Consider the following summary of a data set on the weekly earnings (in hundreds of dollars) of residents of NSW with a colleague degree: Number of observations : 25 Sample mean : 20.5 Sample standard deviation : 8 We assume that the weekly earning in the population is normally distributed with unknown mean and unknown variance. Please select the correct 95% confidence interval for the mean of the weekly earning (in hundreds of dollars) in the population. Group of answer choices [17.2, 23.8] [17.8, 23.2] [19.8, 21.2] [20.0, 21.0] [6.8, 34.2]

A survey of 8 university student revealed a mean income of $180.98 with a standard deviation of $143.042. Calculate a 95% confidence interval for the mean income of all students:

A random sample of 160 households is selected to estimate the mean amount spent on electric service. A 95% confidence interval was determined from the sample results to be ($151, $216). Which of the following is the correct interpretation of this interval? There is a 95% chance that the mean amount spent on electric service is between $151 and $216. We are 95% confident that the mean amount spent on electric service among the 160 households is between $151 and $216. 95% of the households will have an electric bill between $151 and $216. We are 95% confident that the mean amount spent on electric service among all households is between $151 and $216

You would like to construct a 95% confidence interval to estimate the population mean annual income for people over age 40 in your state. You choose a random sample of such incomes and compute the mean of the sample as 40,295.5 dollars with a standard deviation of 7563.4 dollars.(a) What is the best point estimate, based on the sample, to use for the population mean?dollars(b) For each of the following sampling scenarios, determine which distribution should be used to calculate the critical value for the 95% confidence interval for the population mean.(In the table, Z refers to a standard normal distribution, and t refers to a t distribution.)Sampling scenario Z t Could use either Z or t UnclearThe sample has size 15, and it is from a population with a distribution about which we know very little. Zt The sample has size 85, and it is from a non-normally distributed population with a known standard deviation of 7752. Zt The sample has size 20, and it is from a normally distributed population with an unknown standard deviation. Zt CheckSave For LaterSubmit Assignment

A student was asked to find a 95% confidence interval for weight of their backpacks in pounds using data from a random sample of size n = 20. Which of the following is a correct interpretation of the interval 3.1 < μ < 8.6? Assume the population is normally distributed.With 95% confidence, the weight of a randomly selected backpack will be between 3.1 and 8.6 pounds.There is a 95% chance that the mean of a sample of 20 backpacks will weigh between 3.1 and 8.6 pounds.There is a 95% chance that the weight is between 3.1 and 8.6.With 95% confidence, the mean weight of all backpacks is between 3.1 and 8.6 pounds.The sample mean weight of all backpacks is between 3.1 and 8.6 pounds, 95% of the time. We know this is true because the mean of our sample is between 3.1 and 8.6.

A study was conducted to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the population standard deviation is assumed to be σ = 3.6 hours. The 95% confidence interval for the mean, μ, is (7.7, 9.3).Which of the following will provide a more informative (i.e., narrower) confidence interval than the 95% confidence interval? Check all that apply. Using a sample of size 400 (instead of 81) Using a sample of size 36 (instead of 81) Using a different sample of size 81 Using a 90% level of confidence (instead of 95%) Using a 99% level of confidence (instead of 95%)