2. Suppose that for a sample of n=11 measurements, we find that i = 72 and...
(6 pts) Suppose you have selected a random sample of n 7 measurements from a normal distribution. Compare the standard normal z values with the corresponding t values if you were forming the following confidence intervals. (a) 80% confidence interval (b) 90% confidence interval (c) 95% confidence interval
Suppose that we will randomly select a sample of 72 measurements from a population having a mean equal to 18 and a standard deviation equal to 9. (a) Describe the shape of the sampling distribution of the sample mean. Do we need to make any assumptions about the shape of the population? Why or why not? (b) Find the mean and the standard deviation of the sampling distribution of the sample mean. (Round your σx⎯⎯ answer to 1 decimal place.)...
When is unknown and the sample is of size n 230, there are two methods for computing confidence intervals for u. (Notice that, When is unknown and the sample is of size n<30, there is only one method for constructing a confidence interval for the mean by using the student's t distribution with d.f. = n - 1.) Method 1: Use the Student's t distribution with d.f. = n - 1. This is the method used in the text. It...
2. Assume that the observed value of the sample mean X and of the sample variance S2 of a random sample of size n from a normal population is 81.2 and 26.5, respectively Find %90,%95, %99 confidence intervals for the population mean μ
2. Assume that the observed value of the sample mean X and of the sample variance S2 of a random sample of size n from a normal population is 81.2 and 26.5, respectively Find %90,%95, %99 confidence...
When σ is unknown and the sample is of size n ≥ 30, there are two methods for computing confidence intervals for μ. Method 1: Use the Student's t distribution with d.f. = n − 1. This is the method used in the text. It is widely employed in statistical studies. Also, most statistical software packages use this method. Method 2: When n ≥ 30, use the sample standard deviation s as an estimate for σ, and then use the...
2. A simple random sample of size n is drawn. The sample mean I is found to be 53.1, and the sample standard deviation s is found to be 7.8 a) (3 points) Construct a 95% confidence interval for the population mean u if the sample size n is 81. b) (3 points) Construct a 95% confidence interval for the population mean u if the sample size n is 30. c) (3 points) Construct a 90% confidence interval for the...
Suppose you have selected a random sample of n=9 measurements from a normal distribution. Compare the standard normal zz= values with the corresponding t values if you were forming the following confidence intervals. (a) 98% confidence interval z= t= (b) 95% confidence interval z= t= (c) 99% confidence interval z= t=
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11.* A random sample of size n 64 is drawn from a population with mean μ and standard deviation σ. The mean and standard deviation of the sample are X = 308.9 and s 31.9 a. Find a 90%confidence interval for the mean μ. Interpret this interval. b. Find a 95%confidence interval for the mean μ. Interpret this interval. c. Find a 99%confidence interval for the mean μ. Interpret this interval. d. Compare the widths of...
a) Suppose we know the population variance σ2-400, there is a randorm sample with i-135, and the sample size is n the mean. I. 100. Please construction a 95% confidence interval for b) Suppose we do not know the population variance, and only know the sample variable s2-400, there is a random sample with -135, and the sample size is n-25. Please construct a 95% confidence interval for the mean.
In class we had 41 95% confidence intervals that we believe to be calculated correctly. The confidence intervals were collected by taking a sample of 60 data points from the population data. 41 of the confidence intervals appear to be calculated correctly. Of these 4 of them do not have the population mean inside the confidence interval. Based on a 95% confidence, we expected 2 to not contain the population mean. Did this happen by chance alone? To find your...