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 standard normal distribution. This method is based on the fact that for large samples, s is a fairly good approximation for σ. Also, for large n, the critical values for the Student's t distribution approach those of the standard normal distribution. Consider a random sample of size n = 31, with sample mean x = 44.7 and sample standard deviation s = 5.5.
(a) Compute a 99% confidence intervals for μ using Method 1 with a Student's t distribution. Round endpoints to two digits after the decimal. lower limit 41.98 Correct: Your answer is correct. upper limit 47.42 Correct: Your answer is correct.
(b) Compute a 99% confidence interval for μ using Method 2 with the standard normal distribution. Use s as an estimate for σ. Round endpoints to two digits after the decimal.
lower limit
upper limit
(c) Compare intervals for the two methods. Would you say that confidence intervals using a Student's t distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution?
No. The respective intervals based on the t distribution are shorter.
Yes. The respective intervals based on the t distribution are shorter.
Yes. The respective intervals based on the t distribution are longer.
No. The respective intervals based on the t distribution are longer.
(d) Now consider a sample size of 71. Compute a 99% confidence interval for μ using Method 1 with a Student's t distribution. Round endpoints to two digits after the decimal.
lower limit
upper limit
(e) Compute a 99% confidence interval for μ using Method 2 with the standard normal distribution. Use s as an estimate for σ. Round endpoints to two digits after the decimal.
lower limit
upper limit
(f) Compare intervals for the two methods. Would you say that confidence intervals using a Student's t distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution?
No. The respective intervals based on the t distribution are shorter.
Yes. The respective intervals based on the t distribution are shorter.
No. The respective intervals based on the t distribution are longer.
Yes. The respective intervals based on the t distribution are longer.
With increased sample size, do the two methods give respective confidence intervals that are more similar?
As the sample size increases, the difference between the two methods is less pronounced.
As the sample size increases, the difference between the two methods becomes greater.
As the sample size increases, the difference between the two methods remains constant.
When σ is unknown and the sample is of size n ≥ 30, there are two...
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...
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...
Consider a 90% confidence interval for µ not known. For which sample size, n = 10 or n = 20, is the confidence interval longer? Critical Thinking Lorraine computed a confidence interval for µ based on a sample of size 41. Since she did not know α, she used s in her calculations. Lorraine used the normal distribution for the confidence interval instead of a Student's t distribution. Was her interval longer or shorter than one obtained by using an...
Diet Methods: Here we study the effectiveness of two dieting methods. Method 1 is a diet high in protein, fiber, and fat but low in carbohydrates. Method 2 is a diet low in fat but high in fiber and carbohydrates. After one year, a sample of participants from each diet method were checked for the amount of weight lost. The 99% confidence intervals for the mean weight loss of each plan are given below. Method 1: 8.77 < μ <...
If sample size is 15, below please fill the UPPER and LOWER critical values of the standard normal distribution and t distribution under the various confidence levels. (20%) Confidence level standard normal distribution t distribution 80% _______________ _______________ 90% _______________ _______________ 95% _______________ _______________ 98% _______________ _______________ 99% _______________ _______________
99 and standard deviation σ A population whose distribution is unknown has mean μ and a sample of size 26 is drawn from this population, then 1, a. The mean oJ a b. The standard error ot c. The distribution of A population whose distribution is unknown has mean μ = 99 and standard deviation σ = 7 and a sample of size 26 is drawn from this population, then 1, a. b. c. The mean of X= The standard...
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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...
Please show all work!!!
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...