The previous two chapters cure the questions of estimating and also making inferences around a parameter the a solitary population. In this thing we think about a compare of parameters that belong to two various populations. For example, we could wish to to compare the average earnings of all adult in one region of the country with the average earnings of those in an additional region, or we could wish to compare the proportion of all men who are vegetarians with the ratio of all females who room vegetarians.

You are watching: The mean game scores and standard deviations of four seasons of a football team are given below.

We will study construction of to trust intervals and tests the hypotheses in four situations, depending on the parameter the interest, the size of the samples drawn from every of the populations, and the an approach of sampling. We likewise examine sample size considerations.

## 9.1 comparison of Two population Means: Large, elevation Samples

### Learning Objectives

To understand the logical frame for estimating the difference in between the method of two distinct populations and also performing tests of hypotheses concerning those means. To learn just how to build a confidence interval because that the distinction in the method of two distinct populations making use of large, elevation samples. Come learn how to do a check of hypotheses concerning the difference in between the means of two distinctive populations using large, live independence samples.Suppose we wish to to compare the means of two distinctive populations. Number 9.1 "Independent Sampling from 2 Populations" illustrates the theoretical framework the our investigation in this and also the following section. Each populace has a mean and also a traditional deviation. Us arbitrarily label one populace as populace 1 and the other as populace 2, and subscript the parameters through the numbers 1 and 2 to tell them apart. We draw a random sample from population 1 and label the sample statistics the yields through the subscript 1. Without referral to the an initial sample we attract a sample from populace 2 and label the sample statistics v the subscript 2.

Figure 9.1 independent Sampling from 2 Populations

### Definition

*Samples from two distinctive populations are* **independent** *if each one is drawn without recommendation to the other, and also has no link with the other.*

Our goal is to usage the info in the *samples* to calculation the distinction μ1−μ2 in the means of the 2 *populations* and to make statistically valid inferences around it.

## Confidence Intervals

Since the mean x-1 the the sample drawn from population 1 is a good estimator the μ1 and also the median x-2 of the sample attracted from population 2 is a good estimator of μ2, a reasonable allude estimate the the distinction μ1−μ2 is x-1−x-2. In order to widen this point estimate into a to trust interval, we first suppose the both samples space large, the is, that both n1≥30 and also n2≥30. If so, then the adhering to formula for a confidence interval for μ1−μ2 is valid. The icons s12 and also s22 signify the squares of *s*1 and also *s*2. (In the relatively rare instance that both populace standard deviations σ1 and σ2 are recognized they would certainly be used instead of the sample standard deviations.)

### 100(1−α)% trust Interval because that the Difference between Two populace Means: Large, elevation Samples

(x-1−x-2)±zα∕2s12n1+s22n2The samples must be independent, and also *each* sample need to be large: n1≥30 and n2≥30.

### Example 1

To compare customer satisfaction level of two completing cable television companies, 174 client of agency 1 and also 355 customers of firm 2 were randomly selected and were inquiry to rate their cable suppliers on a five-point scale, v 1 being least satisfied and 5 most satisfied. The survey results are summary in the adhering to table:

company 1 firm 2

n1=174 | n2=355 |

x-1=3.51 | x-2=3.24 |

s1=0.51 | s2=0.52 |

Construct a point estimate and also a 99% trust interval for μ1−μ2, the distinction in median satisfaction levels of customers of the two companies as measure up on this five-point scale.

Solution:

The point estimate the μ1−μ2 is

x-1−x-2=3.51−3.24=0.27.In words, we estimate that the median customer satisfaction level for company 1 is 0.27 points greater on this five-point scale than the is for company 2.

To use the formula because that the trust interval, proceed exactly as was done in thing 7 "Estimation". The 99% confidence level means that α=1−0.99=0.01 so that zα∕2=z0.005. From figure 12.3 "Critical worths of " we read straight that z0.005=2.576. Thus

(x-1−x-2)±zα∕2s12n1+s22n2=0.27±2.5760.512174+0.522355=0.27±0.12We space 99% confident that the difference in the population means lies in the interval <0.15,0.39>, in the feeling that in recurring sampling 99% of all intervals created from the sample data in this manner will contain μ1−μ2. In the paper definition of the problem we to speak we are 99% confident that the median level of customer satisfaction for agency 1 is between 0.15 and also 0.39 clues higher, on this five-point scale, 보다 that for firm 2.

## Hypothesis Testing

Hypotheses concerning the loved one sizes the the way of two populaces are tested utilizing the same an important value and also *p*-value actions that were used in the case of a single population. All that is necessary is come know just how to to express the null and alternative hypotheses and to know the formula for the standardized test statistic and the circulation that it follows.

