When σ is unknown and the sample size is n\geq30, there are tow methods for computing confidence intervals for μμ. Method 1: Use the Student's t distr

nagasenaz 2021-05-14 Answered
When σ is unknown and the sample size is n30, there are tow methods for computing confidence intervals for μμ. 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 n30, use the sample standard deviation s as an estimate for σσ, 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 σσ. 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¯=45.2 and sample standard deviation s = 5.3. (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?
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

d2saint0
Answered 2021-05-15 Author has 89 answers
Result part (a):
90%: 43.5846 to 46.8154
95%: 43.2562 to 47.1438
99%: 42.5823 to 47.8177
Result part (b):
90%: 43.6341 to 46.7659
95%: 43.3343 to 47.0657
99%: 42.7441 to 47.6559
The confidence intervals using a Student's t distribution are more conservative, because the confidence intervals are a little wider than the confidence intervals using the standard normal distribution.
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-08-03
A simple random sample of 60 items resulted in a sample mean of 80. The population standard deviation is σ=15
a) Compute the 95% confidence interval for the population mean. Round your answers to one decimal place.
b) Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean. Round your answers to two decimal places.
c) What is the effect of a larger sample size on the interval estimate?
Larger sample provides a-Select your answer-largersmallerItem 5 margin of error.
asked 2021-08-04
A random sample of 100 automobile owners in the state of Virginia shows that an automobile is driven on average 23,500 kilometers per year with a standard deviation of 3900 kilometers.
Assume the distribution of measurements to be approximately normal.
a) Construct a 99% confidence interval for the average number of kilometers an automobile is driven annually in Virginia.
b) What can we assert with 99% confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year?
asked 2021-03-09

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 298 accurate orders and 51 that were not accurate. a. Construct a 90​% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part​ (a) to this 90​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: 0.127<p<0.191. What do you​ conclude? a. Construct a 90​% confidence interval. Express the percentages in decimal form. ___

asked 2021-02-23
Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let j: denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting 95% confidence intervals (7-8, 9.6)
(a) Would 2 90%% confidence interval calculated from this same sample have been narrower or wider than the glven interval? Explain your reasoning.
(b) Consider the following statement: There is 9 95% chance that Is between 7.8 and 9.6. Is this statement correct? Why or why not?
(c) Consider the following statement: We can be highly confident that 95% of al bottles ofthis type of cough syrup have an alcohol content that is between 7.8 and 9.6. Is this statement correct? Why or why not?
asked 2020-11-29
A random sample of size n=25 from a normal distribution with mean μ and variance 36 has sample mean X=16.3
a)
Calculate confidence intervals for μ at three levels of confidence: 80. How to the widths of the confidence intervals change?
b) How would the CI width change if n is increased to 100?
asked 2021-09-18

A local firm manufactures LED products that have a lifespan that is approximately normally distributed with a std. dev. of 30 hours. If a sample of 30 LED products has an average lifespan of 780 hours, find a 96% confidence interval for the population mean of all LED products produced by this firm.
Choose 2 answers in nearest unit (ones) or in whole number.
Example, if your answer is 888.83μ899.56, choose 889 and 900.
775773807797791769789768805763771792

asked 2021-03-05
A random sample of n=400 students is selected from a population of N=4000 students to estimate the average weight of the students. The sample mean and sample variance are found to be x=140lbands2=225. Find the 95%(z=2) confident interval.