Recent questions in Confidence intervals

Confidence intervals
Answered

Quinlan7g
2022-09-25

[126.4, 132.2] is a 95% confidence interval for the mean $\mu $ of a normally distributed random variable with known variance. Find a 98% confidence interval for $\mu $, based on the same sample.

I got gamma $=0.98\text{}gamma/2=0.49\text{}so\text{}Z\ast =2.33$ but then I am lost on what to do next.

Confidence intervals
Answered

Camila Brandt
2022-09-25

Suppose we use a sample mean $\overline{X}$ to construct a 95% confidence interval [a,b].

I was told that it is incorrect to say that there is a 95% probability that the population mean lies between a and b. Because the population mean is a constant and not a random variable. The probability that a constant falls within any given range is either 0 or 1.

However, from the textbook it is said that we expect 95% of the confidence intervals to include the population mean.

If 95% of the confidence intervals are expected to include the population mean, then each confidence interval has 95% probability to include the population mean. Therefore I think it is correct to say there is a 95% probability that the population mean lies between a and b.

Where did I make mistakes?

Confidence intervals
Answered

waldo7852p
2022-09-25

However, to my knowledge, isn't this one of the common misconceptions of a confidence interval? And that the actual interpretation is that if we gather n amount of these confidence intervals, there is a 95% probability that the n collected intervals all contain the true parameter?

My notes give this definition:Let $L:=L({X}_{1},\dots ,{X}_{n})$ and $U:=U({X}_{1},\dots ,{X}_{n})$ be such that for all $\theta \in \mathrm{\Theta}$,

$$\mathbb{P}(L<\theta \le U)\ge 1-\alpha $$

and then it says, that this is the probability $1-\alpha $ that the true parameter lies within this interval . Which one is correct?

Confidence intervals
Answered

Alexus Deleon
2022-09-25

For this question from a large amount of data I have calculated that the mean is 44.22, the sample size is 100 and the standard deviation is 22.0773.

From this I am asked to , make the 98% confidence intervals for the (1) true mean µ of the module mark (2) true variance of the module mark

And for each: (a) Determine what quantity to look at, and which distribution table to use, justifying your choice. (b) Determine the number of degrees of freedom, justifying your answer. (c) Calculate the actual intervals.

So far for 1, I have used the z table to look for 99% as I need 1% to the right of 2.33 and 1% to the left of −2.33, so 98% is between $\pm 2.33$. Giving me

$$\overline{x}\pm 2.33\frac{\sigma}{\sqrt{n}}$$

Which provides me with a 39.08 to 49.36 confidence interval, is this correct? And how would I determine degrees of freedom and go about answering part 2?

Confidence intervals
Answered

Freddy Chaney
2022-09-24

Lets say I measure quantity $\mu $ and compute confidence interval for it $[\mu -{\delta}_{1},\mu +{\delta}_{2}]$ with a 68% confidence limit.

What if I want a confidence interval for 95% CL, can I simply scale errors like this? Let $x=\frac{0.95}{0.68}$

$[\mu -x{\delta}_{1},\mu +x{\delta}_{2}]$

Confidence intervals
Answered

Addyson Bright
2022-09-24

For the formula $\overline{X}\pm {z}_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$, we substitute ${\overline{x}}_{obs}$ for $\overline{X}$, but isn't that observed sample mean ${\overline{x}}_{obs}$ only for one particular sample?

Also why is it incorrect to interpret a confidence interval as the probability that the actual $\mu $ lies in a interval (a,b) is $(1-\alpha )100\mathrm{\%}$

Confidence intervals
Answered

Madelynn Winters
2022-09-23

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.

Confidence intervals
Answered

Hagman7v
2022-09-23

A random sample of six 2009 sports cars is taken and their "in the city" miles per gallon is recorded. The results are as follows: 23, 19, 24, 17, 16, 22. Assuming the population distribution is normal, calculate the 99% confidence interval for μ, the population mean "in city" mpg for 2009 sports cars.