The null and different hypotheses will always be expressed in terms of the distinction of the two population means. Hence the null hypothesis will always be written

H0:μ1−μ2=D0where *D*0 is a number the is deduced native the statement of the situation. As was the situation with a single population the alternate hypothesis deserve to take one of the 3 forms, through the same terminology:

kind of Ha ax

Ha:μ1−μ2D0 | Left-tailed |

Ha:μ1−μ2>D0 | Right-tailed |

Ha:μ1−μ2≠D0 | Two-tailed |

As long as the samples are independent and both are large the adhering to formula because that the standardized test statistic is valid, and also it has the typical normal distribution. (In the reasonably rare instance that both populace standard deviations σ1 and σ2 are recognized they would certainly be used instead of the sample typical deviations.)

### Standardized check Statistic for theory Tests worrying the Difference in between Two population Means: Large, independent Samples

Z=(x-1−x-2)−D0s12n1+s22n2The test statistic has actually the typical normal distribution.

The samples have to be independent, and also *each* sample must be large: n1≥30 and also n2≥30.

### Example 2

Refer to note 9.4 "Example 1" concerning the median satisfaction levels of client of two contending cable television companies. Test at the 1% level of significance whether the data carry out sufficient evidence to break up that agency 1 has a higher mean satisfaction rating 보다 does company 2. Use the vital value approach.

Solution:

Step 1. If the median satisfaction level μ1 and also μ2 are the same then μ1=μ2, however we always express the null theory in terms of the difference in between μ1 and μ2, therefore *H*0 is μ1−μ2=0. To say that the average customer satisfaction for firm 1 is greater than the for firm 2 method that μ1>μ2, i beg your pardon in regards to their distinction is μ1−μ2>0. The check is therefore

α=0.01

Step 2. Due to the fact that the samples room independent and both are huge the check statistic is

Z=(x-1−x-2)−D0s12n1+s22n2Step 3. Inserting the data into the formula because that the check statistic gives

Z=(x-1−x-2)−D0s12n1+s22n2=(3.51−3.24)−00.512174+0.522355=5.684### Example 3

Perform the check of keep in mind 9.6 "Example 2" using the *p*-value approach.

Solution:

The first three steps are the same to those in keep in mind 9.6 "Example 2".

Figure 9.3 *P*-Value for note 9.7 "Example 3"

Step 5. Because 0.0000 p -valueα so the decision is to reject the null hypothesis:

The data administer sufficient evidence, in ~ the 1% level of significance, to conclude that the average customer satisfaction for firm 1 is greater than that for firm 2.

### Key Takeaways

A point estimate for the difference in 2 population means is simply the distinction in the matching sample means. In the context of estimating or trial and error hypotheses worrying two populace means, “large” samples means that*both*samples space large. A trust interval for the difference in 2 population means is computed utilizing a formula in the very same fashion together was done because that a single population mean. The exact same five-step procedure supplied to test hypotheses concerning a single population mean is offered to test hypotheses concerning the difference in between two populace means. The only distinction is in the formula for the standardized test statistic.

### Exercises

### Basic

Construct the to trust interval because that μ1−μ2 because that the level of confidence and the data native independent samples given.

90% confidence,

n1=45, x-1=27, s1=2

n2=60, x-2=22, s2=3

99% confidence,

n1=30, x-1=−112, s1=9

n2=40, x-2=−98, s2=4

Construct the to trust interval for μ1−μ2 because that the level the confidence and also the data from independent samples given.

95% confidence,

n1=110, x-1=77, s1=15

n2=85, x-2=79, s2=21

90% confidence,

n1=65, x-1=−83, s1=12

n2=65, x-2=−74, s2=8

Construct the to trust interval because that μ1−μ2 because that the level of confidence and the data from independent samples given.

99.5% confidence,

n1=130, x-1=27.2, s1=2.5

n2=155, x-2=38.8, s2=4.6

95% confidence,

n1=68, x-1=215.5, s1=12.3

n2=84, x-2=287.8, s2=14.1

Construct the trust interval for μ1−μ2 because that the level of confidence and the data native independent samples given.

99.9% confidence,

n1=275, x-1=70.2, s1=1.5

n2=325, x-2=63.4, s2=1.1

90% confidence,

n1=120, x-1=35.5, s1=0.75

n2=146, x-2=29.6, s2=0.80

Perform the test of hypotheses indicated, utilizing the data native independent samples given. Use the an important value approach. Compute the *p*-value the the test together well.

