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Recent questions in Confidence intervals

michiiiiiakqm 2022-03-25

Confidence Itervals; $Z_{α} and Z_{\frac{α}{2}}$
I'm confused about what exactly $Z_{α}$ is, does there exist a formula for it in terms of $α$ ? IF so, is there also one for $Z_{\frac{α}{2}}$ ?

zatajuxoqj 2022-03-25

Confidence Intervals, Proportion Estimations
In a study of perception, 107 men are tested and 24 are found to have red/green color blindness.
(a) Find a 92% confidence interval for the true proportion of men from the sampled population that have this type of color blindness.
(b) Using the results from the above mentioned survey, how many men should be sampled to estimate the true proportion of men with this type of color blindness to within 2% with 98% confidence?
(c) If no previous estimate of the sample proportion is available, how large of a sample should be used in (b)?
I have already answered (a). However, I am at a complete loss as to how to answer (b) or (c). For one thing, I am not quite sure what (b) is even asking. Advice on this question would be greatly appreciated.

Aryan Salinas 2022-03-25

Confidence interval; exponential distribution (normal or student approximation?)
Let's say we have got a sample of size n from an exponential distribution with an unknown mean $λ$ .
We want to construct a confidence interval and so we can compare this:
$\frac{\overset{―}{X} - λ}{\sqrt{\frac{S^{2}}{n}}}$
to a student t-distribution with $n - 1$ degrees of freedom.
However, as in the case of the exponential distribution, we know that $V a r [X] = {(E [X])}^{2}$ , so rather than introducing an estimator for variance, we can simply use one estimator, i.e:
$\frac{\overset{―}{X} - λ}{\sqrt{\frac{{\overset{―}{X}}^{2}}{n}}} .$
And now the question is:
Do we compare this statistic with normal distribution or again with student t-distribution?

svrstanojpkqx 2022-03-25

Confidence interval of the parameter of exp and normal distribution from MLE?
I have a sample $X_{1}, X_{2}, \dots, X_{n}$
1. If the sample is from exponential distribution, I want to use MLE to estimate the parameter $β$ . I know the result that
$\hat{β} = \frac{X_{1} + X_{2} + \dots + X_{n}}{n}$
But how can I calculate the 95% confidence interval of $β$ ?
2. If the sample is from normal distribution, I want to use MLE to estimate the parameter $μ and σ$ . I also know the result that $\hat{μ} = \frac{X_{1} + X_{2} + \dots + X_{n}}{n}, \hat{σ} = {[\frac{n - 1}{n} \cdot S^{2} (n)]}^{\frac{1}{2}}$
But how can I calculate the 95% confidence interval of $μ and σ$ ?
Step1 to calculate confidence interval,
$\hat{θ} \pm z_{1 - \frac{α}{2}} \sqrt{\frac{δ (\hat{θ})}{n}}$
Step2 to calculate confidence interval
$δ (\hat{θ}) = - n {(E [\frac{d^{2}}{d θ^{2}} \ln L (θ)])}^{- 1}$
I know the equation, but I am still confuzed how to calculate
$(E [\frac{d^{2}}{d θ^{2}} \ln L (θ)])$

Averie Ferguson 2022-03-25

Confidence interval for population proportion
You have a population of 10M people. Suppose you want to determine the share of people with a certain disease. To do so you randomly choose X people and check if they have the disease. You determine that a share P of those people have the disease.
How do I compute the 75% / 90% / 95% confidence intervals for the disease prevalence for the whole population?
e.g. find y so that there is a 95% chance the disease rate in the whole population is in the range $[X - y, X + y]$ .

abitinomaq1 2022-03-25

Confidence Interval for Pareto Distribution
A random variable is said to have probability density function
$f_{X} (x) = \frac{α k^{α}}{x^{α + 1}}, α, k > 0; and; x > k .$

Tony Mccarthy 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?

Marzadri9lyy 2022-03-25

Confidence interval for parameter
$X_{1}, X_{2}, \dots, X_{n}$ - i.i.d observations
$X_{1} = ξ + η$ where $ξ ~ N (θ^{2}, θ^{2} + 1), η = {\begin{array}{ll} 0, & 1 / 2 \\ 4 θ, & 1 / 2 \end{array}, ξ and η$ and $η$ are independent.
Find $α$ -confidence interval.
The first that I need to do is to find some estimate for $θ$ . The only one that I find is $\frac{5 \overset{―}{X} - S^{2} + 1}{10}$ but it is difficult to find distribution.
Is it possible to do something else?

