Assume that a sample is used to estimate a population mean . Use
the given confidence level and sample data to find the margin of
error. Assume that the sample is a simple random sample and the
population has a normal distribution. Round your answer to one more
decimal place than the sample standard deviation. (please show
work) 
Degrees of freedom = n-1
= 51 - 1
= 50
t critical value at 0.05 significance level for 50 df is 2.009
Margin of error = t * S / sqrt(n)
= 2.009 * 202 / sqrt(51)
= 56.8
Margin of error = 56.8 (Rounded one more decimal places than sample standard deviation)
Assume that a sample is used to estimate a population mean . Use the given confidence...
Assume that a sample is used to estimate a population mean μ. Use the given confidence level and sample data to find the margin of error. Assume that the sample is a simple random sample and the population has a normal distribution. Round your answer to one more decimal place than the sample standard deviation. 99% confidence; n = 201; x=276; s = 75 answer options 13.8 12.4 16.0 10.5
4) Assime that a sample is used to estimate a population mean. Us a confidence level of 95%, a sample size of 60, sample mean of 5.4, and sample standard deviation of 0.93 to find the margin of error. Assume that the sample is a simple random sample and the population has a normal distribution. Round your answer to one more decimal place that the sample standard deviation.
Assume that a sample is used to estimate a population mean µ. Use the given confidence level and sample data to find the margin of error. Assume that the sample is a random sample and the population has a normal distribution. 95% confidence; n= 51; sample mean= 240 ; s=242
Assume that a sample is used to estimate a population mean μμ. Find the margin of error M.E. that corresponds to a sample of size 15 with a mean of 37.7 and a standard deviation of 16.5 at a confidence level of 99.9%. Report ME accurate to one decimal place because the sample statistics are presented with this accuracy. Do not round the answer.
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A sample mean, sample size, population standard deviation, and confidence level are provided. Use this information to complete parts (a) through (c) x = 33, n = 25, C = 6, confidence level = 90% Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table a. Use the one-mean z-interval procedure to find a confidence interval for the mean of the population from which the sample...