1. Hypothesis Testing Questions : Ex 7A
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Worked Solutions to
Hypothesis Testing Questions : Ex 7A
Prologue
Hypothesis Testing Procedure
Question 1
Mrs da Silva
Question 2
Passing your driving test first time (1)
Question 3
Synthetic and real coffee
Question 4
Butter side down?
Question 5
Passing your driving test first time (2)
Question 6
Cracked milk bottles
Question 7
Defective mugs
Question 8
Mathematics contest

2. 5 steps in Hypothesis Testing
Establish null (H0) and alternative (H1) hypotheses
Decide on significance level
Collect suitable data, using a random sampling procedure that ensures the data are independent.
Conduct the test, doing the necessary calculations.
Interpret the result in terms of the original claim, theory or problem.

3. Data collection: 9 out of 20 say they support her.1
Let X represent number of people supporting Mrs da Silva.
Let p represent probability that a person supports her.
H0: p = 0.6
H1: p < 0.6
One-tailed test
Significance level = 5%
Data collection: 9 out of 20 say they support her.
Conduct the test: n = 20, p = [np = 12]
P(X ≤ 9) = = % > 5%
Interpret the result:
Since 12.75% > 5%, there is not sufficient evidence, at the 5% level, to reject H0, accept H0.
She is not overestimating her support.

4. Data collection: N out of 20 pupils pass first time.
Let X represent number of pupils passing first time
Let p represent probability that a pupil passes first time.
H0: p = 0.6
H1: p < 0.6
One-tailed test
Significance level = 5%
Data collection: N out of 20 pupils pass first time.
Conduct the test: n = 20, p = 0.6 [np = 12]
P(X ≤ 8) = = 5.65% > 5%
P(X ≤ 7) = = 2.10% < 5%
Interpret the result:
Provided N ≤ 7 there is sufficient evidence, at the 5% level, to reject H0, so accept H1.
Conclude instructor’s claim is exaggerated if N ≤ 7.

5. Data collection: 8 out of 10 say coffee is synthetic3
Let X represent number of people saying coffee is synthetic.
Let p represent probability person says coffee is synthetic.
H0: p = 0.5
H1: p > 0.5
One-tailed test
Significance level = 5%
Data collection: 8 out of 10 say coffee is synthetic
Conduct the test: n = 10, p = 0.5 [np = 5]
P(X ≥ 8) = 1 – P(X ≤ 7) = = = 5.47% > 5%
Interpret the result:
Since 5.47% > 5%, there is not sufficient evidence, at the 5% level, to reject H0, so acceptH0.
People cannot tell the difference.

6. Data collection: 11 out of 18 land butter side down.4
Let X represent number of pieces landing butter side down.
Let p represent probability that a piece lands butter side down.
H0: p = 0.5
H1: p > 0.5
One-tailed test
Significance level = 5%
Data collection: 11 out of 18 land butter side down.
Conduct the test: n = 18, p = [np = 9]
P(X ≥ 11) = 1 – P(X ≤ 10) = 1 – = = % > 10%
Interpret the result:
Since 24.03% > 10%, there is not sufficient evidence, at the 5% level, to reject H0, so accept H0.
Piece of toast not more likely to land butter side.

7. Data collection: 10 out of 20 people pass first time5
Let X represent number of people passing test first time.
Let p represent probability person passes test first time.
H0: p = 0.7
H1: p < 0.7
One-tailed test
Significance level = 5%
Data collection: 10 out of 20 people pass first time
Conduct the test: n = 20, p = [np = 14]
P(X ≤ 10) = = 4.8% < 5%
Interpret the result:
Since 4.8% < 5%, there is sufficient evidence, at the 5% level, to reject H0, so accept H1.
Mr. McTaggart is exaggerating his claim.

8. Data collection: 5 out of 50 cracked bottles in sample6
Let X represent number of cracked bottles in sample.
Let p represent probability that a chosen bottle is cracked.
H0: p = 0.05
H1: p > 0.05
One-tailed test
Significance level = 5%
Data collection: 5 out of 50 cracked bottles in sample
Conduct the test: n = 50, p = [np = 2.5]
P(X ≥ 5) = 1 – P(X ≤ 4) = = = % > 5%
Interpret the result:
Since 10.36% > 5%, there is not sufficient evidence, at the 5% level, to reject H0, so accept H0.
Insufficient evidence that machine needs servicing.

9. Data collection: 1 out of 20 mugs are defective7
Let X represent number of defective mugs.
Let p represent probability that a mug is defective.
H0: p = 0.2
H1: p < 0.2
One-tailed test
Significance level = 5%
Data collection: 1 out of 20 mugs are defective
Conduct the test: n = 20, p = 0.2 [np = 4]
P(X ≤ 1) = = 6.92% > 5%
Interpret the result:
Since 6.92% > 5%, there is not sufficient evidence, at the 5% level, to reject H0, so accept H0.
No improvement in the performance of the machine.

10. Data collection: 8 out of 10 long questions correct
Let X represent number of long questions correct.
Let p represent probability that a long question is correct.
H0: p = 0.5
H1: p > 0.5
One-tailed test
Significance level = 5%
Data collection: 8 out of 10 long questions correct
Conduct the test: n = 10, p = 0.5 [np = 5]
P(X ≥ 8) = 1 – P(X ≤ 7) = 1 – = = 5.47% > 5%
Interpret the result:
Since 5.47% > 5%, there is not sufficient evidence, at the 5% level, to reject H0, so accept H0.
No improvement in the performance on long questions.