Confidence intervals
Answered

Ivan Buckley
2022-09-23

Consider a random sample ${X}_{1},\dots ,{X}_{n}$ from a Uniform($\theta $, $\theta +a$) distribution, where $\theta $ is unknown and a is known. We wish to determine a confidence interval for $\theta $.

The reader may verify the following details: The statistics $Y={\text{min}}_{i}{X}_{i}$ and $Z={\text{max}}_{i}{X}_{i}$ are jointly sufficient for $\theta $. For $\theta \le {c}_{1}\le {c}_{2}\le \theta +a$, $P\{{c}_{1}\le Y\le Z\le {c}_{2}\}=[({c}_{2}-{c}_{1})/a{]}^{n}$. For $0<\gamma <1$, set ${d}_{1}=(1-\sqrt[n]{\gamma})/2$ and ${d}_{2}=(1+\sqrt[n]{\gamma})/2$. Then $\gamma =P\{\theta +a{d}_{1}\le Y\le Z\le \theta +a{d}_{2}\}=P\{Z-a{d}_{2}\le \theta \le Y-a{d}_{1}\}$. Thus, $\gamma =P\{\theta +a{d}_{1}\le Y\le Z\le \theta +a{d}_{2}\}=P\{Z-a{d}_{2}\le \theta \le Y-a{d}_{1}\}$ is a γ confidence interval for $\theta $.

Now here's the difficulty: If we observe $Z-Y>a\sqrt[n]{\gamma}$, then $Z-a{d}_{2}>Y-a{d}_{1}$, so our formula yields a nonsensical answer. Have I made an error in my calculations? Or is this one of those problems with confidence intervals?

Confidence intervals
Answered

2k1ablakrh0
2022-09-22

I'm reviewing some basic statistics, and I'm asking myself questions on things I used to take for granted when I first saw them years ago. I'm going to state things as I understand them, so there might be a mistake in the following.

Consider a random variable X following a distribution of mean $\mu $ and standard deviation $\sigma $. We measure n samples of X, and observe an empirical mean $\overline{x}$ and empirical std s.

Because we're observing samples, $\overline{x}$ and s are random variables themselves. We should therefore not use $\overline{x}$ and $\mu $ interchangeably, and the same goes for s and $\sigma $. Instead, people compute confidence interval on μ based on $\overline{x}$. By the central limit theorem, if n is large enough, we can say that $\overline{x}\sim N(\mu ,\frac{\sigma}{\sqrt{n}})$.

When computing confidence intervals, we usually use ${\sigma}_{\overline{x}}=\frac{s}{\sqrt{n}}$. If all of this is correct, my question is the following: since both s and $\overline{x}$ are random variables, why does it seems to be ok to consider $s=\sigma $ when computing confidence intervals for $\mu $?

Confidence intervals
Answered

Melina Barber
2022-09-21

I'm fairly comfortable calculating the confidence interval. But now I'm seeing a problem where I'm giving a confidence interval CI(27.6621,30.3379) and I'm requested to calculate the confidence level. I'm also given the sample size $n=85$ and a standard deviation $\sigma =7.5$. I can't find a formula for that, I feel like it's something simple I'm not seeing. Any help would be appreciated.

Confidence intervals
Answered

kennadiceKesezt
2022-09-21

Which of the following will result in a wider confidence interval?

Larger sample size

Higher level of confidence

Larger population standard deviation

Larger population mean

Larger sample mean

So after running some numbers, I believe 2,3, and 4 are correct because they appeared to work however apparently that's incorrect. Could anyone please help me?

Confidence intervals
Answered

Colten Andrade
2022-09-21

I am given a sample size of 9 from a normal population, as well as the 95% confidence interval for population mean. I want to calculate the 95% population variance.

