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Recent questions in Confidence intervals

Taliyah Spencer 2022-04-19

Solution verification for hypothesis testing and confidence interval problem
There's a random sample of size $n = 10$ observations over a normally distributed population with standard deviation $σ = 2$ :
8.7, 7, 4, 7.6, 3, 8.1, 6.4, 6.1, 9.4, 6.2
- Find 96% confidence interval for the population mean
- Test the hypothesis $H_{0} : μ = 7.5$ against Ha: $μ \neq 7.5$ with significance level $α = 0.01$ . Find approximation of the observed p value.
My Solution:
We are looking for real numbers l and u s.t.:
$P (l < \hat{θ} < u) = 0.96$ , where $\hat{θ} = \frac{\sum_{i = 1}^{n} X_{i} - n θ}{\sqrt{n σ^{2}}} \sim N (0, 1)$
We are looking for $z_{0.2}$ which from the z-score table is: 2.055. Therefore:
$- 2.055 < \frac{\sum_{i = 1}^{n} X_{i} - n θ}{\sqrt{n σ^{2}}} < 2.055 \equiv - 2.055 \sqrt{40} < \sum_{i = 1}^{n} X_{i} - n θ < 2.055 \sqrt{40} \equiv$
$\equiv 12.996 - 66.5 < - 10 θ < 12.996 - 66.5 \equiv 5.350 < θ < 7.94 .$
For the second part:
We set two z-scores: $z_{1} = - 2.325 and z_{2} = 2.325$ from which we construct a two-tailed test.
Our test statistic is: $t = \frac{(6.65 - 7.5) \sqrt{10}}{2} = - 1.343 \Rightarrow t \geq - 2.235$ from where we see the test statistic t is not in the rejection region, therefore we fail to reject the null hypothesis.

Terrence Riddle 2022-04-16

What am I doing wrong in finding the confidence interval?
In a certain bio-engineering experiment, a successful outcome was achieved 60 times out of 125 attempts.
Construct a 95% confidence interval for the probability, p, of success in a single trial.
My answer:
We know that the confidence interval is:
$(p - k s, p + k s)$
Where s is the standard error.
I found the k value using Matlab code:
$k = ‖ \in ‖ v (0.975) = 1.9600$
Also: $p = \frac{60}{125} = 0.48$
Standard Error:
$s = \sqrt{\frac{p (1 - p)}{n}} = \sqrt{\frac{0.48 \cdot 0.52}{125}} = 0.04469$
We get the Confidence Interval:
$(0.392, 0.568)$
But my answer is wrong. Is there anything I'm missing out here?

Efrain Watkins 2022-04-16

Weaknesses of Wald confidence interval for binomial distribution
A survey samples 1000 people, among which 500 say they will vote for A, 400 for B e 100 for C. Calculate a confidence interval for the proportion of people that will vote for A. What are the major weaknesses of the confidence interval you just calculated?
Let $A_{n} \sim Binom (n, p)$ . By the CLT,
$\sqrt{n} ({\overset{―}{A}}_{n} - p) x \to {d} N (0, p (1 - p))$
hence $\frac{\sqrt{n}}{\sqrt{p (1 - p)}} ({\overset{―}{A}}_{n} - p) x \to {d} N (0, 1)$
Since p is unknown, we approximate it with ${\overset{―}{A}}_{n}$ . It follows that, at $1 - α$ confidence,
$p \in {\overset{―}{A}}_{n} \pm z_{1 - \frac{α}{2}} \sqrt{\frac{{\overset{―}{A}}_{n} (1 - {\overset{―}{A}}_{n})}{n}}$
Since this confidence interval relies on the CLT, it gives poor results when n is small or p is very close to 0 or 1. But this is not the case here, as $n = 1000$ and ${\overset{―}{A}}_{n} = 0.5$ , so I fail to see what weaknesses the question is talking about?

