If the critical t is ±1.796 and the obtained t is -2.09, what decision would you make regarding the null hypothesis?
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If the critical t is ±1.796 and the obtained t is -2.09, what decision would you make regarding the null hypothesis?
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- 10% of all Americans suffer from sleep apnea. A researcher suspects that a higher percentage of those who live in the inner city have sleep apnea. Of the 302 people from the inner city surveyed, 36 of them suffered from sleep apnea. What can be concluded at the level of significance of αα = 0.05? For this study, we should use The null and alternative hypotheses would be: Ho: (please enter a decimal) H1: (Please enter a decimal) The test statistic = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is αα Based on this, we should the null hypothesis. Thus, the final conclusion is that ... The data suggest the population proportion is not significantly larger than 10% at αα = 0.05, so there is sufficient evidence to conclude that the population proportion of inner city residents who have sleep apnea is equal to 10%. The data suggest the population proportion is not…For (15) – (23) identify the IV, DV, and the correct hypothesis test to run. Then conduct the hypothesis test based on the instructions given with each item. 15. A researcher for a cereal company wanted to demonstrate the health benefits of eating oatmeal. A sample of 9 volunteers was obtained and each participant at a fixed diet without any oatmeal for 30 days. At the end of the 30-day period, cholesterol was measured for each individual. Then participants began a second 30-day period in which they repeated exactly the same diet except they added 2 cups of oatmeal each day. After the second 30-day period, cholesterol levels were measured again and the researcher recorded the difference between the two scores for each participant. For this sample, cholesterol scores averated M = 16 points lower with the oatmeal diet with SS = 538 for the difference scores. a. Are the data sufficient to indicate a significant change in cholesterol level? Test %3D %3D with α.01. %3D b. Compute estimated…Imagine you are doing a Related-Samples t-test, and you are on Step Two of the hypothesis test. The sample of difference scores that you calculated is -1, -1, -3, +2, 0, -4. If you are doing the test with α = .01, and two tails, what would he correct t-critical be?
- Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three cell phones. State the two tailed hypothesis test, calculate the p-value for the test and test whether you can reject H0 using α= 0.05.Suppose a group of 900 smokers (who all wanted to give up smoking) were randomly assigned to receive an antidepressant drug or a placebo for six weeks. Of the 356 patients who received the antidepressant drug, 113were not smoking one year later. Of the 544 patients who received the placebo, 203 were not smoking one year later. Given the null hypothesis ?0:(?1−?2)=0H0:(p1−p2)=0 and the alternative hypothesis ??:(?1−?2)≠0H, conduct a test to see if taking an antidepressant drug can help smokers stop smoking. Use ?=0.01(a) The rejection region is |?|>(b) The test statistic is ?=zThe final conclustion isYou run a one sample t-test on your data and get a p value of .003. That means that: you can reject the alternate hypothesis that your sample did not come from this population O you fail to reject the null hypothesis that your sample came from this population O you can reject the null hypothesis that your sample came from this population O you can accept the null hypothesis that your sample came from this population
- Normally only about 19% of people who try to quit smoking succeed, but sellers of a motivational tape claim that listening to their recorded messages can help people quit (i.e., more than 19% of the people who use the tapes will successfully quit). Let p represents the percentage of people who have listened to the recorded messages and have successfully quit. Match the appropriate null and alternative hypotheses, based on the sellers' claim. H,: p = 0.19 p> 0.19 p2 0.19 ps 0.19 p 0.19 p2 0.19 ps 0.19 p < 0.19 Op- 0.19 Do you consider a Type I or Type Il error a more serious mistake here? Why? Your answer should make it clear what Type I and Il errors are in this context.A well-known brokerage firm executive claimed that at least 24 % of investors are currently confident of meeting their investment goals. An XYZ Investor Optimism Survey, conducted over a two week period, found that out of 150 randomly selected people, 42 of them said they are confident of meeting their goals. Suppose you are have the following null and alternative hypotheses for a test you are running: Ho:p = 0.24 Ha:p + 0.24 Calculate the test statistic, rounded to 3 decimal places80% of all supermarket prices end in a 9 or a 5. Suppose you check a random sample of 113 items and find that 84 have prices that end in a 9 or 5. Does that indicate that fewer than 80% of the prices end in a 9 or a 5? Use alpha=0.05 State the null and alternet hypotheses.
- You collect N = 200 frogs from a pond and find that 120 of them are female. You wish to conduct the following hypothesis test (α = 0.05) for p (the proportion of female frogs in the pond): H0 : p = 0.55 HA : p > 0.55 Determine which of the following statements are true. Assume the true p = 0.65. I. The one-sided p-value is 0.1528. II. You reject the null hypothesis H0. III. You fail to reject H0 and make a Type II error. Group of answer choices None of the Above III only I and III only II only I only19% of all college students volunteer their time. Is the percentage of college students who are volunteers smaller for students receiving financial aid? Of the 338 randomly selected students who receive financial aid, 44 of them volunteered their time. What can be concluded at the α = 0.05 level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: (please enter a decimal) Please enter a decimal) c. The test statistic (please show your answer to 3 decimal places.) 2 (Please show your answer to 4 decimal places.) d. The p-value e. The p-value is? a f Based on this, we should Select an answer g. Thus, the final conclusion is that the null hypothesis. The data suggest the population proportion is not significantly lower than 19% at α-0.05, so there is insufficient evidence to conclude that the percentage of financial aid recipients who volunteer is lower than 1996. The data suggest the population proportion is not significantly…Suppose a group of 800800 smokers (who all wanted to give up smoking) were randomly assigned to receive an antidepressant drug or a placebo for six weeks. Of the 500500 patients who received the antidepressant drug, 134134 were not smoking one year later. Of the 300300 patients who received the placebo, 5151 were not smoking one year later. Given the null hypothesis ?0:(?1−?2)=0H0:(p1−p2)=0 and the alternative hypothesis ??:(?1−?2)≠0Ha:(p1−p2)≠0, conduct a test to see if taking an antidepressant drug can help smokers stop smoking. Use ?=0.03α=0.03 (a) The rejection region is |?|>|z|> (b) The test statistic is ?=z=