Let's discuss a little bit about the difference between Z- and t-confidence intervals. First of all, the t-confidence interval is always wider than z confidence interval. This is because we also have to evaluate variance, so compensating for uncertainty increases the tail probabilities for the pivotal quantity used in construction of the the confidence interval and therefore, respectively, the length of the confidence interval. Let's assume below that we would use the same standard error s.e. = 1 for both z- and t-confidence intervals so we can compare this effect slightly. a) How many percent wider is the 95 % symmetric t confidence interval is like a 95 % symmetric z confidence interval if n = 10 and the standard error is 1 each? b) How many percent wider is the 95 % confidence interval of t is like a 95-percent z-confidence interval for n = 30 and the standard error is 1 each? c) What must the sample size n be at least, so the 95 % t confidence intervals would be at most 1% wider than 95% z confidence interval? You can approximate the upper quantile of the t distribution using the upper quantile za of the standard normal distribution (from a asymptotic development) tv (α) ≈ za + α z³ + za Αν You can also search for this numerically either by trying all of the numbers n = 50,..., 150, or by using binary search. This is the answer is slightly different that the asymptotic approximation gives, but it does not matter.
Let's discuss a little bit about the difference between Z- and t-confidence intervals. First of all, the t-confidence interval is always wider than z confidence interval. This is because we also have to evaluate variance, so compensating for uncertainty increases the tail probabilities for the pivotal quantity used in construction of the the confidence interval and therefore, respectively, the length of the confidence interval. Let's assume below that we would use the same standard error s.e. = 1 for both z- and t-confidence intervals so we can compare this effect slightly. a) How many percent wider is the 95 % symmetric t confidence interval is like a 95 % symmetric z confidence interval if n = 10 and the standard error is 1 each? b) How many percent wider is the 95 % confidence interval of t is like a 95-percent z-confidence interval for n = 30 and the standard error is 1 each? c) What must the sample size n be at least, so the 95 % t confidence intervals would be at most 1% wider than 95% z confidence interval? You can approximate the upper quantile of the t distribution using the upper quantile za of the standard normal distribution (from a asymptotic development) tv (α) ≈ za + α z³ + za Αν You can also search for this numerically either by trying all of the numbers n = 50,..., 150, or by using binary search. This is the answer is slightly different that the asymptotic approximation gives, but it does not matter.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:Let's discuss a little bit about the difference between Z- and t-confidence intervals.
First of all, the t-confidence interval is always wider than z confidence interval. This is
because we also have to evaluate variance, so compensating for uncertainty increases
the tail probabilities for the pivotal quantity used in construction of the the confidence
interval and therefore, respectively, the length of the confidence interval.
Let's assume below that we would use the same standard error s.e. = 1 for both z-
and t-confidence intervals so we can compare this effect slightly.
a) How many percent wider is the 95 % symmetric t confidence interval is like a 95
% symmetric z confidence interval if n = 10 and the standard error is 1 each?
b) How many percent wider is the 95 % confidence interval of t is like a 95-percent
z-confidence interval for n = 30 and the standard error is 1 each?
c) What must the sample size n be at least, so the 95 % t confidence intervals would
be at most 1% wider than 95% z confidence interval? You can approximate the
upper quantile of the t distribution using the upper quantile za of the standard
normal distribution (from a asymptotic development)
tv (α) ≈ za +
α
z³ + za
Αν
You can also search for this numerically either by trying all of the numbers
n = 50,..., 150, or by using binary search. This is the answer is slightly
different that the asymptotic approximation gives, but it does not matter.
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