Master Confidence Intervals: Examples, Equations, and Questions

find the critical value za÷2corresponding. In

to the given confidence intervals: 90%, 93

There is a 95% chance that the true mean is between 1 and 3, if our sample's 95% confidence interval is between 1 and 3.

An educational psychologist wishes to know the mean number of words a third grader can read per minute. She wants to make an estimate at the $99\%$ level of confidence. For a sample of $1869$ third graders, the mean words per minute read was $25.3$. Assume a population standard deviation of $5.3$. Construct the confidence interval for the mean number of words a third grader can read per minute. Round your answers to one decimal place.

A survey of several 10$10$ to 11$11$ year olds recorded the following amounts spent on a trip to the mall:

$23.22,$9.71,$14.34,$23.05,$16.61,$7.22,$22.15
 

Construct the 99% confidence interval for the average amount spent by $10$ to 11 year olds on a trip to the mall. Assume the population is approximately normal.

Step 4 of 4 : 

Construct the 99% confidence interval. Round your answer to two decimal places.

Use the table for z or t-distribution to determine the value of z or t to construct a confidence interval for a population mean for each of the following combinations of confidence and sample size:

Confidence interval 90% n=37

For the provided sample​ mean, sample​ size, and population standard​ deviation, complete parts​ (a) through​ (c) below.

x=32​,

n=100​,

σ=3

Question content area bottom

Part 1

a. Find a​ 95% confidence interval for the population mean.

Simple Linear Regression - Difference between predicting and estimating?
Here is what my notes say about estimation and prediction:
Estimating the conditional mean
"We need to estimate the conditional mean ${\displaystyle {\beta }_0+{\beta }_1{x}_0}$ at a value ${\displaystyle x_0}$, so we use ${\displaystyle \widehat{Y_0}=\widehat{{\beta }_0}+\widehat{{\beta }_1}{x}_0}$ as a natural estimator." here we get
$\widehat{Y_0}~N({\beta }_0+{\beta }_1{x}_0,{\sigma }^2{h}_{00}) \text{where }{h}_{00}=\frac{1}{n}+\frac{{(x_0-\bar{x})}^2}{(n-1)s_x^2}$
with a confidence interval for ${\displaystyle E\left(Y_0\right)={\beta }_0+{\beta }_1{x}_0}$ is
${\displaystyle (\widehat{b_0}+\widehat{b_1}{x}_0-cs\sqrt{h_{00}},\widehat{b_0}+\widehat{b_1}{x}_0+cs\sqrt{h_{00}})}$
where ${\displaystyle c=t_{n-2,1-\frac{\alpha }{2}}}$ Where these results are found by looking at the shape of the distribution and at ${\displaystyle E\left(\widehat{Y_0}\right)}$ and ${\displaystyle var\left(\widehat{Y_0}\right)}$
Predicting observations
"We want to predict the observation ${\displaystyle Y_0={\beta }_0+{\beta }_1{x}_0+{ϵ}_0}$ at a value ${\displaystyle x_0}$"
$E(\widehat{Y_0}-Y_0)=0 \text{and }var(\widehat{Y_0}-Y_0)={\sigma }^2(1+h_{00})$
Hence a prediction interval is of the form
${\displaystyle (\widehat{b_0}+\widehat{b_1}{x}_0-cs\sqrt{h_{00}+1},\widehat{b_0}+\widehat{b_1}{x}_0+cs\sqrt{h_{00}+1})}$

If we calculate the probability P(A) that event A will occur and obtain the corresponding confidence interval 
$z_1$. If I have n trials, the expected number of trials that will result in event A is ${\displaystyle n\cdot P\left(A\right)}$. But what then is the confidence interval ${\displaystyle z_2}$ for this expected number of events A?

Given you have an independent random sample ${\displaystyle X_1,X_2,\dots ,X_n}$ 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 ${\displaystyle \hat{P}}$ will be p, and the variance will be:
$Var\left[\hat{P}\right]=\frac{1}{nE\left[\right((\partial /\partial p) \ln \hspace{0.17em}{f}_x\left(X\right){)}^2]}$
Now I'm having some trouble calculating the variance of ${\displaystyle \hat{P}}$, this is what I have so far:
since the probability function of ${\displaystyle \bar{X}}$ is binomial, we have:
$f_x\left(\overline{X}\right)=\left(\genfrac{}{}{0ex}{}{n}{\sum _{i=1}^n{X}_i}\right)*p^{\sum _{i=1}^n{X}_i}*(1-p{)}^{n-\sum _{i=1}^n{X}_i}$
so: $\ln \hspace{0.17em}{f}_X\left(X\right)=\ln \left(\left(\genfrac{}{}{0ex}{}{n}{\sum _{i=1}^n{X}_i}\right)\right)+\sum _{i=1}^n{X}_{i}ln\left(p\right)+\hspace{0.17em}(n-\sum _{i=1}^n{X}_i)\ln (1-p)$
and: $\frac{\partial \ln \hspace{0.17em}{f}_X\left(X\right)}{\partial p}=\frac{{\sum }_{i=1}^n{X}_i}{p}-\frac{(n-{\sum }_{i=1}^n{X}_i)}{(1-p)}=\frac{n\overline{X}}{p}-\frac{(n-n\overline{X})}{(1-p)}$
and: $(\frac{\partial ln\hspace{0.17em}{f}_X\left(X\right)}{\partial p}{)}^2=(\frac{n\overline{X}}{p}-\frac{(n-n\overline{X})}{(1-p)}{)}^2=\frac{n^2{p}^2-2n^{2}p\overline{X}+n^2{\overline{X}}^2}{p^2(1-p{)}^2}$
since ${\displaystyle E\left[{\bar{X}}^2\right]={\mu }^2+\frac{{\sigma }^2}{n}}$, and for a Bernoulli random variable ${\displaystyle E\left[X\right]=\mu =p=E\left[\bar{X}\right]}$ and ${\displaystyle Var\left[X\right]={\sigma }^2=p(1-p)}$:
${\displaystyle E\left[{\left(\frac{\partial \ln :\left\{f_X\left(X\right)\right\}}{\partial p}\right)}^2\right]=\frac{n^2{p}^2-2n^{2}pE\left[\bar{X}\right]+n^{2}E\left[{\bar{X}}^2\right]}{p^2{(1-p)}^2}=\frac{n^2{p}^2-2n^2{p}^2+n^2(p^2+\frac{p(1-p)}{n})}{p^2{(1-p)}^2}=\frac{np(1-p)}{p^2{(1-p)}^2}=\frac{n}{p(1-p)}}$
Therefore, $Var\left[\hat{P}\right]=\frac{1}{nE\left[\right((\partial /\partial p) \ln \hspace{0.17em}{f}_x\left(X\right){)}^2]}=\frac{1}{n\frac{n}{p(1-p)}}=\frac{p(1-p)}{n^2}$
However, I believe the true value I should have come up with is ${\displaystyle \frac{p(1-p)}{n}}$.