siliciooy0j 2022-03-24

Given you have an independent random sample $X_{1}, X_{2}, \dots, X_{n}$ of a Bernoulli random variable with parameter $p$ , estimate the variance of the maximum likelihood estimator of $p$ using the Cramer-Rao lower bound for the variance
So, with large enough sample size, I know the population mean of the estimator $\hat{P}$ will be $p$ , and the variance will be:
$V a r [\hat{P}] = \frac{1}{n E [((\partial / \partial p) \ln f_{x} (X))^{2}]}$
Now I'm having some trouble calculating the variance of $\hat{P}$ , this is what I have so far:
since the probability function of $\overset{―}{X}$ is binomial, we have:
$f_{x} (\bar{X}) = (\binom{n}{\sum_{i = 1}^{n} X_{i}}) * p^{\sum_{i = 1}^{n} X_{i}} * (1 - p)^{n - \sum_{i = 1}^{n} X_{i}}$
so: $\ln f_{X} (X) = \ln ((\binom{n}{\sum_{i = 1}^{n} X_{i}})) + \sum_{i = 1}^{n} X_{i} \ln (p) + (n - \sum_{i = 1}^{n} X_{i}) \ln (1 - p)$
and: $\frac{\partial \ln f_{X} (X)}{\partial p} = \frac{\sum_{i = 1}^{n} X_{i}}{p} - \frac{(n - \sum_{i = 1}^{n} X_{i})}{(1 - p)} = \frac{n \bar{X}}{p} - \frac{(n - n \bar{X})}{(1 - p)}$

and: $(\frac{\partial l n f_{X} (X)}{\partial p})^{2} = (\frac{n \bar{X}}{p} - \frac{(n - n \bar{X})}{(1 - p)})^{2} = \frac{n^{2} p^{2} - 2 n^{2} p \bar{X} + n^{2} {\bar{X}}^{2}}{p^{2} (1 - p)^{2}}$
since $E [{\overset{―}{X}}^{2}] = μ^{2} + \frac{σ^{2}}{n}$ , and for a Bernoulli random variable $E [X] = μ = p = E [\overset{―}{X}]$ and $V a r [X] = σ^{2} = p (1 - p)$ :
$E [(\frac{\partial \ln f_{X} (X)}{\partial p})^{2}] = \frac{n^{2} p^{2} - 2 n^{2} p E [\bar{X}] + n^{2} E [{\bar{X}}^{2}]}{p^{2} (1 - p)^{2}} = \frac{n^{2} p^{2} - 2 n^{2} p^{2} + n^{2} (p^{2} + \frac{p (1 - p)}{n})}{p^{2} (1 - p)^{2}} = \frac{n p (1 - p)}{p^{2} (1 - p)^{2}} = \frac{n}{p (1 - p)}$
Therefore, $V a r [\hat{P}] = \frac{1}{n E [((\partial / \partial p) \ln f_{x} (X))^{2}]} = \frac{1}{n \frac{n}{p (1 - p)}} = \frac{p (1 - p)}{n^{2}}$
However, I believe the true value I should have come up with is $\frac{p (1 - p)}{n}$ .

dotzis16xd 2022-03-24

Let's say I have two samples of results of two bernoulli experiments.
$H_{0} : p_{1} = p_{2}$
$H_{1} : p_{1} \neq p_{2}$
And I want to try to reject $H_{0}$ at a confidence level.
I already know a proper way to solve this, but I was wondering, if I have a confidence interval for $p_{1}$ and $p_{2}$ , at the same level of significance. Can I just check if the intervals overlaps each other to test this?

Leroy Davidson 2022-03-24

Let $x_{1}, \dots, x_{n}$ be a sample from a normal population having unknown mean and variance. Let $\overset{―}{x}$ be the average of the first n of them.
What is the distribution of $x_{n + 1} - \overset{―}{x}$ ?
If $\overset{―}{x} = 4$ , give an interval that, with $90 %$ confidence, will contain the value of $x_{n + 1}$ .
The distribution of $\overset{―}{x}$ would be normal with mean equal to the population mean and standard deviation equal to the population standard deviation divided by $\sqrt{n}$ . But how do we find the distribution of $x_{n + 1}$ ?