Test H0:μ1−μ2=3 vs. Ha:μ1−μ2≠3

α=0.05,

n1=35, x-1=25, s1=1

n2=45, x-2=19, s2=2

Test H0:μ1−μ2=−25 vs. Ha:μ1−μ2−25

α=0.10,

n1=85, x-1=188, s1=15

n2=62, x-2=215, s2=19

Perform the check of hypotheses indicated, making use of the data indigenous independent samples given. Use the vital value approach. Compute the *p*-value that the test as well.

Test H0:μ1−μ2=45 vs. Ha:μ1−μ2>45

α=0.001,

n1=200, x-1=1312, s1=35

n2=225, x-2=1256, s2=28

Test H0:μ1−μ2=−12 vs. Ha:μ1−μ2≠−12

α=0.10,

n1=35, x-1=121, s1=6

n2=40, x-2=135, s2=7

Perform the check of hypotheses indicated, making use of the data indigenous independent samples given. Use the critical value approach. Compute the *p*-value that the test together well.

Test H0:μ1−μ2=0 vs. Ha:μ1−μ2≠0

α=0.01,

n1=125, x-1=−46, s1=10

n2=90, x-2=−50, s2=13

Test H0:μ1−μ2=20 vs. Ha:μ1−μ2>20

α=0.05,

n1=40, x-1=142, s1=11

n2=40, x-2=118, s2=10

Perform the check of hypotheses indicated, making use of the data indigenous independent samples given. Use the an important value approach. Compute the *p*-value the the test as well.

Test H0:μ1−μ2=13 vs. Ha:μ1−μ213

α=0.01,

n1=35, x-1=100, s1=2

n2=35, x-2=88, s2=2

Test H0:μ1−μ2=−10 vs. Ha:μ1−μ2≠−10

α=0.10,

n1=146, x-1=62, s1=4

n2=120, x-2=73, s2=7

Perform the check of hypotheses indicated, making use of the data indigenous independent samples given. Usage the *p*-value approach.

Test H0:μ1−μ2=57 vs. Ha:μ1−μ257

α=0.10,

n1=117, x-1=1309, s1=42

n2=133, x-2=1258, s2=37

Test H0:μ1−μ2=−1.5 vs. Ha:μ1−μ2≠−1.5

α=0.20,

n1=65, x-1=16.9, s1=1.3

n2=57, x-2=18.6, s2=1.1

Perform the test of hypotheses indicated, using the data from independent samples given. Usage the *p*-value approach.

Test H0:μ1−μ2=−10.5 vs. Ha:μ1−μ2>−10.5

α=0.01,

n1=64, x-1=85.6, s1=2.4

n2=50, x-2=95.3, s2=3.1

Test H0:μ1−μ2=110 vs. Ha:μ1−μ2≠110

α=0.02,

n1=176, x-1=1918, s1=68

n2=241, x-2=1782, s2=146

Perform the test of hypotheses indicated, making use of the data from independent samples given. Usage the *p*-value approach.

Test H0:μ1−μ2=50 vs. Ha:μ1−μ2>50

α=0.005,

n1=72, x-1=272, s1=26

n2=103, x-2=213, s2=14

Test H0:μ1−μ2=7.5 vs. Ha:μ1−μ2≠7.5

α=0.10,

n1=52, x-1=94.3, s1=2.6

n2=38, x-2=88.6, s2=8.0

Perform the test of hypotheses indicated, using the data from independent samples given. Usage the *p*-value approach.

Test H0:μ1−μ2=23 vs. Ha:μ1−μ223

α=0.20,

n1=314, x-1=198, s1=12.2

n2=220, x-2=176, s2=11.5

Test H0:μ1−μ2=4.4 vs. Ha:μ1−μ2≠4.4

α=0.05,

n1=32, x-1=40.3, s1=0.5

n2=30, x-2=35.5, s2=0.7

### Applications

In order to investigate the relationship between mean task tenure in years among workers who have actually a bachelor’s level or greater and those who execute not, arbitrarily samples the each kind of worker were taken, through the adhering to results.

*n*x-

*s*

Bachelor’s degree or higher | 155 | 5.2 | 1.3 |

No degree | 210 | 5.0 | 1.5 |

construct the 99% trust interval for the difference in the population way based on these data. Test, at the 1% level that significance, the claim that average job tenure amongst those with greater education is greater than amongst those without, against the default that there is no difference in the means. Compute the observed meaning of the test.

