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Chapter 9 Inferences from Two Samples

In this chapter we will deal with two samples from two populations.

The general goal is to compare the parameters of the two populations.

For the first population we use index 1, for the second population index 2.

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Section 9-2 Two Proportions

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p2, n

2 , x

2 , p

2, and q

2 are used for the second population.^^

Notation for Two ProportionsFor the first population, we let:

p1 = first population proportion

n1 = size of the first sample

x1 = number of successes in the first sample

p̂1 = (the first sample proportion)

q1 = 1 – p

1 ^^

n1

x1

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The pooled sample proportion

is denoted by p and is given by:

Pooled Sample Proportion

=p n1 + n

2

x1 + x

2

We denote q = 1 – p

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Requirements

1. We have two independent random samples.

2. For each of the two samples, the number of successes is at least 5 and the number of failures is at least 5.

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Tests for Two Proportions

H0: p1 = p

2

H1: p1

p2 , H1: p1

< p2 , H1: p

1> p

2

Note: no numerical values for p1 or p

2 are

claimed in the hypotheses.

The goal is to compare the two proportions.

two tails left tail right tail

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Test Statistic for Two Proportions

+

z =( p

1 – p

2 ) – ( p

1 – p

2 )^ ^

n1

pqn2

pq

=p1

^ x1

n1

p2

^ x2

n2

=and

and q = 1 – pn

1 + n

2

p = x

1 + x

2

Note: p1 – p

2 =0 according to H0

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Example:

The table below lists results from a simple random sample of front-seat occupants involved in car crashes. Use a 0.05 significance level to test the claim that the fatality rate of occupants is lower for those in cars equipped with airbags.

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Example:

Requirements are satisfied: two simple random samples, two samples are independent; Each has at least 5 successes and 5 failures.

Step 1: Express the claim as p1 < p2.

Step 2: p1 < p2 does not contain equality so it is the alternative hypothesis. The null hypothesis is the statement of equality.

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Example:

H0: p1 = p2 H1: p1 < p2 (original claim)

Step 3: Significance level is 0.05

Step 4: Compute the pooled proportion:

p

x1 x

2

n1 n

2

41 52

11,541 9,8530.004347

With p 0.004347 it follows q 0.995653

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Example:Step 5: Find the value of the test statistic.

41 520

11 541 9 853

0 004347 0 995653 0 004347 0 995653

11 541 9 853

, ,

. . . .

, ,

z 1.91

z p̂1 p̂2 p1 p2

pqn1

pqn2

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Example:

Left-tailed test. Area to left of z = –1.91 is 0.0281 (Table A-2), so the P-value is 0.0281.

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Example:

Step 6: Because the P-value of 0.0281 is less than the significance level of = 0.05, we reject the null hypothesis of p1 = p2.

Because we reject the null hypothesis, we conclude that there is sufficient evidence to support the original claim.

Final conclusion:the proportion of accident fatalities for occupants in cars with airbags is less than the proportion of fatalities for occupants in cars without airbags.

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Example: Using the Traditional Method

the critical value is z = –1.645. The test statistic of z = –1.91 does fall in the critical region bounded by the critical value of z = –1.645.

We again reject the null hypothesis.

With a significance level of = 0.05 in a left- tailed test,

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Confidence Interval Estimate of p

1 – p

2

n1 n2

p1 q

1p

2 q

2+^ ^^ ^

where E = z

( p1

– p2 ) – E < ( p

1 – p

2 ) < ( p

1 – p

2 ) + E ^ ^ ^^

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Example:

Use the same sample data to construct a 90% confidence interval estimate of the difference between the two population proportions.

Note: 1─ = 0.90, so = 0.10 and = 0.05 .

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Example:

Requirements are satisfied as we saw earlier.

90% confidence interval: z/2 = 1.645Calculate the margin of error, E

1.645

4111,541

11,50011,541

11,541

529,853

98019,853

9,853

0.001507

E z 2

p̂1q̂1

n1

p̂2q̂2

n2

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Example:Construct the confidence interval

0.003553 0.005278 0.001507

p1 p2 0.003553 0.005278 0.001507

p̂1 p̂2 E p1 p2 p̂1 p̂2 E

0.00323 p1 p2 0.000218

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Final note:

The confidence interval limits do not contain 0, implying that there is a significant difference between the two proportions.

Thus the confidence interval, too, suggests that the fatality rate is lower for occupants in cars with air bags than for occupants in cars without air bags.

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• Press STAT and select TESTS• Scroll down to 2-PropZTest press ENTER

• Type in x1: (number of successes in 1st sample)

• n1: (number of trials in 1st sample)

• x2: (number of successes in 2nd sample)

• n2: (number of trials in 2nd sample)

• choose H1: p1 ≠p2 <p2 >p2

(two tails) (left tail) (right tail)• Press on Calculate• Read test statistic z=… and P-value p=…

Two proportions by TI-83/84

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• Press STAT and select TESTS• Scroll down to 2-PropZInt press ENTER

• Type in x1: (number of successes in 1st sample)

• n1: (number of trials in 1st sample)

• x2: (number of successes in 2nd sample)

• n2: (number of trials in 2nd sample)

C-Level: (confidence level)• Press on Calculate• Read the interval (…,…)

Two proportions by TI-83/84

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Section 9-3

Two Means: Independent Samples

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Definitions

Two samples are independent if the sample values selected from one population are not related to or somehow paired or matched with the sample values from the other population.