Ariel Collier 2022-04-14

Find CI for mean of linear regression with variance unknown
For the simple linear regression model:
$Y_{i} = β \cdot X_{i} + ϵ_{i}$
I want to find CI for $β x$ which is $E (Y_{i})$ when $x_{i} = x$ .
I find that $\hat{β} \sim N (β, \frac{σ^{2}}{\sum (X_{i}^{2})})$ , so the distribution of $β$ x is N( $β x, \frac{X^{2} σ^{2}}{\sum (X_{i}^{2})})$ . If the variance is known I can use $P (- z_{\frac{α}{2}} \leq \frac{\hat{β} x - β x}{x \frac{σ}{\sum} (x_{i}^{2})} \leq z_{\frac{α}{2}})$ .
If the variance is unknown, what unbiased estimator should I use for $σ^{2}$ ? Is it the sample variance? What is it in this case?

Eduardo Mclaughlin 2022-04-13

Maximum error in confidence interval
A sample of 40 cows is drawn to estimate the mean weight of a large herd of cattle. If the standard deviation of the sample is 96 kg, what is the maximum error in a 90% confidence interval estimate?

Aldo Fernandez 2022-04-11

Use $2 \sum_{i = 1}^{n} \frac{Y_{i}}{β}$ which is a pivotal quantity to derive a $95 %$ confidence interval for $β$

Boston Fitzgerald 2022-04-11

Two-sided confidence intervals and tests
From a sample of 1751 army hospitals, estimate the mean expenses for a full time equivalent employee for all US army hospitals using a $90 %$ confidence interval given $x = 6563$ and $s = 2484$ .

zdebe5l8 2022-04-11

To calculate the credible region for the odds ratio $\frac{θ}{1 - θ}$ . This is part of a bigger problem where I have already calculated $90 %$ credible region for random variable $θ$ :
$P (θ \leq 0.30) = 0.05$
$P (θ \leq 0.52) = 0.95$
$P (0.30 \leq θ \leq 0.52) = 0.9 .$

Jaylyn Villarreal 2022-04-09

How to keep the information of the confidence interval in the predictive step?
Suppose I have a sample $X_{1}, X_{2}, \dots X_{n}$ . Each $X_{i}$ comes from a poisson distribution $X_{i} \sim P o (λ)$ . The unknown paremeter $λ$ is inferred using the next confidence interval: $λ^{\pm} = \overset{―}{X} \pm t^{\cdot} \frac{\overset{―}{X}}{\sqrt{n}}$ where $λ^{+}$ is the upper bound and $λ^{-}$ is the lower bound, t∗ is the confidence expressed over t-student distribution.
I have two questions. The second question is actually the question I ask, the first is only to check my reasoning is right.
First: Taking into account the data come from Poisson distribution and taking into account the Central Limit Theorem, is it right the inference a did?It means: is it right to use the t-student approach? Or is it another better way to do?
Second: (That's the main question):
As I don't know the $λ$ parameter, I inferred it using confident intervals. Now I want to carry out some predictions of future events and calculate the associated probabilities, it means I wanto to know:
$p r o b {X_{n + 1} = K}$ . To do this: What number in the interval $[λ^{-}, λ^{+}]$ I should take? Obviously, maximize the likelihood is the best option, so I would take $\overset{―}{X}$ . So the question is:
Is there a way to take into account the uncertainty in the probabilities of the prediction? or: Is there a way to keep the knowledge of the confident interval in the probabilities of the new prediction? or: At the moment in that you use an only number(point estimation), how to keep the uncertainty information?
To be clearer take into account these two examples. The same process but with two different interval confidence:
1) $[λ^{-}, λ^{+}] = [6, 8]$
$\overset{―}{X} = 7 \to X_{n + 1} \sim P o (λ = \overset{―}{X}) \to P r o b {X_{n + 1} = 6} = 0.449$
2) $[λ^{-}, λ^{+}] = [1, 13]$
$\overset{―}{X} = 7 \to X_{n + 1} \sim P o (λ = \overset{―}{X}) \to P r o b {X_{n + 1} = 6} = 0.449$
The first contain less uncertainty than the second, so how could I maintain the information of confident interval?
I hope I was clear in the question.

