Practice Final Exam ANSWER KEY Chapters 7-13

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1 Practice Final Exam ANSWER KEY Chapters (normal) (normal) (reverse normal) 4. The critical value for a sample size of n = 6 is Since the value of r (0.974) is greater than the critical value, we can conclude that the data come from a population that is normally distributed. 5. Let x = the time between two eruptions of Old Faithful Let x = the mean time between two eruptions of a random sample of time intervals between eruptions of Old Faithful. a) P(x > 95) = There is a 31.90% chance that a randomly selected time interval between eruptions is longer than 95 minutes. b) P x > 95 = There is a 1.77% chance that a random sample of 20 time intervals between eruptions has a mean longer than 95 minutes. c) P x > 95 = There is a 0.5% chance that a random sample of 30 time intervals between eruptions has a mean longer than 95 minutes. d) Increasing the sample size decreased the probability of observing a mean interval between eruptions of 95 minutes or longer.

2 6. We are 84% confident that the proportion of all Americans who have seen a UFO is between 2.53% and 5.47%. 7. We are 95% confident that the mean age of mean at their first marriage is between and years old. 8. In order to be accurate to within 4% of the true proportion, the sample size should be at least 601 American citizens. 9. In order to be accurate to within 10 points of the true mean combine score, the sample size should contain at least 3510 students. 10. Let μ = the average number of hours that a symphony musician practices each week. H O : μ = 5 hours H V : μ > 5 One Mean Test 1. We have a random sample 2. We can assume the population is normally distributed T=2.50 P- value= Decision: Since P- value<α, we REJECT H0 Conclusion: At the 5% level of significance, there is enough evidence to conclude that the mean time that symphony musicians practice per week is greater than 5 hours per week. 11. Let p = the proportion of Americans that believe that UFO s are real. H O : p =.5 H V : p <.5 One Proportion Test

3 b. np 1 p = = c. 20n = < N Z= P- value= Conclusion: At the 5% level of significance, there is enough evidence to conclude that less than half of all Americans think that UFO s are real. 12. H O : Health is independent of level of education H V : Health and level of education are not independent Test for Independence b. all expected counts are > 5 χ h = P- value< Conclusion: At the 5% level of significance, there is enough evidence to conclude that a person s health and level of education are not independent 13. let p i = proportion of males with at least one tatoo p l = proportion of females with at least one tatoo H O : p i = p l H V : p i p l Two Proportion Test b. males: np i 1 p i = = females: np l 1 p l = = c. males: 20n = < N ; females: 20n = < N Z=1.37

4 P- value= Decision: Since P- value>α, we FAIL TO REJECT H 0 Conclusion: At the 5% level of significance, there is not enough evidence to conclude that the proportion of males that have at least one tattoo is different than the proportion of females that have at least on tattoo. 14. Let: μ V = mean walking speed of departing passengers μ h = mean walking speed of arriving passengers H O : μ V = μ h H V : μ V < μ h not stated so use 5% level of significance Two Mean Test 1. We have a random sample 2. Both samples are greater than 30 T=-.846 P- value= Decision: Since P- value>α, we FAIL TO REJECT H 0 Conclusion: At the 5% level of significance, there is not enough evidence to conclude that the mean time walking speed of departing passengers is faster than the mean walking time of arriving passengers. 15. Let: μ V = mean number of soybean pods from the Liberty plot μ h = mean number of soybean pods from the No till plot μ v = mean number of soybean pods from the Chosel plo ed plot H O : μ V = μ h = µμ v H V : At least one mean differs ANOVA 1. All samples are random and independent 2. The QQ Plots indicate that the samples come from populations that are normally distributed (r V = 0.997, r h = 0.993, r v = The largest sample standard deviation is no more than twice the smallest sample standard deviation.

