Determine the Sample size taken from a normal distribuhion N(M, 25) to get leng th of interval of 4 using 90% Confidence level for the mean Ans. n =17
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- The critical value of a one-tailed greater than, alpha equals 0.05, df=11 One-Sample T-Test for Mean is?You've collected a sample of 14 concrete blocks and weighed them. You want to produce a 99% confidence interval for the variance of the weight of the blocks. Provide the two values of x² that you would use in the calculation of this interval. РAn SRS of 450 high school seniors gained an average of x = 20.80 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation a = 54.81. We want to estimate the mean change in score in the population of all high school seniors. (a) Using the 68-95-99.7 Rule or the z-table (Table A), give a 68% confidence interval (a, b) for u based on this sample. H (Enter your answers rounded to three decimal places. If you are using Crunchlt, adjust the default precision under Preferences as necessary. See the instructional video on how to adjust precision settings.) a: (b) Based on your confidence interval in part (a), how certain are you that the mean change in score u in the population of all high school seniors is greater than 0? The interval does not contain 0, so we are 68% certain that the mean change in score in the population of all high school seniors is greater than 0. We cannot be certain at all. If we want…
- 2) (9.2) Suppose a random sample of 1001 Americans age 15 or older were asked "How much time do you spend eating or drinking per day?" Of the 1001 participants, the mean amount of time spent eating or drinking per day is 1.22 hours with a standard deviation of 0.65 hours. Determine a 99% confidence interval for the mean amount of time Americans age 15 or older spend eating or drinking each day.Find the 95% confidence interval for estimating the population mean μ If sample mean X = 50, sample size n= 60 and population standard deviation σ is known to be 10.The sample mean and standard deviation from a random sample of 32 observations from a normal population were computed asx =25 and s = 10. Calculate the t statistic of the test required to determine whether there is enough evidence to infer at the 8% significance level that the population mean is greater than 22.Test Statistic =
- A random sample of 16 men was selected from the population of ISU students. The sample's average BMI value was 22.5 with a standard deviation of 4.8. Create a 99% confidence interval for the mean of the population of male student. I DFocus WThe additional growth of plants in one week are recorded for 6 plants with a sample standard deviation of 1 inches and sample mean of 12 inches. t∗ at the 0.10 significance level = Margin of error = Confidence interval = [ , ] [smaller value, larger value]A random sample of 100 components is drawn from a population size of 1546. Assume that the population standard deviation is 2.93 mm and the mean measurement of the components in the sample is 67.45mm with. Determine the 95% confidence intervals for an estimate of the mean of the population.
- A sample of size 65 gave a mean 7.4 and standard deviation as 1.6 then 95% confidence interval for population mean will be Select one: O (17.008, 17.792) O [27.008, 30.008] O (7.008, 7.792) O (9,265, 10.735)The additional growth of plants in one week are recorded for 11 plants with a sample standard deviation of 2 inches and sample mean of 12 inches. t* at the 0.05 significance level = Ex: 1.234 Margin of error = Ex: 1.234 Confidence interval = [ Ex: 12.345 E, Ex: 12.345 ]The oxide thickness of semiconductors for a random sample of n = 16 yielded a mean of 424 with a standard deviation of 8. It is assumed that oxide thickness is normally distributed. Calculate the 90% confidence interval for the mean oxide thickness of semiconductors. Round off final calculations to 1 decimal place. 1. What are the Lower limit and Upper limit