WARM – UPAs you prepare for College, one cost you should As you prepare for College, one cost you should consider is your Meal Plan. A friend tells you that you consider is your Meal Plan. A friend tells you that you should budget for $1000 in food cost per semester. You should budget for $1000 in food cost per semester. You feel that the actual figure is something feel that the actual figure is something differentdifferent. What . What can you conclude from an SRS of 10 universities?can you conclude from an SRS of 10 universities?

$1200 $1450 $1284 $920 $780$1200 $1450 $1284 $920 $780

$1526 $1152 $1120 $1760 $1245$1526 $1152 $1120 $1760 $1245

μμ = The true mean cost of College Meal Plans per semester. = The true mean cost of College Meal Plans per semester.

HH00: : μμ = 1000 = 1000

HHaa: : μμ ≠ 1000≠ 1000

txsn

0

t 12437 1000

2859410

.

.t 2 6951.

0.0246P Value Since the P-Value is less than Since the P-Value is less than αα = 0.05 the data IS significant . There = 0.05 the data IS significant . There is strong evidence to REJECT His strong evidence to REJECT H00 . The average college meal plan is . The average college meal plan is

NOT $1000.NOT $1000.1.1. SRS – Stated SRS – Stated 2.2. Approximately Normal Distribution – Graph Approximately Normal Distribution – Graph

One Sample

One Sample

t – Testt – Test

In a Matched Pairs design subjects are matched in Homogeneous pairs or the same subject is used in both treatments. A common type is a Before-and-After Design. This leads to two Dependent data sets of which you subtract.

MATCHED PAIRS t–TESTS

Objective: Use the One sample t procedures to interpret a Matched Pairs t-test and be able to make inferences about the differences in the two treatments which share a common characteristic.

CHAPTER 25

The Matched Pairs t-Procedures

The Parameters and StatisticsThe Parameters and Statistics::

μμdd = the true mean difference in the two populations. = the sample mean of all the differences of each

individual pairing. = the sample standard deviation of all the differences

xdsd

0d

d

xts

n

HH00: : μμdd = = 00

HHaa: : μμdd ≠, <, ≠, <, oror > > 00

2. T-Test:T-Test: Precede with a ONE SAMPLE t-Test for the mean difference with

μd always equal to Zero.

1. Confidence Interval:Confidence Interval: * dd

sx t

n

P-VALUESμd<0: tcdf(-E99, t, df)μd>0: tcdf(t, E99, df)μd≠0: 2·tcdf(|t|, E99, df)

EXAMPLE 1EXAMPLE 1::

Many drivers of cars that can run on regular gas actually buy premium in the belief that they will get better gas mileage. To test this, 10 cars are randomly selected. Each car is run using both regular and premium gas. The mileage is recorded. Is there sufficient evidence at the 0.05 level to support the belief?

μμdd = The true mean DIFFERENCE in gas = The true mean DIFFERENCE in gas

mileage (Premium – Regular)mileage (Premium – Regular)

HH00: : μμdd = 0 = 0

HHaa: : μμdd > 0> 0

0d

d

xts

n

t

2 01414

10.

t 4 4721. P Value 00008.Since the P-Value < Since the P-Value < αα = 0.05 - REJECT H = 0.05 - REJECT H00

STRONG evidence that Premium Gas does STRONG evidence that Premium Gas does improve gas mileage.improve gas mileage.

1.1. SRS – Stated SRS – Stated 2.2. Approximately Normal Distribution – Graph Approximately Normal Distribution – Graph

Car #Car # Reg.Reg. Prem.Prem.

1 16 19

2 20 22

3 21 24

4 22 24

5 23 25

6 22 25

7 27 26

8 25 26

9 27 28

10 28 32

Prem.Prem.–Reg. –Reg.

3

2

3

2

2

3

-1

1

1

4

x sd d 2 1414.

Matched Pairs

Matched Pairs

1-Sample1-Sample

t – Testt – Test

EXAMPLE 2EXAMPLE 2::

“Freshman – 15” Many people believe that students gain a significant amount of weight their freshman year of college. Is there enough evidence at the 5% level to support that there is a weight increase? Use the weights of 6 randomly chosen students. Student weights were measure at the beginning and at the end of the fall semester.

μμdd = The true mean DIFFERENCE in Weight = The true mean DIFFERENCE in Weight

in pounds (End of Semester – Beginning)in pounds (End of Semester – Beginning)

HH00: : μμdd = 0 = 0

HHaa: : μμdd > 0> 0txs

n

d d

d

1.833 0

2.7876

t

1.611t 0.0840P Value

Since the P-Value is NOT less than Since the P-Value is NOT less than αα = 0.05 = 0.05 Fail to REJECT HFail to REJECT H00 . The belief that Freshman . The belief that Freshman

experience weight increase `is not supported.experience weight increase `is not supported.1.1. SRS – Stated SRS – Stated 2.2. Approximately Normal Distribution – Graph Approximately Normal Distribution – Graph

BeginBegin EndEnd

1 171 168

2 110 111

3 134 136

4 115 119

5 150 155

6 104 106

End–End–Begin Begin

-3

1

2

4

5

2

1.833 2.787d dx s

EXAMPLE 3EXAMPLE 3::

The maker of a new tire claims that his Tires are superior in all road conditions. He claims that with his tires there is no difference in stopping distance between dry or wet pavement. To test this you select an SRS of 9 cars and at 60 mph you slam on the breaks. Estimate the Mean difference in stopping distance in feet with a 90% Confidence Interval.

μμdd= The true mean DIFFERENCE in stopping = The true mean DIFFERENCE in stopping

distance in feet (Wet – Dry Pavement)distance in feet (Wet – Dry Pavement)

x t snd

d FH IK*

10.21055 1.869

We are 90% Confident that the True mean We are 90% Confident that the True mean difference in stopping distance in feet difference in stopping distance in feet between Wet Pavement – Dry Pavement is between Wet Pavement – Dry Pavement is between 48.671 ft and 61.329 ft. between 48.671 ft and 61.329 ft.

1.1. SRS – Stated SRS – Stated 2.2. Approximately Normal Distribution – Graph Approximately Normal Distribution – Graph

Car #Car # WetWet DryDry

1 201 150

2 220 147

3 192 136

4 182 130

5 173 134

6 202 134

7 180 128

8 192 136

9 206 158

Wet – Wet – Dry Dry

51

73

56

52

39

68

52

56

48

x sd d 55 10 210.

55 7 848 .

Matched Pairs Matched Pairs 1-Sample1-Sample

t – Intervalt – Interval

1000 Chips in 1000 Chips in Every Bag!Every Bag!

Chips Ahoy used to advertise “1000 Chips in Every Bag!1000 Chips in Every Bag!”

1. How would YOU do a significance test to test this?

2. What issues do you think would arise and how would you over come them?

1000 Chips in 1000 Chips in Every Bag!Every Bag!

Chips Ahoy used to advertise “1000 Chips in Every Bag!1000 Chips in Every Bag!” With 38 cookies in each bag, the true mean number of chips in each cookie should be 26.3 chips per cookie. Because there was a suspicion of LESS than 26.3 chip per cookie the company disregarded the ad. Test this claim by crumbling cookies and counting the number of chips in each.a.) Is there evidence to support the company’s decision.

Gather evidence and Conduct a Significance test.

b.) Estimate the true # of chips in the bag by finding a 95% Confidence Interval for the true # of chips per cookie and then multiply it by 38.

c.) Eat your evidence.