[2] Hypothesis Testing on Two Populations

•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

•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.

• 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

© 2002 The Wadsworth Group

• Test Statistic

[x – x ] – [m – m ] 1 20 t= 1 2      

s p2 n1 + n1 1 2

     

where

(n –1)s 2 + (n –1)s 2 1 2 2 s p2 = 1 n +n –2 1 2 and df = n1 + n2 – 2; © 2002 The Wadsworth Group

• 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

• 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.

• 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

© 2002 The Wadsworth Group

• Test Statistic ◦ where

t=s d d

n 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



A study is conducted to whether different training methods have an effect on the productivity of employees in a company manufacturing electronic equipment. Twelve recently hired employees were divided into two groups of 6. The first group received a computer-assisted, individual-based training program, and the other received a collaborative team-based training program. After the training, the employees were evaluated on the time (in seconds) it takes to assemble an electronic part. The data from the study are tabulated below.

Team

Assembly Time (in seconds)

Computer- 19.4 assisted individualbased

19.4

20.7

21.8

19.3

18.5

Teambased program

15.6

16.0

21.7

30.7

20.8

22.4

Is there a sufficient evidence to conclude that employees under computer-assisted individual-based program have significantly faster assembly time than those employees under team-based program? Use 5% level of significance. (a) Assume that the variances of the assembly of training methods are equal. (b) Assume that the variances of the assembly of training methods are equal.

•Problem : 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? x 1 Videotape 1: x = 77.1, s1 = 7.8, n1 = 25 Videotape 2: 2 = 80.0, s2 = 8.1, n2 = 25

• I. H0:

µ1 – µ2 = 0 The two videotapes are

equally effective. There is no difference in student performance.

H1:

µ1 – µ2  0 The two videotapes are not

equally effective. There is a difference in student performance.

• II. Rejection Region a = 0.05 Reject H 0 df = 25 + 25 – 2 = 48 0.025 Reject H0 if t > 2.011 or t < –2.011

Do Not Reject H

0.95

t=-2.011

0

Reject H 0

0.025 t=2.011

© 2002 The Wadsworth Group

• Test Statistic 2 + 24(8.1)2 1460.16 + 1564.64  24 ( 7 . 8 ) = = 63.225 s p2 = 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 

• 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



A taxi company is trying to decide whether the use of radial tires instead of regular belted tires improves fuel economy. Twelve cars were equipped radial tires and driven over a prescribed test course. Without changing drivers, the same cars were then equipped with regular belted tires and driven once again over the test course. The gasoline consumption, in kilometers per liter, was recorded as follows:

Car 1 2 3 4 5 6 7 8 9 10 11 12 Radial Tires 4.2 4.7 6.6 7.0 6.7 4.5 5.7 6.0 7.4 4.9 6.1 5.2 Belted Tires 4.1 4.9 6.2 6.9 6.8 4.4 5.7 5.8 6.9 4.7 6.0 4.9 At the 0.01 level, can we conclude that cars equipped with radial tires give better fuel economy than those equipped with belted tires?

• Test Statistic

◦ where n p + p = 1 1 n + 1

n p 2 2 n 2



Suppose that in a poll survey, 925 out of 2500 respondents would like candidate A to be elected as the president of the country, and 840 out of 2500 would like candidate B to succeed as the president. Do we have reason to believe that the candidate A would win over candidate B as the president? Use 5% level of significance.



In a test of the quality of two television commercials, each commercial was shown in a separate test area six times over a one-week period. The following week a telephone survey was conducted to identify individuals who had seen the commercials. Those individuals were asked to state the primary message in the commercials. The following results were recorded: 

 

Number Who Saw Commercial Number Who Recalled Message

Commercial A

150 63

Commercial B

200 60



Use 5% level of significance and test the hypothesis that there is no difference in the recall proportions for two commercials.







The Bureau of Transportation tracks the flight arrival performances of the 10 biggest airlines in the United States (Wall Street Journal, 2003). Flights that arrive within 15 minutes of schedule are considered on time. Using sample data below: January 2001: A sample of 924 flights showed 742 on time January 2002: A sample of 841 flights showed 714 on time



State the hypotheses that could be tested to determine whether the major airlines improved on-time flight performance during the one-year period. What is your conclusion at 5% level of significance.

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

    

A firm is studying the delivery time of two raw material suppliers. The firm is basically satisfied with supplier A and is prepared to stay with that supplier if the mean delivery time is the same or less than that of supplier B. However, if the firm finds that the mean delivery time of supplier B is less that that of supplier A, it will begin making raw material purchases from supplier B. Supplier A Supplier B n1 = 50 n2 = 31 mean= 14 days mean = 12.5 days s1 = 3 days s2 = 2 days

What are the null and alternative hypotheses? With 5% level of significance, what action do you recommend in terms of supplier selection?