I am looking at the specification for my exam that is coming up and these are three sections: A) Construct symmetric confidence intervals for the mean of a normal distribution with known variance. B) Construct symmetric confidence intervals from large samples, for the mean of a normal distribution with unknown variance. C) Construct symmetric confidence intervals from small samples, for the mean of a normal distribution with unknown variance using the t -distribution.

Madelynn Winters 2022-09-23 Answered
Confidence Intervals (A level)
Firstly let me apologise for asking this question on here - I see this as getting a sledgehammer to crack a nut, but I have nobody else that I can ask for advice on this topic.
I'm an A level Mathematics student so please forgive me for lack of/poor notation that you may normally come to expect/be familiar with.
I am looking at the specification for my exam that is coming up and these are three sections:
A) Construct symmetric confidence intervals for the mean of a normal distribution with known variance
B) Construct symmetric confidence intervals from large samples, for the mean of a normal distribution with unknown variance.
C) Construct symmetric confidence intervals from small samples, for the mean of a normal distribution with unknown variance using the t -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

Answers (1)

Marvin Hughes
Answered 2022-09-24 Author has 6 answers
Step 1
The difference between B and C is in the choice of critical value. A typical symmetric confidence interval for a location parameter has the form
point estimate ± critical value × standard error ,
where in turn
standard error = standard deviation sample size ,
and
critical value × standard error = margin of error .
The choice of critical value is informed by the nature of the sampling distribution. When a normally distributed population has unknown variance, the sampling distribution of the mean is Student's t-distributed; however, when the sample size is large, the difference in critical values is negligible; i.e.,
lim ν t ν , α = z α ,
where t ν , α is the upper α quantile of the Student's t distribution with ν degrees of freedom, and z α is the upper α quantile of the standard normal distribution. So for scenario B, you will use z α / 2 for a two-sided CI with large sample size as an approximation; for scenario C, you will use t ν , α / 2 for the critical value.
Step 2
The difference between A and B lies in the standard deviation. In scenario A, it is presumed to be known, thus the standard error will use the population standard deviation σ in the numerator. In scenario B, it is unknown, thus you will use the unbiased estimator of the standard deviation
s = 1 n 1 i = 1 n ( x i x ¯ ) 2 .
In practice, the factor of n 1 rather than n makes little difference when n is large as in this case. But in scenario C, as σ is also unknown and must be estimated from the sample, you must use n 1, otherwise your CI will not have the required coverage probability even if you correctly use the t-distribution quantile for the critical value.
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 2021-08-04

The average annual mileage of a car is 23,500 kilometers, with a standard deviation of 3900 kilometers, according to a sample of 100 Virginians who own cars.
Assume that the measurement distribution is roughly normal.
a) Create a 99 percent confidence interval for the typical Virginian vehicle's annual mileage in kilometers.
b) If we assume that Virginians drive an average of 23,500 kilometers per year, what can we say with 99 percent certainty about the potential extent of our error?

asked 2021-01-10

The average zinc concentration recovered from a sample of measurements taken in 36 different locations in a river is found to be 2.6 grams per liter. Find the 95% confidence intervals for the mean zinc concentration in the river. Assume that the population standard deviation is 0.3 gram per liter. (Z0.025=1.96,Z0.005=2.575)

asked 2020-12-27

Consider the next 1000 98% Cis for mu that a statistical consultant will obtain for various clients. Suppose the data sets on which the intervals are based are selected independently of one another. How many of these 1000 intervals do you expect to capture the corresponding value of μ? 
What is the probability that between 970 and 990 of these intervals conta the corresponding value of Y =the number among the 1000 intervals that contain What kind of random variable is Y) (Use the normal approximation to the binomial distribution)

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 2022-09-24
How we interpret confidence intervals.
For the formula X ¯ ± z α 2 σ n , we substitute x ¯ o b s for X ¯ , but isn't that observed sample mean x ¯ o b s only for one particular sample?
Also why is it incorrect to interpret a confidence interval as the probability that the actual μ lies in a interval (a,b) is ( 1 α ) 100 %
asked 2022-04-21
We are given a 95% confidence interval with some range - in this particular case (4.0; 11.0) - to find the 90% and 99% confidence intervals for the same parameter, assuming normal distribution, without any aditional information.
asked 2022-03-15
A t - distribution's convergence to a Normal distribution
I'm currently messing around with confidence intervals and I can't really understand how a t distribution converges to a normal distribution for large n.
For example, suppose we want to construct a 95% confidence interval when we have a sample mean X=74.8 and sample variance S=1.23 with n=143.
I would construct the confidence interval using,
(Xz1α2Sn,X+z1α2Sn)
since n>30. If n<30 I would have used a t-distribution.
My question is why does the t - distribution approach a normal distribution for relatively large n?

New questions