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# Ch11 Hypothesis Testing Involving Two Sample Means or Proportions(商务统计导论-英文版)

CHAPTER 11: Hypothesis Testing Involving Two Sample Means or Proportions
to accompany

fourth edition, by Ronald M. Weiers

Presentation by Priscilla Chaffe-Stengel Donald N. Stengel

Chapter 11 - Learning Objectives
? Select and use the appropriate hypothesis test in comparing
– – – – Means of two independent samples Means of two dependent samples Proportions of two independent samples Variances of two independent samples

? Construct and interpret the appropriate confidence interval for differences in
– Means of two independent samples – Means of two dependent samples – Proportions of two independent samples

Chapter 11 - Key Terms
? Independent vs dependent samples ? Pooled estimate of the
– common variance – common standard deviation – population proportion

? Standard error of the estimate for the
– difference of two population means – difference of two population proportions

? Matched, or paired, observations ? Average difference

Independent vs Dependent Samples
? Independent Samples:
Samples taken from two different populations, where the selection process for one sample is independent of the selection process for the other sample.

? Dependent Samples:

Samples taken from two populations where either (1) the element sampled is a member of both populations or (2) the element sampled in the second population is selected because it is similar on all other characteristics, or “matched,” to the element selected from the first population

Examples: Independent versus Dependent Samples
? Independent Samples:
– Testing a company’s claim that its peanut butter contains less fat than that produced by a competitor.

? Dependent Samples:
– Testing the relative fuel efficiency of 10 trucks that run the same route twice, once with the current air filter installed and once with the new filter.

Identifying the Appropriate Test Statistic

? Are the data from measurements (continuous variables) or counts (discrete variables)? ? Are the data from independent samples? ? Are the population variances approximately equal? ? Are the populations approximately normally distributed? ? What are the sample sizes?

Test of (μ1 – μ2), s1 = s2, Populations Normal
? Test Statistic
[x – x ] – [m – m ] t = 1 2 ? 1 ?2 0 ? ÷ 2?? 1 + 1 ÷ sp n n ÷ ? ÷ ? ? 1 ? 2÷ (n –1)×2 + (n –1)× 2 s s 1 2 2 where s p2 = 1 n +n – 2 1 2 and df = n1 + n2 – 2

Example: Equal-Variances t-Test
? Problem 11.2: An educator is considering two different
videotapes for use in a half-day session designed to introduce students to the basics of economics. Students have been randomly assigned to two groups, and they all take the

same written examination after viewing the videotape. The scores are summarized below. Assuming normal populations with equal standard deviations, does it appear that the two videos could be equally effective? What is the most accurate statement that could be made about the p-value for the test? Videotape 1: x1= 77.1, s1 = 7.8, n1 = 25 Videotape 2: x = 80.0, s2 = 8.1, n2 = 25 2

t-Test, Two Independent Means
? I. H0: μ1 – μ2 = 0 H1: μ1 – μ2 ? 0
The two videotapes are equally effective. There is no difference in student performance. The two videotapes are not equally effective. There is a difference in student performance.

? II. Rejection Region

Reject H 0

Do Not Reject H

a = 0.05 ???? ????? ????? df = 25 + 25 – 2 = 48 t=-2.011 t=2.011 Reject H0 if t > 2.011 or t < –2.011

0

Reject H 0

t-Test, Problem 11.2 cont.
? III. Test Statistic
24× .8)2 + 24×.1)2 = 1460.16 + 1564.64 = 63.225 (8 s p2 = (7 25 + 25 – 2 48 t= x –x 1 2 s p2 1 + 1 n n 1 2
? ? ? ? ? ? ? ? ? ÷ ÷ ÷ ÷ ÷ ÷ ?

=

77.1– 80.0 = –1.289 ? ? 1 1 ?÷ 63.225?? + ÷ ? 25 ? 25 ÷

t-Test, Problem 11.2 cont.
? IV. Conclusion:
Since the test statistic of t = – 1.289 falls between the critical bounds of t = ± 2.011, we do not reject the null hypothesis with at least 95% confidence.

? V. Implications:
There is not enough evidence for us to conclude that one videotape training session is more effective than the other.

