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Worked Solutions to Hypothesis Testing Questions : Ex 7A PrologueHypothesis Testing Procedure Question 1Mrs da Silva Question 2Passing your driving test.

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Presentation on theme: "Worked Solutions to Hypothesis Testing Questions : Ex 7A PrologueHypothesis Testing Procedure Question 1Mrs da Silva Question 2Passing your driving test."— Presentation transcript:

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

2 Statistics 1Hypothesis Testing Solutions 1Establish null (H 0 ) and alternative (H 1 ) hypotheses 2Decide on significance level 3Collect suitable data, using a random sampling procedure that ensures the data are independent. 4Conduct the test, doing the necessary calculations. 5Interpret the result in terms of the original claim, theory or problem. 5 steps in Hypothesis Testing

3 Statistics 1Hypothesis Testing Solutions 1 Let X represent number of people supporting Mrs da Silva. Let p represent probability that a person supports her. 5Interpret the result: Since 12.75% > 5%, there is not sufficient evidence, at the 5% level, to reject H 0, accept H 0. She is not overestimating her support. 1H 0 : p = 0.6 H 1 : p < 0.6 One-tailed test 2Significance level = 5% 3Data collection: 9 out of 20 say they support her. 4Conduct the test: n = 20, p = 0.6 [np = 12] P(X ≤ 9) = = 12.75% > 5%

4 Statistics 1Hypothesis Testing Solutions 2 Let X represent number of pupils passing first time Let p represent probability that a pupil passes first time. 5Interpret the result: Provided N ≤ 7 there is sufficient evidence, at the 5% level, to reject H 0, so accept H 1. Conclude instructor’s claim is exaggerated if N ≤ 7. 1H 0 : p = 0.6 H 1 : p < 0.6 One-tailed test 2Significance level = 5% 3Data collection: N out of 20 pupils pass first time. 4Conduct the test: n = 20, p = 0.6 [np = 12] P(X ≤ 8) = = 5.65% > 5% P(X ≤ 7) = = 2.10% < 5%

5 Statistics 1Hypothesis Testing Solutions 3 Let X represent number of people saying coffee is synthetic. Let p represent probability person says coffee is synthetic. 5Interpret the result: Since 5.47% > 5%, there is not sufficient evidence, at the 5% level, to reject H 0, so acceptH 0. People cannot tell the difference. 1H 0 : p = 0.5 H 1 : p > 0.5 One-tailed test 2Significance level = 5% 3Data collection: 8 out of 10 say coffee is synthetic 4Conduct the test: n = 10, p = 0.5 [np = 5] P(X ≥ 8) = 1 – P(X ≤ 7) = = = 5.47% > 5%

6 Statistics 1Hypothesis Testing Solutions 4 Let X represent number of pieces landing butter side down. Let p represent probability that a piece lands butter side down. 5Interpret the result: Since 24.03% > 10%, there is not sufficient evidence, at the 5% level, to reject H 0, so accept H 0. Piece of toast not more likely to land butter side. 1H 0 : p = 0.5 H 1 : p > 0.5 One-tailed test 2Significance level = 5% 3Data collection: 11 out of 18 land butter side down. 4Conduct the test: n = 18, p = 0.5 [np = 9] P(X ≥ 11) = 1 – P(X ≤ 10) = 1 – = = 24.03% > 10%

7 Statistics 1Hypothesis Testing Solutions 5 Let X represent number of people passing test first time. Let p represent probability person passes test first time. 5Interpret the result: Since 4.8% < 5%, there is sufficient evidence, at the 5% level, to reject H 0, so accept H 1. Mr. McTaggart is exaggerating his claim. 1H 0 : p = 0.7 H 1 : p < 0.7 One-tailed test 2Significance level = 5% 3Data collection: 10 out of 20 people pass first time 4Conduct the test: n = 20, p = 0.7 [np = 14] P(X ≤ 10) = = 4.8% < 5%

8 Statistics 1Hypothesis Testing Solutions 6 Let X represent number of cracked bottles in sample. Let p represent probability that a chosen bottle is cracked. 5Interpret the result: Since 10.36% > 5%, there is not sufficient evidence, at the 5% level, to reject H 0, so accept H 0. Insufficient evidence that machine needs servicing. 1H 0 : p = 0.05 H 1 : p > 0.05 One-tailed test 2Significance level = 5% 3Data collection: 5 out of 50 cracked bottles in sample 4Conduct the test: n = 50, p = 0.05 [np = 2.5] P(X ≥ 5) = 1 – P(X ≤ 4) = = = 10.36% > 5%

9 Statistics 1Hypothesis Testing Solutions 7 Let X represent number of defective mugs. Let p represent probability that a mug is defective. 5Interpret the result: Since 6.92% > 5%, there is not sufficient evidence, at the 5% level, to reject H 0, so accept H 0. No improvement in the performance of the machine. 1H 0 : p = 0.2 H 1 : p < 0.2 One-tailed test 2Significance level = 5% 3Data collection: 1 out of 20 mugs are defective 4Conduct the test: n = 20, p = 0.2 [np = 4] P(X ≤ 1) = = 6.92% > 5%

10 Statistics 1Hypothesis Testing Solutions 8 Let X represent number of long questions correct. Let p represent probability that a long question is correct. 5Interpret the result: Since 5.47% > 5%, there is not sufficient evidence, at the 5% level, to reject H 0, so accept H 0. No improvement in the performance on long questions. 1H 0 : p = 0.5 H 1 : p > 0.5 One-tailed test 2Significance level = 5% 3Data collection: 8 out of 10 long questions correct 4Conduct the test: n = 10, p = 0.5 [np = 5] P(X ≥ 8) = 1 – P(X ≤ 7) = 1 – = = 5.47% > 5%


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