Hypothesis Testing Questions : Ex 7A The Further Mathematics Network 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

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.

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 = 0.6 [np = 12] P(X ≤ 9) = 0.1275 = 12.75% > 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.

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) = 0.0565 = 5.65% > 5% P(X ≤ 7) = 0.0210 = 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.

Data collection: 8 out of 10 say coffee is synthetic 3 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) = 1 - 0.9453 = 0.0547 = 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.

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 = 0.5 [np = 9] P(X ≥ 11) = 1 – P(X ≤ 10) = 1 – 0.7597 = 0.2403 = 24.03% > 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.

Data collection: 10 out of 20 people pass first time 5 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 = 0.7 [np = 14] P(X ≤ 10) = 0.0480 = 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.

Data collection: 5 out of 50 cracked bottles in sample 6 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 = 0.05 [np = 2.5] P(X ≥ 5) = 1 – P(X ≤ 4) = 1 - 0.8964 = 0.1036 = 10.36% > 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.

Data collection: 1 out of 20 mugs are defective 7 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) = 0.0692 = 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.

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 – 0.9453 = 0.0547 = 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.