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Claims about two large means

• E.g., “South students have a higher average SAT score that Monmouth Regional Students”

• The two samples are independent. The samples from one population are not related to the other samples

• Both samples are simple random samples.

• Large and small samples

• Same general procedure– Test statistic requires two samples from each population

– Classic method

– P-value (calculator)

2

Hypothesis Testing: Claims

1. State the claim: South students have the same mean SAT scores as North students

μ1 = μ2

2. State the opposite claim: μ1 μ2

3. Pick your hypothesis

4. Sample:X S n

South 1710 200 60

North 1680 280 110

3

Hypothesis Tests: Test Statistic

5. Test statistic

– (μ1 – μ2) will almost always be 0

6. Critical value:– Z table if both samples are 30 or more,

– T table otherwise

1 2 1 2

2 21 2

1 2

X Xz

n n

Testing two sample means

Original claim Pick null and alternate hypothesisOpposite claim

Degree of confidence

First sample X-bar = s = n =

Second sample X-bar = s = n =

Test statistic

Two tailed (=)

Right tailed ()

Left tailed

Critical value

P-value

Conclusion

4

5

Using the Calculator

• Finding the p-value• STAT→TEST→2-SampleZTest• STAT→TEST→2-SampleTTest• Enter the mean, standard deviation, and sample size

for both samples

• Select H1

• Select Calculate• If p-value is less than alpha, reject.

Testing two sample means

Original claim Pick null and alternate hypothesisOpposite claim

Degree of confidence

First sample X-bar = s = n =

Second sample X-bar = s = n =

Test statistic

Two tailed (=)

Right tailed ()

Left tailed

Critical value

P-value

Conclusion

6

7

Live example/Your turn

• It rains more on Saturday than Monday

• Books are more expensive at UMass than DCC

X s n

Monday 0.1 0.26 25

Saturday 0.14 0.29 42

X s n

UMass 65.12 28.03 40

DCC 57.62 14.68 35

Testing two sample means

Original claim Pick null and alternate hypothesisOpposite claim

Degree of confidence

First sample X-bar = s = n =

Second sample X-bar = s = n =

Test statistic

Two tailed (=)

Right tailed ()

Left tailed

Critical value

P-value

Conclusion

8

Testing two sample means

Original claim Pick null and alternate hypothesisOpposite claim

Degree of confidence

First sample X-bar = s = n =

Second sample X-bar = s = n =

Test statistic

Two tailed (=)

Right tailed ()

Left tailed

Critical value

P-value

Conclusion

9

Homework

1. Use a 0.05 significance level to test the claim that the two populations have the same mean:

– Treatment group: n = 50, x-bar = 7.00, s = 1.00

– Placebo group: n = 100, x-bar = 6.00, s = 2.00

2. Use a 90% degree of confidence to test the claim that the two populations have different means:

– Math majors: n = 40, x-bar = 75.0, s = 15.0

– English majors: n = 60, x-bar = 70.0, s = 14.00

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More Homework

3. A sample of 290 children not wearing seatbelts during car accidents resulted in an average of 1.39 days in the ICU (s = 3.06). On the other hand, 123 children wearing seat belts averaged 0.83 days in the ICU (s = 1.77). Test the claim that wearing seatbelts results in less severe injuries during car accidents.

4. Test the claim that American cars are as reliable as Japanese cars. 35 Japanese cars had an average repair bill of $650 with a standard deviation of $250. 41 American cards has a mean repair bill of $720 with a standard deviation of $390. 11

Still More Homework

5. A recent poll at a local college showed that the average age of 217 students’ cars was 7.89. (s = 3.67). The average age of 152 faculty cars was 5.99(s = 3.65) . Using a 0.05 significance level to test the claim that student cars are older than faculty cars.

6. Thirty five pennies minted before 1983 weighed an average of 3.0748 grams (s = 0.0391). Thirty five pennies minted after 1983 weighed an average of 2.5002 grams (s = 0.0149). Test the claim that the weight of pennies did not change after 1983.

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Homework #1

Original claim μ1 = μ2 H0 Pick null and alternate hypothesis

Opposite claim μ1 ≠ μ2 H1

Degree of confidence 95%

First sample X-bar = 7.00 s = 1.00 n = 50

Second sample X-bar = 6.00 s = 2.00 n = 100

Test statistic z = 4.08

Two tailed (=)

Right tailed ()

Left tailed

Critical value

P-value 0.000044 < 0.05

Conclusion Reject null, reject original claim

13

Homework #2

Original claim μ1 ≠ μ2 H1 Pick null and alternate hypothesis

Opposite claim μ1 = μ2 H0

Degree of confidence 90%

First sample X-bar = s = n =

Second sample X-bar = s = n =

Test statistic

Two tailed (=)

Right tailed ()

Left tailed

Critical value

P-value

Conclusion

14

Homework #3

Original claim μ1 < μ2 H1 Pick null and alternate hypothesis

Opposite claim μ1 ≥ μ2 H0

Degree of confidence 95%

First sample (seat belts) X-bar = s = n =

Second sample (not) X-bar = s = n =

Test statistic

Two tailed (=)

Right tailed ()

Left tailed

Critical value

P-value

Conclusion

15

Homework #4

Original claim μ1 = μ2 H0 Pick null and alternate hypothesis

Opposite claim μ1 ≠ μ2 H1

Degree of confidence 95%

First sample (USA) X-bar = s = n =

Second sample (Japan) X-bar = s = n =

Test statistic

Two tailed (=)

Right tailed ()

Left tailed

Critical value

P-value

Conclusion

16

Homework #5

Original claim μ1 > μ2 H1 Pick null and alternate hypothesis

Opposite claim μ1 ≤ μ2 H0

Degree of confidence 95%

First sample (students) X-bar = s = n =

Second sample (faculty) X-bar = s = n =

Test statistic

Two tailed (=)

Right tailed ()

Left tailed

Critical value

P-value

Conclusion

17

Homework #6

Original claim μ1 = μ2 H0 Pick null and alternate hypothesis

Opposite claim μ1 ≠ μ2 H1

Degree of confidence 95%

First sample X-bar = s = n =

Second sample X-bar = s = n =

Test statistic

Two tailed (=)

Right tailed ()

Left tailed

Critical value

P-value

Conclusion

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