95% confidence interval for the average increase of height.
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A scientist wants to establish if the Mississippi’s river height has increased using data from 1970 and 2020. Historically, it’s known that the variance of the height within any given year is 4.5ft. The sample from 1970 has 80 measurements with ?1μ1 = 30ft and the sample from 2020 has 260 measurements with ?2μ2 = 31.5ft. Find a 95% confidence interval for the average increase of height.
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- The table below lists measured amounts of redshift and the distances (billions of light-years) to randomly selected astronomical objects. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. There is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For the prediction interval, use a 90% confidence level with a redshift of 0.0126. Redshift Distance a. Find the explained variation. 0.0237 0.34 (Round to six decimal places as needed.) 0.0541 0.75 0.0723 0.98 C 0.0397 0.57 0.0444 0.62 0.0103 0.13Suppose the length of a particular type of snake is normally distributed with a mean of 50 inches and a standard deviation of 2.2 inches. You took a sample of 25 such snakes and the mean length for these snakes is 53 inches. a. Test the hypothesis that this sample came from a population with an average length of 50 inches. What is your conclusion if α = .05. Show your calculations including Excel functions used to do the calculations. b. Your friend believes that you took the sample from a population of snakes whose length is below 50 inches. Is your friend correct? What is your conclusion? Use α = .05 (5+5)The accompanying table lists systolic blood pressures (mm Hg) and diastolic blood pressures (mm Hg) of adult females. Find the (a) explained variation, (b) unexplained variation, and (c) prediction interval for a systolic blood pressure of 120 mm Hg using a 99% confidence level. There is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. Click the icon to view the blood pressure data. Blood Pressures a. The explained variation is 650 (Round to two decimal places as needed.) b. The unexplained variation is (Round to two decimal places as needed.) Systolic Diastolic 126 69 102 67 128 75 107 63 c. The 99% prediction interval for a systolic bloo (Round to one decimal place as needed.) 156 74 95 52 154 89 110 71 120 69 116 75 103 60 126 69 Print Done X
- The table below lists measured amounts of redshift and the distances (billions of light-years) to randomly selected astronomical objects. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. There is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For the prediction interval, use a 90% confidence level with a redshift of 0.0126. D Redshift Distance 0.0237 0.32 0.0545 0.75 0.0724 1.02 0.0397 0.56 0.0442 0.61 0.0103 0.16 a. Find the explained variation. (Round to six decimal places as needed.) b. Find the unexplained variation. (Round to six decimal places as needed.) c. Find the indicated prediction interval. billion light-yearsA science teacher claims that the mean scores on a science assessment test for fourth grade boys and girls are equal. The mean score for 18 randomly selected boys is 153 with a standard deviation of 23 and the mean score for 20 randomly selected girls is 149 with a standard deviation of 30. At α=0.01, can you reject the teacher’s claim? Assume the populations are normally distributed and the variance are not equal. Include a filled out standardized test statistic formula. You may not use the following df formula for this problemFifty students take two midterm exams. On the first exam, the mean score is 65.0 and the standard deviation is 7.00. On the second exam, the mean score is 55.0 with a standard deviation of 10.00. The correlation coefficient of the two score is 0.70. Obtain the least squares regression line using the second exam score to predict the first exam score.In the 1800s, German physician Carl Reinhold, took millions of axillary (i.e. armpit) temperatures from soldiers. This study established that body temperature is normally distributed and the standard normal human body temperature is 98.6°F with a standard deviation of 0.72 °F. In a recent study, American researchers obtained 5,000 axillary temperatures from a Los Angeles hospital. The mean of these temperature readings was 97.9 °F. Assuming a Type I error risk of no more than 5%, did the findings support the theory that human, body temperature has decreased since the 1800s? What is the Z crit?A researcher speculates that because of differences in diet, Japanese children have a lower mean blood cholesterol level than U.S. children do. Suppose that the mean level for U.S. children is known to be 165. Let μ represent the true mean blood cholesterol level for Japanese children. What hypotheses should the researcher test? Ho: μ = 165 versus H₂: μ> 165 a O Ho: μ = 165 versus H₂: μ 165 O Ho: μ = 165 versus H: μ > 1553.2/2The table below lists measured amounts of redshift and the distances (billions of light-years) to randomly selected astronomical objects. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. There is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For the prediction interval, use a 90% confidence level with a redshift of 0.0126. Redshift Distance 0.0238 0.31 a. Find the explained variation. 0.0543 0.74 (Round to six decimal places as needed.) b. Find the unexplained variation. (Round to six decimal places as needed.) c. Find the indicated prediction interval. 0.0722 1.02 billion light-yearsIt has been reported that the variance of the speeds of drivers on the Turnpike near Somerset, PA is 64 for all vehicles. A driver who was pulled over near Somerset on the pike feels that value is incorrect and conducts a 4. survey of speeds. He finds that the standard deviation of 50 vehicles is 10.5 mph. At a 0.05, is the driver correct? %3DA math teacher claims that the mean scores on a math assessment test for fourth grade boys and girls are equal. The mean score for 13 randomly selected boys is 151 with a standard deviation of 36, and the mean score for 15 randomly selected girls is 149 with a standard deviation of 34. At α = 0.01, can you reject the teacher's claim? Assume the populations are normally distributed and the population variances are equal.SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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