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Worked Solutions to Hypothesis Testing Questions : Ex 7A

The Further Mathematics Network. Worked Solutions to Hypothesis Testing Questions : Ex 7A. 5 steps in Hypothesis Testing. Establish null (H 0 ) and alternative (H 1 ) hypotheses Decide on significance level

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Worked Solutions to Hypothesis Testing Questions : Ex 7A

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  1. The Further Mathematics Network Worked Solutions to Hypothesis Testing Questions : Ex 7A

  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. Let X represent number of people supporting Mrs da Silva. Let p represent probability that a person supports her. 1 H1: p < 0.6 One-tailed test • H0: p = 0.6 • 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.

  4. Let X represent number of pupils passing first time Let p represent probability that a pupil passes first time. 2 H1: p < 0.6 One-tailed test • H0: p = 0.6 • 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 ≤7there is sufficient evidence, at the 5% level, to reject H0, so accept H1. • Conclude instructor’s claim is exaggerated if N ≤7.

  5. Let X represent number of people saying coffee is synthetic. Let p represent probability person says coffee is synthetic. 3 H1: p > 0.5 One-tailed test • H0: p = 0.5 • 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.

  6. Let X represent number of pieces landing butter side down. Let p represent probability that a piece lands butter side down. 4 H1: p > 0.5 One-tailed test • H0: p = 0.5 • 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.

  7. Let X represent number of people passing test first time. Let p represent probability person passes test first time. 5 H1: p < 0.7 One-tailed test • H0: p = 0.7 • 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.

  8. Let X represent number of cracked bottles in sample. Let p represent probability that a chosen bottle is cracked. 6 H1: p > 0.05 One-tailed test • H0: p = 0.05 • 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.

  9. Let X represent number of defective mugs. Let p represent probability that a mug is defective. 7 H1: p < 0.2 One-tailed test • H0: p = 0.2 • 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.

  10. Let X represent number of long questions correct. Let p represent probability that a long question is correct. 8 H1: p > 0.5 One-tailed test • H0: p = 0.5 • 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.

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