5 F= P- value= Decision: Since P- value< α, we REJECT H 0 Conclusion: At the 5% level of significance, there is enough evidence to conclude that the mean number of pods is not the same for all three plot types. 16. Let: p V = proportion of brown m&ms p h = proportion of yellow m& s p v = proportion of red m&ms p { = proportion of blue m&ms p = proportion of orange m&ms p } = proportion of green m&ms H O : p V =.13, p h = 0.14, p v = 0.13, p { =.24, p = 0.20, p } = 0.16 H V : At least two proportions differ Goodness- of- Fit 1. The sample is random. 2. All expected counts are all greater than 5 χ h = P- value= Decision: Since P- value< α, we REJECT H 0 Conclusion: At the 5% level of significance, there is enough evidence to conclude that the m&m s do not follow the distribution stated by M&M/Mars company. 17. Let: μ ~ = father s height son s height H O : μ ~ = 0 H V : μ ~ < 0 Paired difference test 1. Men s heights are normally distributed, so the differences will be normally distributed T=-.393 P- value=0.3508

6 Decision: Since P- value>α, we FAIL TO REJECT H 0 Conclusion: At the 5% level of significance, there is not enough evidence to conclude that sons are taller than their fathers. 18. H O : political affiliation s independent of income level H V : political affiliation and income level are not independent Test for Independence b. all expected counts are > 5 χ h = P- value< Conclusion: At the 5% level of significance, there is enough evidence to conclude that a person s political affiliation and their level of income are not independent 19. Let: p V = proportion ofabsences on Monday p h = proportion of absences of absences on Tuesday p v = prop rtion of absences of Wednesday p { = proportion of absences on Thursday p = proportion of absences on Friday H O : p V = p h = p v = p { = p = 0.20 H V : At least two proportions differ Goodness- of- Fit 1. The sample is random. 2. All expected counts are all greater than 5 χ h = P- value= Decision: Since P- value> α, we FAIL TO REJECT H 0

7 Conclusion: At the 5% level of significance, there is enough evidence to conclude that the proportion of absences is the same for each day of the week. 20. Let: μ V = mean assembly time for Method A μ h = mean mean assembly time for Method B H O : μ V = μ h H V : μ V μ h α = 0.01 Two Mean Test 1. We have a random sample 2. We will assume both populations have a normal distribution T=0.797 P- value= Decision: Since P- value>α, we FAIL TO REJECT H 0 Conclusion: At the 1% level of significance, there is not enough evidence to conclude that the mean assembly time differs for the two methods. 21. Let p = the proportion of Americans that smoke cigarettes. H O : p =.2 H V : p >.2 One Proportion Test b. np 1 p = = c. 20n = < N Z= 1.58 P- value= Decision: Since P- value>α, we FAIL TO REJECT H 0 Conclusion: At the 5% level of significance, there is not enough evidence to conclude that more than 20% of all Americans smoke cigarettes.

8 Let μ = average monthly long distance bill H O : μ = $20 H V : μ < $20 α = 0.10 One Mean Test 1. We have a random sample 2. n>30 T= P- value=0.065 Conclusion: At the 10% level of significance, there is enough evidence to conclude that the average monthly long distance bill is less than $20. let p V = proportion of people getting off a plane who stated they were afraid of flying p l = proportion of people getting on a plane who stated they ere afraid of flying H O : p V = p h H V : p V p h α = 0.01 Two Proportion Test b. off: np V 1 p V = = on: np h 1 p h = = c. off: 20n = < N ; on: 20n = < N Z=2.88 P- value=0.004 Conclusion: At the 1% level of significance, there is enough evidence to conclude that the population proportions are not equal.

9 Let: μ ~ = # home runs with aluminum bat #of home runs with a wood bat H O : μ ~ = 0 H V : μ ~ > 0 Paired difference test 1. The QQ Plot indicates that the population of differences is normally distributed n<N T=1.844 P- value= Conclusion: At the 5% level of significance, there is enough evidence to conclude that more home runs are hit with an aluminum bat than a wood bat. Let: μ V = mean yield from Fertilizer A μ h = mean yield from Fertilizer B μ v = mean yield from Fertilizer C μ { = mean yield from ertilizer D H O : μ V = μ h = μ v = μ { H V : At least one mean differs ANOVA 1. All samples are random and independent 2. The QQ Plots indicate that the samples come from populations that are normally distributed (r V = 0.957, r h = , r v = 0.964, r { = The largest sample standard deviation is no more than twice the smallest sample standard deviation. F= P- value= Decision: Since P- value< α, we REJECT H 0 Conclusion: At the 5% level of significance, there is enough evidence to conclude that at least one fertilizer produces a different mean yield.