A genetic experiment with peas resulted in one sample of offspring that consisted of 421 green peas and 154 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 from the sample is not 25%, do the results contradict expectations? a. Construct a 95% confidence interval. Express the percentages in decimal form. 234
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![A genetic experiment with peas resulted in one sample of offspring that consisted of 421 green peas and 154 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 from the sample is not 25%, do the results
contradict expectations?
a. Construct a 95% confidence interval. Express the percentages in decimal form.
234 <p<.305 (Round to three decimal places as needed.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1178bd61-fcb4-4065-9a76-5568b10c7a22%2Fd3c8c718-7791-4d86-a6b0-1f10d059c2e5%2F8x3uj7_processed.jpeg&w=3840&q=75)
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- A 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?The Athletics Department at a high school is interested in the difference in the run times between 11th and 12th grade students. Suppose the 11th grade students make up the first sample while the 12th grade students make up the second.Using a confidence level of 95%, the confidence interval found was (−1.8,2.9)(-1.8,2.9).Which of the following is the correct interpretation? We are 95% confident that the run times of the 11th grade students are, on average, faster than those of the 12th grade students. We are 95% confident that the run times of the 11th grade students are, on average, slower than those of the 12th grade students. We are 95% confident that the run times of the 11th grade students are, on average, about the same as those of the 12th grade students.A statistics instructor randomly selected four bags of oranges, each bag labeled 10 pounds, and weighed the bags. They weighed 10.3, 10.8, 10.4, and 10.6 pounds. Assume that the distribution of weights is Normal. Find a 95% confidence interval for the mean weight of all bags of oranges. Use technology for your calculations. Answer parts a and b below. a. Choose the correct interpretation of the confidence interval below and, if necessary, fill in the answer boxes to complete your choice. A. We are 95% confident the population mean is between and B. There is a 95% chance that all intervals will be between and C. We are 95% confident that the sample mean is between and D. The requirements for constructing a confidence interval are not satisfied. (Type integers or decimals rounded to the nearest thousandth as needed. Use ascending order.)
- In a survey of 3242 adults aged 57 through 85 years, it was found that 82.9% of them used at least one prescription medication. Complete parts (a) through (c) below. a. How many of the 3242 subjects used at least one prescription medication? ____ (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. ____%<p<____% (Round to one decimal place as needed.) c. What do the results tell us about the proportion of college students who use at least one prescription medication? A. The results tell us that there is a 90% probability that the true proportion of college students who use at least one prescription medication is in the interval found in part (b). B. The results tell us that, with 90% confidence, the probability that a college student uses at least one prescription medication is in the interval found in part (b).…A genetic experiment with peas resulted in one sample of offspring that consisted of 425 green peas in a 162 yellow peas. a. Construct a 95% confidence interval to estimate of the percentage of yellow peas. b. it was expected at 25% of the offspring peas would be yellow. Given the percentage of offspring yellow peas is not 25%, do the results contradict expectations?A genetic experiment with peas resulted in one sample of offspring that consisted of 408 green peas and 153 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? a. Construct a 95% confidence interval. Express the percentages in decimal form. |A genetic experiment with peas resulted in one sample of offspring that consisted of 405 green peas and 156 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. nothing<p<nothing (Round to three decimal places as needed.) b. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations? No, the confidence interval includes 0.25, so the true percentage could easily equal 25% Yes, the confidence interval does not include 0.25, so the true percentage could not equal 25% Click to select your answer(s).A genetic experiment with peas resulted in one sample of offspring that consisted of 446 green peas and 161 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. OYou want to estimate the average difference in the number of complaints filed by women at work and the number of complaints filed by men at work. You took an SRS of 7 businesses and collected the following data for number of complaints in a year: men 1 3 2 4 6 3 5 women 8 7 6 3 5 12 15 Create a 90% confidence interval for the average difference in the number of complaints filed by women at work and the number of complaints filed by men at work (in a year). You must do this using table C. If this is a 2 sample interval, be sure to specify who is population 1 and who is population 2. If this is a matched pairs interval, be sure to state the order you subtracted your data. Once you create your interval, give a sentence explaining the meaning of the interval (what the numbers in the interval mean, NOT the meaning of 95%) in context of this problem. Round to any appropriate number of decimals.A genetic experiment with peas resulted in one sample of offspring that consisted of 448 green peas and 171 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? Express the percentages in decimal form.A genetic experiment with peas resulted in one sample of offspring that consisted of 431 green peas and 151 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.A survey was conducted to determine the proportion of adults who approve of attempts to clone a human. Based on the response of 1000 adults a 90% confidence interval was calculated to be (0.09, 0.13). A.What is the point estimate of the true proportion of adults who approve of attempts to clone a human? 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