The Excel file STATISTICSSTUDENTSSURVEYFORR contains the column YRAGE (a numerical variable that measures student age at the time of completing an online survey). For education purposes, consider this dataset to be a sample of size 60 taken from a much wider population for statistics students. Use R/Rcmdr to find a 97% confidence interval for the mean of the population. Choose the most correct (closest) statement below? a. Your confidence interval is (19.01790, 20.74877) yrs and it wider than a 90% confidence interval created from this data. b. Your confidence interval is (18.92152, 20.84515) yrs and is wider than a 90% confidence interval created from this data. c. Your confidence interval is (18.92152,20.84515)) yrs and is narrower than a 90% confidence interval created from this data. d. Your confidence interval is (19.01790, 20.74877) yrs and is narrower than a 90% confidence interval created from this data. YRAGE 25 18 19 17 20 19 18 21 18 17 18 19 21 22 19 18 17 22 18 21 18 18 18 18 17 19 18 19 18 17 18 18 18 20 21 17 17 17 19 18 18 19 20 19 25 20 21 33 28 30 25 24 27 17 17 18 19 22 21 20

If n=11, 7(x-bar)=45, and s=5, construct a confidence interval at a 95% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place.

You measure 40 dogs' weights, and find they have a mean weight of 71 ounces. Assume the population standard deviation is 9.6 ounces. Based on this, construct a 95% confidence interval for the true population mean dog weight. Give your answers as decimals, to two places ounces

If n=470 and p̂ (p-hat) = 0.65, construct a 95% confidence interval.

Please help with 1 a,b,c,d,e, and f.

Please help with question 5 a-b

If n = 110 and X = 88, construct a 99% confidence interval for the population proportion, p.Give your answers to three decimals <             p             <

A random sample of 64 shoppers showed that they spend an average of $27.31 per visit at the Sunday Mornings Bookstore. The standard deviation of the population is $2.14. Find the 90% confidence interval of the true mean. Please include all work showing how you found your answer.

If n=22, (x-bar)=34, and s=17, construct a confidence interval at a 99% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place.

If you are given summary statistics for a sample and want to construct a confidence interval, this can be done through the Probability Calculator in Geogebra, or Stat>Basic Stats>1-sample Z in Minitab(web). For Minitab (web), see Tutorials in D2L. For Geogebra, follow the directions below: Open it and click the "Statistics" tab just above the graph. For a confidence interval for a population mean, with unknown o, click the drop-down menu near the top (default is "Z Test of a Mean") and choose "T Estimate of a Mean." Enter the confidence level (in decimal form), sample mean, sample standard deviation, and sample size. We'll use a sample with = 65.17, s = 3.654, n = 26. Based on this sample data, a 95% confidence interval for the population mean is: <μ< The margin of error for this 95% confidence interval is Based on this sample data, a 99% confidence interval for the population mean is: <μ< The margin of error for this 99% confidence interval is -