To find sample size (n) such that
Margin of error = 33
ans-> D) 588
Find an estimate of the sample size needed to obtain a margin of error of 33...
What sample size is needed to obtain a 95% confidence interval whose margin of error is no more than 1.7 for the mean of a normal population with standard deviation 4.5?
Find the margin of error for a 95% confidence interval for estimating the population mean when the sample standard deviation equals 92, with a sample size of (a) 400,(b) 1800. What is the effect of the sample size? 2. The margin of error for a 95% confidence interval with a sample size of 400 is (Round to the nearest tenth as needed.) b. The margin of error for a 90% confidence interval with a sample size of 1600 is (Round...
Assume that you want to construct a 95% confidence interval estimate of a population mean. Find an estimate of the sample size needed to obtain the specified margin of error for the 95% confidence interval. The sample standard deviation is given below. Margin of error =$5,standard deviation=$25 The required sample size is ????? (Round up to the nearest whole number as needed.)
For the provided sample mean, sample size, and population standard deviation, complete parts (a) through (c) below. Assume that x is normally distributed x= 27, n=9, 0 = 6 a. Find a 95% confidence interval for the population mean The 95% confidence interval is from to (Round to two decimal places as needed.) b. Identify and interpret the margin of error. The margin of error is (Round to two decimal places as needed.) Interpret the margin of error. Choose the...
Assume that you want to construct a 95% confidence interval estimate of a population mean. Find an estimate of the sample size needed to obtain the specified margin of error for the 95% confidence interval. The sample standard deviation is given below. Margin of error equals=$66, standard deviation equals=$2222 The required sample size is _____. (Round up to the nearest whole number as needed.)
Assume that you want to construct a 95% confidence interval estimate of a population mean. Find an estimate of the sample size needed to obtain the specified margin of error for the 95% confidence interval. The sample standard deviation is given below. Margin of errorequals $3, standard deviationequals $23 The required sample size is nothing . (Round up to the nearest whole number as needed.)
Assume that you want to construct a 95% confidence interval estimate of a population mean. Find an estimate of the sample size needed to obtain the specified margin of error for the 95% confidence interval. The sample standard deviation is given below. Margin of error equals$5, standard deviation equals$19 The required sample size is __.
For the provided sample mean, sample size, and population standard deviation, complete parts (a) through (c) below. x= 23, n= 36, 3 = 3 a. Find a 95% confidence interval for the population mean. The 95% confidence interval is from to (Round to two decimal places as needed.) b. Identify and interpret the margin of error. The margin of error is (Round to two decimal places as needed.) Interpret the margin of error. Choose the correct answer below. O A....
33. Find the z-score used in the formula to construct a 92% confidence interval for a population proportion: O a. 1.4051 Ob. 1.5548 Oc. 1.7507 Od. 1.96 34. All of the following are TRUE about 95% confidence intervals for a population mean except ::* O a. The population mean may or may not be in the confidence interval. Ob. The value of T varies depending on sample size. Oc. If the sample size is large, the Central Limit Theorem says...
Determine the sample size needed to construct a 95% confidence interval to estimate the average GPA for the student population at a college with a margin of error equal to 0.2. Assume the standard deviation of the GPA for the student population is 25 The sample size needed is (Round up to the nearest integer.) Determine the sample size n needed to construct a 90% confidence interval to estimate the population proportion for the following sample proportions when the margin...