Shaylee House 2022-04-09

The times that a cashier spends processing each person’s transaction are independent and identically distributed random variables with a mean of µ and a variance of $σ^{2}$ . Thus, if Xi is the processing time for each transaction, $E (X_{i}) = μ$ and $Var (X_{i}) = σ^{2}$ . Let Y be the total processing time for 100 orders: $Y = X_{1} + X_{2} + \dots + X_{100}$
a) What is the approximate probability distribution of Y, the total processing time of 100 orders?
b) Suppose for $Z \sim N (0, 1)$ , a standard normal random variable: $P (a < Z < b) = 100 (1 - α) %$ . Using your distribution from part (a), show that an approximate $100 (1 - α) %$ confidence interval for the unknown population mean $μ$ is: $(\frac{Y - 10 b σ}{100}) < μ < (\frac{Y - 10 a σ}{100})$
I have done part (a) and ended up with $Y \sim N (100 μ, σ^{2})$ using the central limit theorem, but I am unsure of where to start in part (b).

alahovom7ahc 2022-04-08

Finding upper confidence limit without mean and sd
So the question I'm trying to answer looks like this:
We are interested in p, the population proportion of all people who are currently happy with their cell phone plans. In a small study done in 2012, it was found that in a sample of 150 people, there were 90 who were happy with their cell phone plans.
Find the upper confidence limit of an 82% confidence interval for p.
I know the formula is $estimate \pm (criticalvalue) (standard error)$

painter555ui8n 2022-04-08

How to get the population standard deviation from a sample standard deviatoin
I am trying to find the confidence interval, and I need to know the population standard deviation. If I am given the sample standard deviation, how can I get the population one? Thanks!

Kendal Day 2022-04-08

Trouble Understanding the Formal Definition of a Confidence Interval
A $1 - α$ confidence interval for a paramater $θ$ is an interval $C_{n} = (a, b)$ where $a = a (X_{1}, \dots, X_{n})$ and $b = b (X_{1}, \dots, X_{n})$ are functions of the data such that
$P_{θ} (θ \in C_{n}) > 1 - α, for all θ \in Θ$
If $θ$ is a vector then we use a confidence set (such as a sphere or an ellipse) instead of an interval.
Question:
While I understand conceptually what a confidence interval is (i.e., a 95% CI means that 95% of experiments will trap the paramater in the interval), I don't understand how this formality is capturing this concept.
In particular, I don't understand what is meant by the notion of $P_{θ} (θ \in C_{n})$ . What is the sample space which P is drawing from? What is the set $θ \in C_{n}$ ? It seems here $θ$ is being treated both as a fixed value (from the notation Pθ) and as a random variable (by the notation $θ \in C_{n}$ ).

Alessandra Carrillo 2022-04-07

For fixed k, the variables $(2 k + 1) {\hat{f}}_{k} (λ)$ are asymptotically distributed according the Gamma distribution with shape parameter $2 k + 1$ and mean $(2 k + 1) f_{X} (λ)$ . It turns out that this suggests a confidence interval of the form
$(\frac{(4 k + 2) {\hat{f}}_{k} (λ)}{χ_{{4 k + 2, 1 - α}}^{2}}, \frac{(4 k + 2) {\hat{f}}_{k} (λ)}{χ_{{4 k + 2, α}}^{2}})$
where $χ_{{k, α}}^{2}$ the $α$ -quantile of the chi-square distribution with k degrees of freedom.
To show that this interval is asymptotically of level $1 - 2 α$

