Conclusion for confidence interval. If I got, let's say, a 95 % confidence interval for the mean and a 95 % confidence interval for the variance.

samuelaplc 2022-10-02 Answered
Conclusion for confidence interval
If I got, let's say, a 95 % confidence interval for the mean and a 95 % confidence interval for the variance.
Would it then be wrong to conclude:
The 95 % confidence interval for the mean contains with at least 95 % probability the true mean?
and
The 95 % confidence interval for the variance contains with at least 95 % probability the true variance?
What would be a more correct/precise way to express what the confidence intervals stand for?
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

Answers (1)

Quinn Alvarez
Answered 2022-10-03 Author has 13 answers
Step 1
The 'meaning' of interval estimates is a controversial topic on applied statistics. So there is no universally accepted answer to your important question.
Let's just use a proposed sample of size n = 31 from a normal population with unknown population mean μ and unknown variance σ 2 ..
Then T = X ¯ μ S / n T ( n 1 ) ,, so that P ( 2.042 T = X ¯ μ S / n 2.042 ) = 0.95.. Here X ¯ and S are the sample mean and variance, respectively.
Manipulating inequalities in the event, we get P ( X ¯ 2.042 S n μ X ¯ + 2.042 S n ) = 0.95.. This is purely a probability statement. Specifically, it is a probability statement about the behavior of the random variable X ¯ and S: the random interval ( X ¯ 2.042 S n , X ¯ + 2.042 S n ) has a 95% probability of covering (including) the unknown constant μ ..
Step 2
Now suppose we take the sample and obtain X ¯ = 21.3 and S 2 = 1.44.. Then the random interval becomes ( 21.3 2.042 ( 0.1249 ) , ( 21.3 + 2.042 ( 0.1249 ) ) , ( 21.3 2.042 ( 0.1249 ) , ( 21.3 + 2.042 ( 0.1249 ) ) or (20.860,21.740).
But now we are dealing with observed quantities. According to the usual frequentist interpretation of probability, this is no longer a probability statement: Either the interval (20.860,21.740) includes μ or it does not. Accordingly, the interval (20.860,21.740) is called a 95% confidence interval.
The confidence interval is a statement about the data. Over the long run, we will obtain data so that the manipulation in the emphasized paragraph will produce an interval that includes the true population μ in 95% of such experiments.
The reason for calling the interval estimate a 'confidence' interval instead of a 'probability' interval has to do with a strict interpretation by frequentist statisticians of the word 'probability'.
Bayesian statisticians treat μ as a random variable, begin with a 'prior' distribution on μ, combine the data with the prior distribution to get a 'posterior' distribution, and use the posterior distribution to get a probability interval for μ (some say a credible interval). If the prior distribution is "flat" (containing little information), then the Bayesian and frequentist interval estimates will be numerically very similar. But philosophies as to the "meaning" of the interval estimate differ.
Both frequentists and Bayesians have their critics. Strictly speaking, frequentists are are not saying anything about the experiment at hand--only about what 'works' over the long run. A Bayesian is addressing the experiment at hand, but needs to explain how the prior distribution was obtained and what effect it has on the interval estimate.
Did you like this example?
Subscribe for all access

Expert Community at Your Service

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

You might be interested in

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 2021-08-09

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment, 13 subjects had a mean wake time of 101.0 min. After​ treatment, the 13 subjects had a mean wake time of 94.6 min and a standard deviation of 24.9 min. Assume that the 13 sample values appear to be from a normally distributed population and construct a 95% confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 101.0 min before the​ treatment? Does the drug appear to be​ effective?
Construct the 95% confidence interval estimate of the mean wake time for a population with the treatment.
min<μ<min ​(Round to one decimal place as​ needed.)
What does the result suggest about the mean wake time of 101.0 min before the​ treatment? Does the drug appear to be​ effective?
The confidence interval ▼ does not include| includes the mean wake time of 101.0 min before the​ treatment, so the means before and after the treatment ▼ could be the same |are different. This result suggests that the drug treatment ▼ does not have | has a significant effect.

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 2021-03-09

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 2021-08-07
Find tα2, n=15 for the 98% confidence interval for the mean.
asked 2022-04-24
Point and interval estimation of normal variance θ when the mean μ is known.
Let X1,,Xn are independent observations over the random variable XiN(μ,θ) where μ is known and θ>0. We are given the following estimator for tehta:
θ^=i=1n{(Xiμ)2}θ
asked 2022-03-25
Confidence in sample mean, given sample variance?
Let's say that I have an large population of data, but that I have a sample mean and sample variance calculated from a subset of that data.
Can I use my sample variance (or standard deviation) to know how confident I should be in my sample mean being close to the population mean?
It seems like I should because a low variance seems to indicate that there isn't very far that the mean could move, but on the other hand, taking more samples isn't going to make the variance approach zero.
Is there some other calculation I should be using for getting a confidence amount in my sample mean?

New questions