I see that the confidence interval for an unknown population mean is $\overline{Y}\pm {t}_{\frac{\alpha}{2}}\frac{S}{\sqrt{n}}$. Then since the midpoint of the confidence interval for population mean is the sample mean, I can plug that into $\overline{Y}$, 9 into n, and 0.025 into $\alpha $. From there, I can solve for S. How can I find the confidence interval for population variance from there?

Confidence intervals
Answered

kennadiceKesezt
2022-09-20

I was curious as to what the relationship between probabilistic values, estimators and confidence intervals are. I was wondering, if I have an estimator of some parameter $\lambda $, and a probability value that depended on $\lambda $ and the estimator, what would be the relationship between those two and a confidence interval for an arbitrary distribution?

Confidence intervals
Answered

mangicele4s
2022-09-20

I have read two books that explicitly state that the $(1-\alpha )$% confidence interval should be interpreted as:

If you construct 100 such confidence intervals, $\alpha $ of them are expected to not contain the true population statistic and $(1-\alpha )$ of them are expected to contain the true population statistic.

and not as

There is a $(1-\alpha )$ probability that the true population statistic is contained in the confidence interval.

In my view, they both equivalent: If you make the first statement, you implicitly make the second statement. You are looking at any one arbitrary confidence interval, which in itself is a random variable, the generic confidence interval should, therefore, be subject to the second statement. Do things change when this random variable is actually realized?

Confidence intervals
Answered

Jazmyn Pugh
2022-09-20

Consider a random sample ${X}_{1},{X}_{2},..,{X}_{n}$ from a variable with density function

$${f}_{X}(x)=2\lambda \pi x{e}^{-\lambda \pi {x}^{2}}\text{}\text{}\text{}\text{}\text{}\text{}\text{}x0$$

A useful estimator of $\lambda $ is

$$\hat{\lambda}=\frac{n}{\pi \sum _{i=1}^{n}{X}_{1}^{2}}$$

Derive a 95% confidence interval for $\lambda $ (a hint is provided suggesting to consider the distribution of $\frac{\lambda}{\hat{\lambda}}{\textstyle )}.$).

Confidence intervals
Answered

Lyla Carson
2022-09-20

Anyway I thought that it may have been due to the fact that a lot of outliers affected the regression line calculated, and so a confidence interval formed from a bad regression line would be bad - resulting in 95% of observations falling outside the 95% CI.

Confidence intervals
Answered

Spactapsula2l
2022-09-17

I am reviewing the construction of confidence intervals for a random sample with Bernoulli distribution. The book uses the statistics of the central limit theorem that distributes N(0,1) to estimate the interval :

$${Z}_{n}=\frac{{X}_{1}+{X}_{2}+\cdots +{X}_{n}-n\mu}{\sigma \sqrt{n}}$$

Why are the intervals constructed from these statistics symmetrical around the origin?

The book says: "Since it is desirable that the length of the interval be as small as possible and since the standard normal distribution is symmetrical around the origin, it turns out that the minimum length interval must also be symmetric around the origin", but I don't understand this.

Confidence intervals
Answered

Liam Keller
2022-09-17

The question looks pretty simple but I can't get my hands on it:Say I have a probability which is the product of two other independent probabilities $p={p}_{1}{p}_{2}$.

I have estimated each probability ${p}_{1}$ and ${p}_{2}$ and found some 95% confidence interval for each. How do I obtain a 95% confidence interval for p?

Taking the product of the bounds of the interval won't work as I would be taking 95% of a 95% confidence interval resulting in an approximately 90% which is not what I want.So conversely I would be taking 97.5% confidence interval for each and by multiplying the bounds I will obtain a 95% confidence interval, is that right?

I feel like something is going wrong. In my situation I deal with probabilities but it could be anything so this question can be generalised to any type of confidence intervals.

If my reasoning is correct, could someone convince me that's the correct way of doing so?

Confidence intervals
Answered

varice2r
2022-09-16

And I know this statement is false, but I want to know exactly why.

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