Jazmyn Holden 2022-04-04

Find a 95 confidence interval for $θ$ based on inverting the test statistic statistic $\hat{θ}$ .
For our data we have $Y_{i} \sim N (θ x_{i}, 1) for i = 1, \dots, n .$
Therefore it can be proven that the MLE for $θ$ is given by
$\hat{θ} = \frac{\sum x_{i} Y_{i}}{\sum x_{i}^{2}}$
To find the confidence interval I should invert the test statistic $\hat{θ}$ .
The most powerful unbiased size $α = 0, 05$ test for testing
$H_{0} : μ = μ_{0} vs. H_{1} : μ \neq μ_{0}$
where $X_{1}, \dots, X_{n} \sim iid n (μ, σ^{2})$ has acceptance region
$A (μ_{0}) = \{x : | \bar{x} - μ_{0} | \leq 1, 96 σ / \sqrt{n}\} .$
Substituting my problem (I think) we get that the most powerful unbiased size $α = 0, 05$ test for testing
$H_{0} : θ = \hat{θ} vs. H_{1} : θ \neq \hat{θ}$
has acceptance region ${A (\hat{θ}) = {y : | \overset{―}{y} - \hat{θ} | \leq 1, \frac{96}{\sqrt{n}}}$
or equivalently, $A (\hat{θ}) = \{y : \frac{\sqrt{n} \bar{y} - 1, 96}{\sqrt{n} x_{i}} \leq \hat{θ} \leq \frac{\sqrt{n} \bar{y} + 1, 96}{\sqrt{n} x_{i}}\}$
Substituting $\hat{θ} = \sum x_{i} Y \frac{i}{\sum} x_{i}^{2}$ we obtain
$A (\hat{θ}) = \{y : \frac{\sqrt{n} \bar{y} - 1, 96}{\sqrt{n} x_{i}} \leq \frac{Σ x_{i} Y i}{Σ x_{i}^{2}} \leq \frac{\sqrt{n} \bar{y} + 1, 96}{\sqrt{n} x_{i}}\}$
This means that my $1 - 0, 05 = 0, 95 (95 %)$ confindence interval is defined to be
$C (y) = {\hat{θ} : y \in A (\hat{θ})}$
But I can't seem to find anything concrete and I feel that I've made mistakes somewhere. What to do?

r1fa8dy5 2022-04-03

Estimate of Proportion
An airline is interested in determining the proportion of its customers who are flying for reasons of business. If they want to be 90 percent certain that their estimate will be correct to within two percent, how large a random sample should they select?
This involves using the Z-distribution to find the 90% confidence interval of the actual value of the proportion. But for this to be done, there needs to be an estimate of the proportion, which is not given. How do we find this out?ie
$(p) \cdot (1 - p) \cdot z_{\frac{a}{2} ⁄ n} = 0.0004$ . How do we find the value of p?

Petrolovujhm 2022-04-03

Statistics finding the confidence level of the interval estimate medical expenses
A confidence interval for the true mean of the annual medical expenses of a middle-class American family is given as (738, 777). If this interval is based on interviews with 110 families and a standard deviation of $ 120 is assumed. Suppose all annual medical expenses of middle-class American families follow an approximately normal distribution.
(a) What is the sample mean of annual medical expenses?
Sample mean $= 777 + \frac{738}{2} = 757.5$
(b) What is the confidence level of the interval estimate (as a decimal)
$C I = sample mean + z \frac{α}{2} \cdot (\frac{standard deviation}{square root n})$
Isolate for z $\frac{α}{2}$
$777 = 757.5 + z \frac{α}{2} \cdot (\frac{120}{square root} 110)$
Rearrange
$\frac{777 \cdot square root 110}{757.5 \cdot 120} = z \frac{α}{2}$
$0.089650657 = z \frac{α}{2}$
What are the correct solutions and answers?

Coradossi7xod 2022-04-01

Statistics range rule of thumb confidence interval
Suppose that the minimum and maximum ages for typical textbooks currently used in college courses are 0 and 8 years. Use the range rule of thumb to estimate the standard deviation.
Standard deviation = I have gotten max - min/4
$= 8 - \frac{0}{4}$ S
$= 2$
Find the size of the sample required to estimage the mean age of textbooks currently used in college courses. Assume that you want 98% confidence that the sample mean is within 0.4 year of the population mean.
Required sample size =
I have no clue how to get the required sample size

jncuenodd4nf 2022-04-01

Rewrite the probability statement
$P (0 < R < k (X_{1} + X_{2})) = 1 - S$

alparcero97oy 2022-04-01

$s_{p}^{2}$ is an unbiased estimator for $σ^{2}$
We have, $\frac{(n_{1} - 1) s_{1}^{2}}{σ^{2}} \sim X_{(n_{1} - 1)}^{2}$
$\frac{(n_{2} - 1) s_{2}^{2}}{σ^{2}} \sim x_{(n_{2} - 1)}^{2}$
$\Rightarrow \frac{(n_{1} - 1) s_{1}^{2}}{σ^{2}} + \frac{(n_{2} - 1) s_{2}^{2}}{σ^{2}} \sim x_{(n_{1} + n_{2} - 2)}^{2}$
$\Rightarrow V a r (s_{p}^{2}) = \frac{2 σ^{4}}{n_{1} + n_{2} - 2}$

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