Express the confidence interval 367.1<μ<452.5367.1<μ<452.5 in the form of ¯x±MEx¯±ME
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Express the confidence interval 367.1<μ<452.5367.1<μ<452.5 in the form of ¯x±MEx¯±ME.
¯x±ME=x¯±ME= ±±
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- Let Y be a continuous random variable that is normally distributed with mean, µ = 30 and standard deviation, o = 2.4, find P( 31.2 < y < 33.6) Answer should be accurate to 4 decimal places and entered in the format 0.XXXXA random sample of 18 size AA batteries for toys yield a mean of 2.6 hours with standard deviation, 1.42 hours. (a) Find the critical value, t*, for a 99% CI. t* = (b) Find the margin of error for a 99% CI. =Express the confidence interval −15.5<μ<99.9-15.5<μ<99.9 in the form of ¯x±MEx¯±ME.
- you measure 22 turtles' weights, and find they have a mean weight of 31 ounces. Assume the population standard deviation is 14.4 ounces. Based on this, construct a 90% confidence interval for the true population mean turtle weight.Give your answers as decimals, to two places ±± ounces please show me what function to use on my ti 84 so I can understand how you got the answer. Im struggling with all of these problems and havent been taught how to solve it by handWhat is the interval ( x-1.645 sigma/n, infinite) for the true average porosity of a certain seam of coal samples that is normally distributed with true standard deviation 0.71 if the average porosity for 20 specimens from the seam was 4.95?A simple random sample from a population with a normal distribution of 108 body temperatures has x= 98.50 F and s 0.62°F. Construct a 98% confidence interval estimate of the standard deviation of body temperature of all healthy humans. Click the icon to view the table of Chi-Square critical values. O'FSolve this problem to find standard deviation to the nearest 10thA simple random sample from a population with a normal distribution of 104 body temperatures has x= 98.20°F and s= 0.69°F. Construct a 95% confidence interval estimate of the standard deviation of body temperature of all healthy humans. Click the icon to view the table of Chi-Square critical values. O'FA simple random sample of size n = 40 is drawn from a population. The sample mean is found to be 108.6, and the sample standard deviation is found to be 19.6. Is the population mean greater than 100 at the x = 0.01 level of significance? Determine the null and alternative hypotheses. Ho: μ = 100 H₁: μ>100 Compute the test statistic. = 2.78 (Round to two decimal places as needed.) Determine the P-value. The P-value is 0.004. (Round to three decimal places as needed.) What is the result of the hypothesis test? the null hypothesis because the P-value is the level of significance. At the α = 0.01 level of significance, the population mean6.1.5 Find the critical value z, necessary to form a confidence interval at the level of confidence shown below. c = 0.94 (Round to two decimal places as needed.)A sample of 55 measurements resulted in a sample mean, ?⎯⎯⎯=9.3x¯=9.3 and sample standard deviation ?=1.47s=1.47. Using ?=0.05α=0.05, test the null hypothesis that the mean of the population is 10.510.5 or less against the alternative hypothesis that the mean of the population, ?>10.5μ>10.5 by giving the following:(a) the degree of freedom (b) the test statistic The final conclusion is A. There is not sufficient evidence to reject the null hypothesis that ?=10.5μ=10.5. B. We can reject the null hypothesis that ?=10.5μ=10.5A random sample of 90 observations produced a mean of x = 34.6 from a population with a normal distribution and a standard deviation o = 4.77. (a) Find a 95% confidence interval for μSEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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