Records of 40 offered passenger cars and also 40 used pickup trucks (none provided commercially) to be randomly selected to investigate whether there was any kind of difference in the median time in years the they were maintained by the original owner before being sold. For cars the mean was 5.3 years v standard deviation 2.2 years. Because that pickup trucks the mean was 7.1 years v standard deviation 3.0 years.

build the 95% trust interval because that the distinction in the means based on these data. Test the theory that over there is a distinction in the way against the null hypothesis that over there is no difference. Usage the 1% level the significance. Compute the observed meaning of the check in component (b).In previous years the average number of patients every hour at a hospital emergency room ~ above weekends surpassed the mean on day of the week by 6.3 visits per hour. A hospital administrator believes the the present weekend typical exceeds the weekday average by fewer than 6.3 hours.

Construct the 99% trust interval for the difference in the population method based ~ above the following data, acquired from a research in which 30 weekend and also 30 weekday one-hour durations were randomly selected and also the number of brand-new patients in each recorded.

*n*x-

*s*

Weekends | 30 | 13.8 | 3.1 |

Weekdays | 30 | 8.6 | 2.7 |

check at the 5% level of significance whether the present weekend typical exceeds the weekday average by fewer 보다 6.3 patients every hour. Compute the observed significance of the test.

A sociologist surveys 50 randomly selected citizens in each of two countries to to compare the mean number of hours of volunteer work-related done by adult in each. Among the 50 occupants of Lilliput, the mean hrs of volunteer occupational per year to be 52, with standard deviation 11.8. Among the 50 residents of Blefuscu, the mean variety of hours that volunteer work-related per year to be 37, through standard deviation 7.2.

construct the 99% to trust interval for the difference in mean variety of hours volunteered through all residents of Lilliput and also the mean number of hours volunteered by all residents of Blefuscu. Test, in ~ the 1% level the significance, the claim that the mean variety of hours volunteered through all citizens of Lilliput is much more than ten hours greater than the mean variety of hours volunteered through all residents of Blefuscu. Compute the observed significance of the test in component (b).A college administrator asserted that upperclassmen spend more time examining than underclassmen.

Test this claim versus the default that the average number of hours of study per main by the two groups is the same, using the complying with information based on random samples from each group of students. Test at the 1% level the significance.

*n*x-

*s*

Upperclassmen | 35 | 15.6 | 2.9 |

Underclassmen | 35 | 12.3 | 4.1 |

Compute the observed meaning of the test.

An kinesiologist cases that the resting heart rate of males aged 18 to 25 that exercise frequently is more than five beats every minute less than that of guys who perform not practice regularly. Men in each classification were selected at random and their relaxing heart prices were measured, through the outcomes shown.

*n*x-

*s*

Regular exercise | 40 | 63 | 1.0 |

No continual exercise | 30 | 71 | 1.2 |

perform the pertinent test of hypotheses in ~ the 1% level of significance. Compute the observed significance of the test.

Children in 2 elementary school classrooms were provided two version of the very same test, however with the order of questions arranged from much easier to more complicated in variation *A* and in turning back order in variation *B*. Randomly selected students indigenous each class were provided Version *A* and the rest Version *B*. The outcomes are displayed in the table.

*n*x-

*s*

Version A | 31 | 83 | 4.6 |

Version B | 32 | 78 | 4.3 |

construct the 90% trust interval for the difference in the method of the populations of all children taking version

*A*of such a test and of all youngsters taking variation

*B*of such a test. Test at the 1% level of significance the theory that the

*A*variation of the check is much easier than the

*B*version (even though the inquiries are the same). Compute the observed meaning of the test.

The Municipal Transit Authority desires to understand if, ~ above weekdays, much more passengers journey the northbound blue heat train towards the city center that departs at 8:15 a.m. Or the one that departs in ~ 8:30 a.m. The following sample statistics room assembled by the Transit Authority.

*n*x-

*s*

8:15 a.m. Train | 30 | 323 | 41 |

8:30 a.m. Train | 45 | 356 | 45 |

build the 90% trust interval because that the difference in the mean number of daily tourists on the 8:15 train and also the mean variety of daily visitors on the 8:30 train. Test at the 5% level of significance whether the data provide sufficient proof to conclude that more passengers drive the 8:30 train. Compute the observed definition of the test.

In compare the scholastic performance of college students who room affiliated with fraternities and those masculine students who room unaffiliated, a arbitrarily sample of student was attracted from every of the two populations on a college campus. Summary statistics top top the college student GPAs are provided below.

