A report states that based on the information obtained from the Centers for Disease Control and Prevention, 20% of young people ages 2-19 are obese. To test the claim, a researcher randomly selected 200 people ages 2-19 and found that 50 were obese. Test the claim that the percentage of obese people ages 2-19 is greater than 20?
(a) State Null and Alternate Hypothesis. Clearly define the parameters.
(b) State Test Statistic and find its value. Check conditions/ state assumptions (if any required).
(c) Find the critical value(s).
d) Carry out test of hypotheses at 1% level of significance. State statistical conclusion.
(e) State the conclusion in context of problem.
(f) Suppose you rejected Null hypothesis when the actual percentage of obese people ages 2-19 is 30, state what type of error did you make, else write no error.
(g) Calculate the probability of not concluding that the percentage of obese people is greater than 20% when the actual percentage of obese individuals is 30% for a = 0.05. (Write down the formula first and then calculate)
(h) Calculate 99% confidence interval for true average percentage of obese people ages 2-19. Clearly write down the formula first and then calculate.
Step by stepSolved in 5 steps with 3 images
- In a science fair project, Emily conducted an experiment in which she tested professional touch therapists to see if they could sense her energy field. She flipped a coin to select either her right hand or her left hand, and then she asked the therapists to identify the selected hand by placing their hand just under Emily's hand without seeing it and without touching it. Among285 trials, the touch therapists were correct 130 times. Construct a 95% confidence interval estimate of the proportion of correct responses made by touch therapists.arrow_forwardResearchers from a national research organization investigated which platform (television, social media, newspaper, Internet, etc.) United States adults prefer to use to obtain their news. In 2016, of the 200 randomly selected United States adults that were surveyed, 114 selected television as their preferred platform. In 2018, of the 200 randomly selected United States adults that were surveyed, 98 selected television as their preferred platform. The conditions for inference are met. A 95 percent confidence interval for the difference in the population proportions of United States adults that use television as their preferred platform for obtaining news between 2016 and 2018 (2016 minus 2018) was found to be (-0.02, 0.18). Based on the interval, is it reasonable for the researchers to claim that there is convincing statistical evidence that the population proportion of United States adults that use television as their preferred method to obtain news differed between 2016 and 2018? (A)…arrow_forwardA random sample of n1 = 299 voters registered in the state of California showed that 140 voted in the last general election. A random sample of n2 = 220registered voters in the state of Colorado showed that 130 voted in the most recent general election. Do these data indicate that the population proportion of voter turnout in Colorado is higher than that in California? Use a 10% level of significance. What is the level of significance? What is the value of the sample test statistic? (Test the difference p1 − p2. Do not use rounded values. Round your final answer to two decimal places.) Find (or estimate) the P-value. (Round your answer to four decimal places.)arrow_forward
- Independent random samples are taken at a university to compare the average GPA of seniors to the average GPA of sophomores. Given a 90% confidence interval for the difference between the true average GPAs (seniors - sophomores) of (-0.05, 0.78), what can you conclude? Question 8 options: 1) There is no significant difference between the true average GPA for seniors and sophomores. 2) We do not have enough information to make a conclusion. 3) We are 90% confident that the average GPA of seniors is less than the average GPA of sophomores. 4) We are 90% confident that the average GPA of seniors is greater than the average GPA of sophomores. 5) We are 90% confident that the difference between the two sample GPAs falls within the interval.arrow_forwardA simple random sample of size n=350 individuals who are currently employed is asked if they work at home at least once per week. Of the 350 employed individuals surveyed, 42 responded that they did work at home at least once per week. Construct a 99% confidence interval for the population proportion of employed individuals who work at home at least once per week.arrow_forwardRecently, a large academic medical center determined that 13 of 20 employees in a particular position were female, whereas 41% of the employees for this position in the general workforce were female. At the 0.01 level of significance, is there evidence that the proportion of females in this position at this medical center is different from what would be expected in the general workforce? Question content area bottom Part 1 What are the correct hypotheses to test to determine if the proportion is different? A. H0: π≤0.41; H1: π>0.41 B. H0: π≥0.41; H1: π<0.41 C. H0: π=0.41; H1: π≠0.41 D. H0: π≠0.41; H1: π=0.41 Part 2 Calculate the test statistic. ZSTAT=enter your response here (Type an integer or a decimal. Round to two decimal places as needed.) Part 3 What is the p-value? The p-value is enter your response here. (Type an integer or a decimal. Round to three decimal places as needed.) Part 4 State…arrow_forward
- If 10 parts per million is the threshold for poor air quality, based on the sample above, with 90% confidence, could you conclude that the air quality is poor? (a) Yes we can conclude that the air quality is poor (b) No we cannot conclude that the air quality is poor (c) We do not have enough information to come to a conclusion (d) None of the above answers are correctarrow_forwardA published report claims that 35% of college students have used an online dating site or app. Believing this claimed value is too low, a researcher surveys a random sample of n = 420 college students about their experiences with online dating sites and apps. A total of 166 of the surveyed students indicate that they have used online dating sites or apps. Use this information to conduct a hypothesis test at a significance (or alpha) level of 0.05. 1.Based on the test statistic you computed to answer Question 11, along with what you see in Table B, what should the P-value be? 2. Based on the P-value you obtained and how it compares to the significance level of 0.05, what is your conclusion? 3. Look again at your answer to Question 13. If a smaller significance level had been chosen—like 0.01—would you have reached a different conclusion? Please explain. 4. Again, look back at how you answered Question 13. If a larger significance level had been chosen—like 0.10—would you have…arrow_forwardA study done by researchers at a university concluded that 90% of all student athletes in a country have been subjected to some form of hazing. The study is based on responses from 1700 athletes. What is the margin of error and the 95% confidence interval for the study? The margin of error is (Round to four decimal places as needed.)arrow_forward
- Two different simple random samples are drawn from two different populations. The first sample consists of 40 people with 20 having a common attribute. The second sample consists of 1800 people with 1291 of them having the same attribute. Compare the results from a hypothesis test p1=p2 (with a 0.05 significance level) and a 95% confidence interval estimate of p1-p2. What are the null and alternative hypotheses for the hypothesis test?arrow_forwardFor a one-tailed (upper tail) hypothesis test with a sample size of 26 and a .01 level of significance, the critical value of the test statistic t isarrow_forwardAccording to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Calculate a 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data. Also, conduct a hypothesis, with alpha = 0.05, test to determine if these data provide strong evidence the rate of sleep deprivation is different for the two states. Construct a 95% confidence interval Which of the following correctly displays the standard error? 0.0046 0.009 0.001 0.00002 0.008arrow_forward
- MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
- Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman