17 One of the different statistics reported by the Centers for Disease Control r

Question

17 One of the different statistics reported by the Centers for Disease Control regarding incidence of obesity among adults in the United States provides that 27.4% of men with college degree are obese. The study also reports that 32.1% of men without a college degree are obese. Assume the latter statistic is based on a sample of 800 men without a college degree. Does the data provide statistically significant evidence that the incidence of obesity among men without a college degree is greater than among those with a college degree? Compute the p-value for this hypothesis test. p-value = ______. a 0.0015 Reject H?. The evidence is statistically significant. b 0.0015 Reject H?. The evidence is not statistically significant. c 0.0185 Do not reject H?. The evidence is statistically significant. d 0.0185 Do not reject H?. The evidence is not statistically significant. 18 We want to test the hypothesis that at least 85% of drivers on a freeway violate the speed limit. In a random sample of n = 1,000 vehicles, 832 violated the speed limit. Compute the sample proportion. State the null and alternative hypotheses and the decision rule. a Reject H? at ? = 5%. Conclude less than 85% of drivers violate the speed limit. b Reject H? at ? = 10% and conclude that less than 85% of drivers violate the speed limit. But, do not reject H? at ? = 5% and conclude 85% or more of drivers violate the speed limit. c Reject H? at ? = 5%. Conclude more than 85% of drivers violate the speed limit. d Reject H? at ? = 1%. Conclude less than 85% of drivers violate the speed limit. 19 In a sample of 845 coffee growers from southern Mexico, 494 growers were certified to sell to organic coffee markets. Is there evidence to indicate that fewer than 60% of the coffee growers in southern Mexico are organic certified? State your conclusion so that there is only 5% chance of making Type I error. Round your p? to three decimal points. a p-value = 0.1867 At a 5% level of significance, the evidence that fewer than 60% are organic certified is not statistically significant. b p-value = 0.1867 At a 10% level of significance, evidence that fewer than 60% are organic certified is statistically significant. c p-value = 0.0867 At a 5% level of significance, the evidence that fewer than 60% are organic certified is not statistically significant. d p-value = 0.0867 At a 10% level of significance, evidence that fewer than 60% are organic certified is statistically significant. 20 In the previous question, the margin of error for rejecting the null hypothesis is. a 0.041 b 0.037 c 0.033 d 0.028 17 One of the different statistics reported by the Centers for Disease Control regarding incidence of obesity among adults in the United States provides that 27.4% of men with college degree are obese. The study also reports that 32.1% of men without a college degree are obese. Assume the latter statistic is based on a sample of 800 men without a college degree. Does the data provide statistically significant evidence that the incidence of obesity among men without a college degree is greater than among those with a college degree? Compute the p-value for this hypothesis test. p-value = ______. a 0.0015 Reject H?. The evidence is statistically significant. b 0.0015 Reject H?. The evidence is not statistically significant. c 0.0185 Do not reject H?. The evidence is statistically significant. d 0.0185 Do not reject H?. The evidence is not statistically significant. 18 We want to test the hypothesis that at least 85% of drivers on a freeway violate the speed limit. In a random sample of n = 1,000 vehicles, 832 violated the speed limit. Compute the sample proportion. State the null and alternative hypotheses and the decision rule. a Reject H? at ? = 5%. Conclude less than 85% of drivers violate the speed limit. b Reject H? at ? = 10% and conclude that less than 85% of drivers violate the speed limit. But, do not reject H? at ? = 5% and conclude 85% or more of drivers violate the speed limit. c Reject H? at ? = 5%. Conclude more than 85% of drivers violate the speed limit. d Reject H? at ? = 1%. Conclude less than 85% of drivers violate the speed limit. 19 In a sample of 845 coffee growers from southern Mexico, 494 growers were certified to sell to organic coffee markets. Is there evidence to indicate that fewer than 60% of the coffee growers in southern Mexico are organic certified? State your conclusion so that there is only 5% chance of making Type I error. Round your p? to three decimal points. a p-value = 0.1867 At a 5% level of significance, the evidence that fewer than 60% are organic certified is not statistically significant. b p-value = 0.1867 At a 10% level of significance, evidence that fewer than 60% are organic certified is statistically significant. c p-value = 0.0867 At a 5% level of significance, the evidence that fewer than 60% are organic certified is not statistically significant. d p-value = 0.0867 At a 10% level of significance, evidence that fewer than 60% are organic certified is statistically significant. 20 In the previous question, the margin of error for rejecting the null hypothesis is. a 0.041 b 0.037 c 0.033 d 0.028

Explanation / Answer

17)

Ho: p =.274

Ha: p > .274

p is the proportion of men without college degrees who are obese

Normal distribution with mean .274 and standard deviation sqrt(.274(1-.274)/800) = .01577

z= .321-.274 / .01577

z= 2.98

P(z>2.98) = .0015

Thus the answers is A. Since the p-value is extremely low, the results ARE statistically significant

18)

Ho: p> or equal to .85

Ha: p<.85

p is the proportion of freeway drivers that violate the speed limit

Normal model, mean = .85, standard deviation = sqrt(.85(1-.85)/1000) = .01129

z= .832 - .85 / .01129

z= -1.594

P(z<-1.594) = .0559

The answer is B. We can reject the null hypothesis at 10% because the p-value is less than .1 but not at 5% because the p-value is greater than .05

19)

Ho: p=.60

Ha: p<.60

Mean = .60, standard deviation = sqrt(.60(1-.60)/845) = .01685

z = .5846 - .60 / .01685

z = -.913

P(z<-.913) = .1814

The answer is A. Since the p-value is greater than .05, the results are not statistically significant.

20)

Since the alpha level was 5%, this corresponds to a confidence interval of 90% so

margin of error = 1.68 (Standard deviation)

= 1.68(.01685) = .283

Thus the answer is D.

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