? p-value:
Using Microsoft Excel, type in a cell: =TDIST(1.289,48,2) The answer: p-value = 0.203576

Test of (μ1 – μ2), Unequal Variances, Independent Samples
? Test Statistic
( x1 ? x2 ) ? ( m1 ? m 2 )0 t= s12 s22 + n1 n2

?(s where df =

n1 ) + ( s n2 )? ( s n1 ) 2 ( s22 n2 ) 2 + n1 ? 1 n2 ? 1
2 1 2 1 2 2 2

Example, Unequal-Variances t-Test, Independent Samples
? Suppose analysis of two independent samples from normally distributed populations reveal the following values: x1 = 120, s1 = 16, n1 = 35 x2 = 114, s2 = 12, n2 = 42 What degrees of freedom should be used on the unequal-variances t-test of the differences in their means?

Example, Calculation of the Degrees of Freedom for the t-Test
?( s df =
35) + (12 42 )? 2 2 2 2 = 62 .04 (16 35) (12 42 ) + 34 41 So we would use a t-test with 62 degrees of freedom to test the differences in the means of the two populations.
2 2 2

?(16 =

n1 ) + ( s n2 )? ( s n1 ) 2 ( s22 n2 ) 2 + n1 ? 1 n2 ? 1
2 1 2 1 2 2 2

Test of Independent Samples (μ1 – μ2), s1 ?s2, n1 and n2 ? 30
? Test Statistic
[x – x ]–[m – m ] 1 20 z = 1 2 s2 s 2 1 + 2 n n 1 2
– with s12 and s22 as estimates for s12 and s22

? Test Statistic

Test of Dependent Samples (μ1 – μ2)

= μd
t=s d d n

– where d = (x1 – x2) d = Sd/n, the average difference n = the number of pairs of observations sd = the standard deviation of d df = n – 1

? Test Statistic

Test of (p1 – p2), where n1p1?5, n1(1–p1)?5, n2p2?5, and n2 (1–p2 )??
p ?p 1 2 z= ? ? ? 1 1 p ?(1? p) ? ?n + n ?? ? 2 ?? ? 1

– where p1 = observed proportion, sample 1 p2 = observed proportion, sample 2 n1 = sample size, sample 1 n2 = sample size , sample 2 n p + n p 2 2 p = 1 1 n + n 1 2 ? 2002 The Wadsworth Group

Testing for Equal Variances
? Pooled-variances t-test assumes the two population variances are equal. ? The F-test can be used to test that assumption. ? The F-distribution is the sampling distribution of s12/s22 that would result if two samples were repeatedly drawn from a single normally distributed population.

Test of s12 = s22
? If s12 = s22 , then s12/s22 = 1. So the hypotheses can be worded either way.
s2 s 2 ? Test Statistic: F = 1 or 2 whichever is larger s 2 s2 2 1 ? The critical value of the F will be F(a/2, n1, n2)
– where a = the specified level of significance n1 = (n – 1), where n is the size of the sample with the larger variance n2 = (n – 1), where n is the size of the sample with the smaller variance

Returning to Problem 11.2, let us test with 95% confidence whether it was reasonable for us to assume that the two population variances were approximately equal. I. H0: s22/s12 = 1 H1: s22/s12 ? 1 II. Rejection Region Do Not Reject H a/2 = 0.025 Reject H 0 0 0.975 numerator df = 24 ????? denominator df = 24 F=2.27 If F > 2.27, reject H0, meaning it was not reasonable for us to assume the population variances were approximately equal.

Testing for Equal Variances An Example

III. Test Statistic s 2 2 F = 2 = 8.1 = 1.0784 s2 7.82 1 IV. Conclusion Since the test statistic of F = 1.078 falls below the critical value of F = 2.27, we do not reject H0 with at most 5% error. V. Implications There is not enough evidence to support a conclusion that the two populations have different variances. The pooled variances t-test can be used in analyzing these data.

Testing for Equal Variances An Example, cont.

Confidence Interval for (μ1 – μ2)
? The (1 – a)% confidence interval for the difference in two means:
– Equal-variances t-interval
1 1 ( x – x ) ± a ×s 2 + t p 1 2 2 n n 1 2
? ? ? ? ? ? ? ? ? ? ÷ ÷ ÷ ÷ ÷ ÷ ÷ ?

– Unequal-variances t-interval
s2 s 2 ( x – x ) ±t a × 1 + 2 1 2 n 2 n 1 2

Confidence Interval for (μ1 – μ2)
? The (1 – a)% confidence interval for the difference in two means:
– Known-variances z-interval

( x1 ? x2 ) ? za 2

s

2 1

n1

+

s

2 2

n2

Confidence Interval for (p1 – p2)
? The (1

– a)% confidence interval for the difference in two proportions:
p (1– p ) p (1– p ) 1 + 2 2 (p – p ) ±z a × 1 1 2 n n 2 1 2
– when sample sizes are sufficiently large.