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ONE-PROPORTION Z-TESTS CHAPTER 20 PART 3

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4 Steps : 1)State the hypotheses 2)Check conditions and model (Normal model) 3)Mechanics (Find z-score and P-value) 4)State your conclusion

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A 1996 report from the U.S. Consumer Product Safety Commission claimed that at least 90% of all American homes have at least one smoke detector. A city’s fire department has been running a public safety campaign about smoke detectors consisting of posters, billboards, and ads on radio and TV and in the newspaper. The city wonders if this concerted effort has raised the local level above the 90% national rate. Building inspectors visit 400 randomly selected homes and find that 376 have smoke detectors. Is this strong evidence that the local rate is higher than the national rate? Step 1: State the hypotheses

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Step 2: Check the conditions and model Random sample 400 is less than 10% of all homes in the city np = 400(0.9)=360; nq = 400(0.1) = 40 Use the Normal model N(0.9, 0.015)

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Step 3: Mechanics (z-score and P-value) Look to the right of the z-score P-value: P(z>2.67) = 1 – P(z<2.67) = 1 – 0.9962 = 0.0038

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The P-value is very low, making it unlikely that the observed results can be explained by sampling error. We reject the null hypothesis. There is strong evidence that in this city more than 90% of homes have smoke detectors. Step 4: State your conclusion

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It is claimed that 20% of M&M’s are orange. Suppose a bag of 122 has only 21 orange ones. Does this contradict the company’s 20% claim? Step 1: State the hypotheses

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It is claimed that 20% of M&M’s are orange. Suppose a bag of 122 has only 21 orange ones. Does this contradict the company’s 20% claim? Step 2: Check conditions and model Random sample, independent 122 is less than 10% of all M&M’s np = 122(0.2)=48.8; nq = 122(0.8) = 97.6

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It is claimed that 20% of M&M’s are orange. Suppose a bag of 122 has only 21 orange ones. Does this contradict the company’s 20% claim? Step 3: Mechanics (z-score, P-value) This indicates a two-tailed test. P-value: P(z 0.77) = 0.44 + (1 - 0.56) = 0.44 + 0.44 = 0.88

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It is claimed that 20% of M&M’s are orange. Suppose a bag of 122 has only 21 orange ones. Does this contradict the company’s 20% claim? Step 4: State your conclusion With a P-value of 0.88, we fail to reject the null hypothesis. This means there is no evidence to suggest that the proportion of orange M&M’s differs from the advertised 20%.

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Today’s Assignment: You should still be reading the book! Homework # 11: p.477 #11-14, 17, 18

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