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- Use the following scenario for the exercises that follow: In the game of Keno, a player starts by selecting 20 numbers from the numbers 1 to 80. After the player makes his selections, 20 winning numbers are randomly selected from numbers 1 to 80. A win occurs if the player has correctly selected 3,4, or 5 of the 20 winning numbers. (Round all answers to the nearest hundredth of a percent.) 51. What is the percent chance that a player selects exactly 3 winning numbers?Use the following scenario for the exercises that follow: In the game of Keno, a player starts by selecting 20 numbers from the numbers 1 to 80. After the player makes his selections, 20 winning numbers are randomly selected from numbers 1 to 80. A win occurs if the player has correctly selected 3,4, or 5 of the 20 winning numbers. (Round all answers to the nearest hundredth of a percent.) 53. What is the percent chance that a player selects all 5 winning numbers?Assume that the probability that an airplane engine will fail during a torture test is 12and that the aircraft in question has 4 engines. Construct a sample space for the torture test. Use S for survive and F for fail.
- Use the following scenario for the exercises that follow: In the game of Keno, a player starts by selecting 20 numbers from the numbers 1 to 80. After the player makes his selections, 20 winning numbers are randomly selected from numbers 1 to 80. A win occurs if the player has correctly selected 3,4, or 5 of the 20 winning numbers. (Round all answers to the nearest hundredth of a percent.) 55. How much less is a player’s chance of selecting 3 winning numbers than the chance of selecting either 4 or 5 winning numbers?In recent years, approximately 55% of eligible voters take the time to vote in presidential elections. A poll based on a random sample of 250 eligible voters finds that 112 plan to vote in the next presidential election. Does this data provide convincing evidence at the a=0.01 level that the proportion of eligible voters who will take time to vote in the next presidential election differs from 0.55? STATE: H₁: P. 0.55 H: 0.55 P where P- - the proportion of all eligible voters who will take the time to vote in the next presidential election. The evidence for His > 0.55 PLAN: Drag each statement from the answer bank to the appropriate box. True Statements Name of text: One-sample-test for p False Statements This is a random sample of 250 eligible voters. Name of text: Two-sample z text for p - P The Large Counts condition is met. Answer Bank (-) 113210 The random condition is not mat -13810 The Large Counts condition is not met. -11210 (A) 138210 (Enter 3 decimal places) z= (Round to…A political geographer is interested in the spatial voting pattern during the last presidential election. She suspects that university professors in her state were more likely than the statewide population to vote for candidate A. The statewide percentage of the population voting for candidate A was 0.38. She takes a random sample of 45 professors in the state, and finds that 20 voted for candidate A. Is there sufficient evidence to support her hypothesis? Use α = 0.05. What is the p-value?
- A researcher is investigating the effect of acupuncture treatment for chronic back pain. A sample of n = 4 participants is obtained from a pain clinic. Each individual ranks the current level of pain and then begins a 8-week treatment. At the end of the program, the pain level is rated again. For this sample, pain level decreased by an average MD = -4.5 points with SS = 27. Can you conclude that acupuncture treatment significantly reduces back pain? Use α = .05.Automobiles Purchased An automobile owner found that 20 years ago, 73% of Americans said that they would prefer to purchase an American automobile. He believes that the number is much less than 73% today. He selected a random sample of 45 Americans and found that 30 said that they would prefer an American automobile. Can it be concluded that the percentage today is less than 73% ? At α=0.05, is he correct? Use the P -value method with tables (a) State the hypotheses and identify the claim. :H0 ▼(Choose one) :H1 ▼(Choose one) This hypothesis test is a ▼(Choose one) test. (b) Find the critical value(s). Round the answer to two decimal places. If there is more than one critical value, separate them with commas. Critical value(s): (c) Compute the test value. Always round z score values to at least two decimal places. =z (d) Make the decision. ▼(Choose…In a recent survey, the average amount of moneystudents have in their pockets is $5.40 with a standarddeviation of $1.20. A teacher feels that the averageamount is actually higher. She surveys 80 randomlyselected students and finds the average amount is$5.50. At α = 0.05, test the teacher’s claim that theaverage amount of money is actually higher than $5.40.
- Are you an impulse shopper? A survey of 600 grocery shoppers indicated that 57% of males and 44% of females make an impulse purchase every time they shop. Assume that the survey consisted of 300 males and 300females. a. At the 0.05level of significance, is there evidence of a difference in the proportion of males and females who make an impulse purchase every time they shop? b. Find the p-value and interpret its meaning.Automobiles Purchased An automobile owner found that 20 years ago, 76% of Americans said that they would prefer to purchase an American automobile. He believes that the number is much greater than 76% today. He selected a random sample of 58 Americans and found that 50 said that they would prefer an American automobile. Can it be concluded that the percentage today is greater than 76% ? At α=0.10 , is he correct? Use the P -value method with tables. (a) State the hypotheses and identify the claim. :H0 ▼(Choose one) :H1 ▼(Choose one) This hypothesis test is a ▼(Choose one) test. (b) Find the critical value(s). Round the answer to two decimal places. If there is more than one critical value, separate them with commas. Critical value(s): (c) Compute the test value. Always round z score values to at least two decimal places. =z (d) Make the decision.…Previously, 4% of mothers smoked more than 21 cigarettes during their pregnancy. An obstetrician believes that the percentage of mothers who smoke 21 cigarettes or more is less than 4% today. She randomly selects 115 pregnant mothers and finds that 3 of them smoked 21 or more cigarettes during pregnancy. Test the researcher's statement at the α=0.1 level of significance. find p value