,- t theFrom

1015,175

nsx

EXAMPLE: Consumer Protection Agency

Selected random sample of 16 packages of a product whose packages are marked as weighing 1 pound.

From the 16 packages: a. find a 95% CI for the mean weight

of the 1-pound packages b. should the company’s claim that the

mean weight is 1 pound be challenged ?

1.10pounds, .36 poundx s

EXAMPLE

95% CI, n=16, df=15, x=1.10

s=.36

critical value of t is 2.1315

becomes

.361.10 (2.1315) 1.10 .19 .91, 1.29

16

Since 1 pound is in the interval, the company's

claim appears reasonable.

t

sx t

n

1.. nfdn

stx 1.. nfd

n

stx

Chapter 20Testing Hypotheses

about Means

22

Sweetness in cola soft drinksCola manufacturers want to test how much the sweetness of cola drinks is affected by storage. The sweetness loss due to storage was evaluated by 10 professional tasters by comparing the sweetness before and after storage (a positive value indicates a loss of sweetness):

Taster Sweetness loss

1 2.0 2 0.4 3 0.7 4 2.0 5 −0.4 6 2.2 7 −1.3 8 1.2 9 1.1 10 2.3

We want to test if storage results in a loss of sweetness, thus:

H0: = 0 versus HA: > 0

where is the mean sweetness loss due to storage.

We also do not know the population parameter , the standard deviation of the sweetness loss.

The one-sample t-test

As in any hypothesis tests, a hypothesis test for requires a few steps:

1. State the null and alternative hypotheses (H0 versus HA)

a) Decide on a one-sided or two-sided test

2. Calculate the test statistic t and determining its degrees of

freedom

3. Find the area under the t distribution with the t-table or

technology

4. State the P-value (or find bounds on the P-value) and interpret

the result

The one-sample t-test; hypotheses

Step 1:

1. State the null and alternative hypotheses (H0 versus HA)

a) Decide on a one-sided or two-sided test

H0: = versus HA: > (1 –tail test)

H0: = versus HA: < (1 –tail test)

H0: = versus HA: ≠ –tail test)

The one-sample t-test; test statistic

We perform a hypothesis test with null hypothesisH : = 0 using the test statistic

where the standard error of is .

When the null hypothesis is true, the test statistic follows a t distribution with n-1 degrees of freedom. We use that model to obtain a P-value.

0

( )

yt

SE y

y

( )s

SE yn

27

The one-sample t-test; P-Values

Recall:The P-value is the probability, calculated assuming the null hypothesis H0 is true, of observing a value of the test statistic more extreme than the value we actually observed.

The calculation of the P-value depends on whether the hypothesis test is 1-tailed(that is, the alternative hypothesis isHA : < 0 or HA : > 0)or 2-tailed(that is, the alternative hypothesis is HA : ≠ 0).

28

P-Values

If HA: > 0, then P-value=P(t > t0)

Assume the value of the test statistic t is t0

If HA: < 0, then P-value=P(t < t0)

If HA: ≠ 0, then P-value=2P(t > |t0|)

Sweetening colas (continued)

Is there evidence that storage results in sweetness loss in colas?

H0: = 0 versus Ha: > 0 (one-sided test) Taster Sweetness loss 1 2.0 2 0.4 3 0.7 4 2.0 5 -0.4 6 2.2 7 -1.3 8 1.2 9 1.110 2.3___________________________Average 1.02Standard deviation 1.196Degrees of freedom n − 1 = 9

Conf. Level 0.1 0.3 0.5 0.7 0.8 0.9 0.95 0.98 0.99Two Tail 0.9 0.7 0.5 0.3 0.2 0.1 0.05 0.02 0.01

One Tail 0.45 0.35 0.25 0.15 0.1 0.05 0.025 0.01 0.005

df Values of t      9 0.1293 0.3979 0.7027 1.0997 1.3830 1.8331 2.2622 2.8214 3.2498

2.2622 < t = 2.70 < 2.8214; thus 0.01 < P-value < 0.025.

Since P-value < .05, we reject H0. There is a significant loss

of sweetness, on average, following storage.

9( 2.70)P value P t

0 1.02 02.70

1.196 10

yt

s n

Finding P-values with ExcelTDIST(x, degrees_freedom, tails)

TDIST = P(t > x) for a random variable t following the t distribution (x positive). Use it in place of t-table to obtain the P-value.

– x  is the absolute value of the test statistic.

– Deg_freedom   is an integer indicating the number of degrees of freedom.

– Tails   specifies the number of distribution tails to return. If tails = 1, TDIST returns the one-tailed P-value. If tails = 2, TDIST returns the two-tailed P-value.

Sweetness in cola soft drinks (cont.)

31

2.2622 < t = 2.70 < 2.8214; thus 0.01 < p < 0.025.0 1.02 0

2.701.196 10

yt

s n

New York City Hotel Room Costs

The NYC Visitors Bureau claims that the average cost of a hotel room is $168 per night. A random sample of 25 hotels resulted in

y = $172.50 and

s = $15.40.

H0: μ= 168

HA: μ

168

n = 25; df = 24

New York City Hotel Room Costs

Do not reject H0: not sufficient evidence that true mean cost is different than $168

.079

0

.079

y μ 172.50 168t 1.46

s 15.40

n 25

1. 46

H0: μ= 168

HA: μ

168

-1. 46

$172.50, $15.40y s

t, 24 df

2 ( 1.46)P value P t Conf. Level 0.1 0.3 0.5 0.7 0.8 0.9 0.95 0.98 0.99Two Tail 0.9 0.7 0.5 0.3 0.2 0.1 0.05 0.02 0.01One Tail 0.45 0.35 0.25 0.15 0.1 0.05 0.025 0.01 0.005

df Values of t  24 0.1270 0.3900 0.6848 1.0593 1.3178 1.7109 2.0639 2.4922 2.7969

P-value = .158

0.1 ≤ P-value ≤ 0.2

Microwave PopcornA popcorn maker wants a combination of

microwave time and power that delivers high-quality popped corn with less than 10% unpopped kernels, on average. After testing, the research department determines that power 9 at 4 minutes is optimum. The company president tests 8 bags in his office microwave and finds the following percentages of unpopped kernels: 7, 13.2, 10, 6, 7.8, 2.8, 2.2, 5.2.

Do the data provide evidence that the mean percentage of unpopped kernels is less than 10%?

H0: μ= 10HA: μ 10where μ is true unknown mean percentage of unpopped kernels

n = 8; df = 7

Microwave Popcorn

Reject H0: there is sufficient evidence that true mean percentage of unpopped kernels is less than 10%

.02

0

6.775 102.51

3.64

8

yt

s

n

H0: μ= 10

HA: μ 10

-2. 516.775, 3.64y s

t, 7 df

( 2.51)P value P t Exact P-value = .02

Conf. Level 0.1 0.3 0.5 0.7 0.8 0.9 0.95 0.98 0.99Two Tail 0.9 0.7 0.5 0.3 0.2 0.1 0.05 0.02 0.01One Tail 0.45 0.35 0.25 0.15 0.1 0.05 0.025 0.01 0.005

df Values of t  7 0.1303 0.4015 0.7111 1.1192 1.4149 1.8946 2.3646 2.9980 3.4995