A genetic experiment with peas resulted in one sample of offspring that consisted of 439 green peas and 169 yellow peas. a. Construct a 95% confidence interval to estimate of the percentage of yellow peas. b. It was expected that 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations?
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- A standardized exam's scores are normally distributed. In a recent year, the mean test score was 1482 and the standard deviation was 315. The test scores of four students selected at random are 1910, 1200, 2210, and 1390. Find the z-scores that correspond to each value and determine whether any of the values are unusual. The z-score for 1910 is (Round to two decimal places as needed.)In a past presidential election, 37,979,113 people voted for Candidate A 37,758,117 for Candidate B; and 206,521 for third-party candidates. a. What percentage of voters chose Candidate A? b. Would it be appropriate to find a confidence interval of voters choosing Candidate A? Why or why not?A genetic experiment with peas resulted in one sample of offspring that consisted of 407 green peas and 153 yellow peas. a. Construct a 90% confidence interval to estimate of the percentage of yellow peas. b. It was expected that 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations? a. Construct a 90% confidence interval. Express the percentages in decimal form. b. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations?
- In a survey of 3492 adults aged 57 through 85 years, it was found that 84.6% of them used at least one prescription medication. Complete parts (a) through (c) below. a. How many of the 3492 subjects used at least one prescription medication? 2954 (Round to the nearest integer as needed.) b. Construct a 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication.In a survey of 3092 adults aged 57 through 85 years, it was found that 88.5% of them used at least one prescription medication. Complete parts (a) through (c) below. K @ a. How many of the 3092 subjects used at least one prescription medication? 2736 (Round to the nearest integer as needed.) b. Construct a 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication. View an example Get more help. 2 %In a random sample of 800 men aged 25 to 35 years, 24% said they live with one or both parents. In another sample of 850 women of the same age group, 18% said that they live with one or both parents.A. Construct a 95% confidence interval for the difference between the proportions of all men and all women aged 25 to 35 years who live with one or both parents.B. Test at the 2% significance level whether the two population proportions are different.C. Repeat the test of part b using the p-value approach.A genetic experiment with peas resulted in one sample of offspring that consisted of 445 green peas and 168 yellow peas. a. Construct a 90% confidence interval to estimate of the percentage of yellow peas. b. It was expected that 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations? a. Construct a 90% confidence interval. Express the percentages in decimal form. (Round to three decimal places as needed.)A sample of 100 shoppers chosen at random from all shoppers in a given province indicated that 55% of them preferred a particular store (Shoprite).i. Find a 95% confidence interval for the proportion of all the shoppers that preferred Shoprite.ii. How large a sample of shoppers should we take in order to be 95% confident that the shoppers will buy from Shoprite?A survey was conducted to determine what percentage of college seniors would have chosen to attend a differentcollege if they had known then what they know now. In a random sample of 100 seniors, 34 percent indicated thatthey would have attended a different college. A 90 percent confidence interval for the percentage of all seniors whowould have attended a different college isA researcher surveyed college students in the United States on the typical amount of time each day that they spend interacting with different types of media (television, social media, Internet-connected devices, game consoles, etc.) The researcher found that the mean amount of time that college students spent watching television each day is 135 minutes with a 95% confidence interval of (105, 165). a. State the conclusion the researcher can make from this confidence interval. b. What is the margin of error for the confidence interval?In a survey funded by the UW school of medicine, 750 of 1000 adult Seattle residents said they did not believe they could come down with a sexually transmitted infection (STI). Construct a 95% confidence interval estimage of the proportion of adult Seattle residents who don't believe they can contract an STI. (Use a z score of 1.96 for your computations.) (.728, .772) (.723, .777) (.718, .782) (.713, .878) (.665, .835)Suppose that a manager is interested in estimating the average amount of money customers spend in her store. After sampling 36 transactions at random, she found that the average amount spent was $32.15. She then computed a 90% confidence interval to be between $29.18 and $35.12. Which statement gives a valid interpretation of the interval? O There is a 90% chance that the mean amount spent by all customers is between $29.18 and $35.12. O There is a 90% chance that a randomly selected customer will spend between $29.18 and $35.12. O The store manager is 90% confident that the average amount spent by the 36 sampled customers is between $ 29.18 and $35.12. The store manager is 90% confident that the average amount spent by all customers is between $29.18 and $35.12.SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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