Hugh Soto 2022-03-24

Let $f : Ω \to R$ be a Borel-measurable function, X a random variable with values in $Ω$ and $X_{i} \in R, i \in N$ realizations of X.
Literature
In their book on Monte Carlo Methods (Simulation and the Monte Carlo method, Second Edition) Rubinstein and Kroese give in Section 4.2.1 an approximate confidence interval for the mean E[f(X)]:
$(\hat{μ} \pm z_{1 - \frac{α}{2}} \frac{\hat{σ_{f}}}{\sqrt{N}}),$
where we write N for the number of Monte Carlo samples, $z_{1 - \frac{α}{2}}$ for the $1 - \frac{α}{2}$ quantile of the standard normal distribution, $\hat{μ} = \frac{1}{N} \sum_{i = 1}^{N} f (X_{i})$ for the unbiased estimator of the mean and ${\hat{σ}}_{f} = \sqrt{\frac{1}{N - 1} \sum_{i = 1}^{N} {(f (X_{i}) - \hat{μ})}^{2}}$ for the empirical standard deviation derived from the unbiased estimator of the variance.
Extension to variance
In my opinion we can derive in a similar way a confidence interval for the variance var[f(X)] if we define $g (X) = {(f (X) - \hat{μ})}^{2}$ and repeat the construction. This leads to the interval
$({\hat{σ}}_{f}^{2} \pm z_{1 - \frac{α}{2}} \frac{{\hat{σ}}_{g}}{\sqrt{N}}),$
where ${\hat{σ}}_{g} = \frac{1}{N - 1} \sum_{i = 1}^{N} {(g (X) - \frac{1}{N} \sum_{i = 1}^{N} g (X_{i}))}^{2}$
How can I transfer this to give a confidence interval for the standard deviation? Surely I can't just take the square root of the boundaries. Any literature, help and thoughts are welcome. Thanks.

Harley Ayers 2022-03-24

Given an sample of data set X which is normal-distributed to $N (μ = 240, σ)$ , I want to find the $95 %$ confidence interval of $max (μ - 200, 0)$ . As the $max (μ - 200, 0)$ cut the normal-distribution curve at the point of 200, the sample of $max (μ - 200, 0)$ is not more normal distributed. Therefore what is a reasonable $95 %$ confidence interval?

afasiask7xg 2022-03-23

Let X be a Poisson ( $θ$ ) random variable. Unbiased estimator
Show that ${(- 1)}^{X}$ is an unbiased estimator for $e^{- 2 θ}$ This is a fairly bad estimator for a number of reasons - so this exercise helps show why unbiasedness is not the most important criterion for an estimator.

Deegan Chase 2022-03-23

Let $X_{1}, \dots, X_{n}$ be a random sample from a normal distribution with known mean $μ$ and unknown variance $σ^{2}$ . Three possible confidence intervals for $σ^{2}$ are
a) $(\sum_{i = 1}^{n} \frac{{(X_{i} - \overset{―}{X})}^{2}}{a_{1}}, \sum_{i = 1}^{n} \frac{{(X_{i} - \overset{―}{X})}^{2}}{a_{2}})$
b) $(\sum_{i = 1}^{n} \frac{{(X_{i} - μ)}^{2}}{b_{1}}, \sum_{i = 1}^{n} \frac{{(X_{i} - μ)}^{2}}{b_{2}})$
c) $(\frac{n {(\overset{―}{X} - μ)}^{2}}{c_{1}}, \frac{n {(\overset{―}{X} - μ)}^{2}}{c_{2}})$
where $a_{1}, a_{2}, b_{1}, b_{2}, c_{1}, c_{2}$ are constants.
Find values of these six constants which give confidence level 0.90 for each of the three intervals when $n = 10$ and compare the expected widths of the tree intervels in this case
With $σ^{2} = 1$ , what value of n is required to achieve a $90 %$ confidence interval of expected width less than 1 in cases (b) and (c) above?

navantegipowh 2022-03-23

Given $X_{1}, \dots, X_{100}$ and $Y_{1}, \dots, Y_{50}$ which are independent random samples from identical distribution $N (μ, 1)$ . Each of two statisticians is trying to build confidence interval for $μ$ on a confidence level of 0.8 $1 - α = 0.8$ . You have to find probability that those intervals are disjoint. I know i have to find $P (| \overset{―}{X} - \overset{―}{Y} | > \frac{1 \cdot 1.28}{\sqrt{50}} + \frac{1 \cdot 1.28}{\sqrt{100}})$ which is $1 - P (- 0.309 \leq \overset{―}{X} - \overset{―}{Y} \leq 0.309)$
Question is what should i do next how to calculate this probability?

Decreasing the sample size, while holding the confidence level the same, will do what to the length of the confidence interval?

Addison Fuller 2022-03-22

Let $X_{1}, \dots, X_{n}$ random sample of $U [θ - \frac{1}{2}; θ + \frac{1}{2}]$ .Consider $[X_{(1)}; X_{(n)}]$ a confidence interval for $θ$ . Find their confidence level and show that result is valid for any distribution symmetric around $θ$

svezic0rn 2022-03-21

Let $X_{1}, \dots, X_{n}$ be a random sample from $f (x ∣ θ) = θ x^{θ - 1}$ for $0 < x < 1$ . Find a confidence interval for $θ$ using as pivotal quantity a function of the maximum likelihood estimator for $θ$ .

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