*n*x-

*s*

Fraternity | 645 | 2.90 | 0.47 |

Unaffiliated | 450 | 2.88 | 0.42 |

Test, at the 5% level that significance, whether the data provide sufficient evidence to conclude that there is a distinction in mean GPA in between the population of fraternity students and also the population of unaffiliated masculine students ~ above this college campus.

In compare the academic performance of university students who space affiliated through sororities and also those mrs students who space unaffiliated, a random sample of college student was attracted from every of the two populaces on a university campus. Summary statistics ~ above the college student GPAs are given below.

*n*x-

*s*

Sorority | 330 | 3.18 | 0.37 |

Unaffiliated | 550 | 3.12 | 0.41 |

Test, at the 5% level the significance, even if it is the data carry out sufficient evidence to conclude the there is a difference in median GPA in between the populace of sorority students and the populace of unaffiliated female students on this university campus.

The owner of a experienced football team believes that the league has become much more offense oriented due to the fact that five year ago. To examine his belief, 32 randomly selected games from one year’s schedule were compared to 32 randomly selected gamings from the schedule five years later. Since more offense produces more points per game, the owner analyzed the complying with information on points per game (ppg).

*n*x-

*s*

ppg previously | 32 | 20.62 | 4.17 |

ppg recently | 32 | 22.05 | 4.01 |

Test, at the 10% level that significance, even if it is the data on points per game carry out sufficient proof to conclude that the video game has become an ext offense oriented.

The owner of a expert football team believes that the league has become much more offense oriented due to the fact that five year ago. To check his belief, 32 randomly selected gamings from one year’s schedule were compared to 32 randomly selected gamings from the schedule five years later. Since more offense produces an ext offensive yards per game, the owner analyzed the adhering to information on offensive yards per game (oypg).

*n*x-

*s*

oypg previously | 32 | 316 | 40 |

oypg recently | 32 | 336 | 35 |

Test, at the 10% level that significance, whether the data on attack yards every game administer sufficient evidence to conclude that the video game has become more offense oriented.

### Large Data collection Exercises

Large Data set 1A and 1B perform the satellite scores for 1,000 randomly selected students. Represent the population of all masculine students as populace 1 and the populace of every female students as populace 2.

http://www.gone.2012books.lardbucket.org/sites/all/files/data1A.xls

http://www.gone.2012books.lardbucket.org/sites/all/files/data1B.xls

Restricting fist to simply the males, uncover*n*1, x-1, and also

*s*1. Restricting attention to just the females, discover

*n*2, x-2, and also

*s*2. Permit μ1 signify the mean SAT score for every males and μ2 the mean SAT score for all females. Use the results of part (a) to construct a 90% trust interval for the difference μ1−μ2. Test, at the 5% level of significance, the hypothesis that the mean SAT scores among males exceeds that of females.

Large Data to adjust 1A and 1B perform the GPAs for 1,000 randomly selected students. Represent the population of all male students as population 1 and the population of every female student as population 2.

See more: For An Object To Be In Static Equilibrium, The Net Force Must Be Equal To Zero And

http://www.gone.2012books.lardbucket.org/sites/all/files/data1A.xls

http://www.gone.2012books.lardbucket.org/sites/all/files/data1B.xls

Restricting fist to just the males, uncover*n*1, x-1, and also

*s*1. Restricting fist to just the females, uncover

*n*2, x-2, and

*s*2. Permit μ1 denote the typical GPA for all males and also μ2 the median GPA for every females. Use the results of part (a) to build a 95% confidence interval for the distinction μ1−μ2. Test, in ~ the 10% level that significance, the hypothesis that the average GPAs amongst males and also females differ.

Large Data to adjust 7A and 7B list the survival times for 65 male and 75 female activities mice v thymic leukemia. Signify the population of every such male mice as population 1 and the population of every such female mouse as populace 2.

http://www.gone.2012books.lardbucket.org/sites/all/files/data7A.xls

http://www.gone.2012books.lardbucket.org/sites/all/files/data7B.xls

Restricting fist to simply the males, find*n*1, x-1, and also

*s*1. Restricting fist to simply the females, find

*n*2, x-2, and also

*s*2. Allow μ1 represent the typical survival for all males and also μ2 the average survival time for every females. Usage the outcomes of component (a) to construct a 99% confidence interval because that the distinction μ1−μ2. Test, at the 1% level of significance, the theory that the median survival time because that males exceeds that for females by an ext than 182 job (half a year). Compute the observed definition of the check in component (c).

### Answers

(4.20,5.80), (−18.54,−9.46)