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1 = population mean

σ1 = population standard deviation

n1 = size of the first sample

= sample mean

s1 = sample standard deviation

Corresponding notations for 2, σ2, s2, and n2 apply to the second population.

x1

x2

Notation for the first population:

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Requirements

1. σ1 an σ2 are unknown and no assumption is made about the equality of σ1 and σ2 .

2. The two samples are independent.

3. Both samples are random samples.

4. Either or both of these conditions are satisfied: The two sample sizes are both large (with n1 > 30 and n2 > 30) or both populations have normal distributions.

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Tests for Two Means

H0: 1 =

2

H1: 1

2 , H1: 1

< 2 , H1:

1>

2

Note: no numerical values for 1 or

2 are

claimed in the hypotheses.

The goal is to compare the two means.

two tails left tail right tail

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Hypothesis Test for Two Means with Independent Samples:

Test Statistic is

t x

1 x

2 1

2 s

12

n1

s

22

n2

Note: 1 –

2 =0 according to H0

Degrees of freedom: df = smaller of n1 – 1 and n2 – 1.

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Example:A headline in USA Today proclaimed that “Men, women are equal talkers.” That headline referred to a study of the numbers of words that men and women spoke in a day, see below.

Use a 0.05 significance level to test the claim that men and women speak the same mean number of words in a day.

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Example:Requirements are satisfied: two population standard deviations are not known and not assumed to be equal, independent samples, both samples are large.

Step 1: Express claim as 1 = 2.

Step 2: If original claim is false, then 1 ≠ 2.

Step 3: Alternative hypothesis does not contain equality, null hypothesis does.

H0 : 1 = 2 (original claim) H1 : 1 ≠ 2

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Example:Step 4: Significance level is 0.05

Step 5: Use a t distribution

Step 6: Calculate the test statistic

t x

1 x

2 1

2 s

12

n1

s

22

n2

15,668.5 16,215.0 0

8632.52

186

7301.22

210

0.676

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Example:Use Table A-3: area in two tails is 0.05, df = 185, which is not in the table, the closest value is

t = ±1.972

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Example:Step 7: Because the test statistic does not fall

within the critical region, fail to reject the null hypothesis:1 = 2 (or 1 – 2 = 0).

Final conclusion:There is sufficient evidence to support the claim that men and women speak the same mean number of words in a day.

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Confidence Interval Estimate of1 – 2: Independent Samples

df = smaller n1 – 1 and n2 – 1

(x1 – x2) – E < (µ1 – µ2) < (x1 – x2) + E

+n1 n2

s1 s2where E = t

22

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Example:

Using the given sample data, construct a 95% confidence interval estimate of the difference between the mean number of words spoken by men and the mean number of words spoken by women.

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Example:

Find the margin of Error, E; use t/2 = 1.972

Construct the confidence interval use E = 1595.4 and

E t 2

s12

n1

s

22

n2

1.9728632.52

186

7301.22

2101595.4

x1 x

2 E 1

2 x1 x

2 E

2141.9 1

2 1048.9

x115,668.5 and x

216,215.0.

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• Press STAT and select TESTS• Scroll down to 2-SampTTest press ENTER• Select Input: Data or Stats. For Stats:

• Type in x1: (1st sample mean)

• sx1: (1st sample st. deviation)

• n1: (1st sample size)

• x2: (2nd sample mean)

• sx2: (2nd sample st. deviation)

• n2: (2nd sample size)

choose H1: 1 ≠2 <2 >2

(two tails) (left tail) (right tail)

Tests about two means by TI-83/84

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• choose Pooled: No or Yes (always No)

Tests about two means (continued)

• Press on Calculate• Read the test statistic t=…• and the P-value p=…• Note: the calculator gives a more accurate

P-value than the book does, because it uses a more accurate formula for degrees of freedom (see the line df=… in the calculator). The book adopts a simple but inaccurate rule df=smaller of n1-1 and n2-1.

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• Press STAT and select TESTS• Scroll down to 2-SampTInt press ENTER• Select Input: Data or Stats. For Stats:

• Type in x1: (1st sample mean)

• sx1: (1st sample st. deviation)

• n1: (1st sample size)

• x2: (2nd sample mean)

• sx2: (2nd sample st. deviation)

• n2: (2nd sample size)

• C-Level: confidence level

Intervals for two means by TI-83/84

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• Press on Calculate• Read the confidence interval (…,…)• Note: the calculator gives a more accurate

confidence interval than the book does, because it uses a more accurate formula for degrees of freedom (see the line df=… in the calculator). The book adopts a simple but inaccurate rule df=smaller of n1-1 and n2-1.

• choose Pooled: No or Yes (always No)

Intervals for two means (continued)