Master Confidence Intervals: Examples, Equations, and Questions

The Level of Significance of a Statistical Test The ${\displaystyle R}$ and $D$ department of a paint company has developed an additive that it hopes will increase the ability of the company’s stain for outdoor decks to resist water absorption. The current formulation of the stain has a mean absorption rate of 35 units. Before changing the stain, a study was designed to evaluate whether the mean absorption rate of the stain with the additive was decreased from the current rate of 35 units. The stain with the additive was applied to 50 pieces of decking material. The resulting data were summarized to ${\displaystyle \bar{x}=33.6}$ and ${\displaystyle s=9.2}$
a) Is there substantial evidence ${\displaystyle (\alpha =0.01)}$ that the additive reduces the mean absorption from its current value?
b) What is the level of significance (p -value) of your test results?
c) What is the probability of a Type II error if the stain with the additive in fact has a mean absorption rate of 30?
d) Estimate the mean absorption using a 99% confidence interval. Is the confidence interval consistent with your conclusions from the test of hypotheses?

In a survey of 2695 adults, 1446 say they have started paying bills electronically in the last year.
Construct a ${\displaystyle 99\mathrm{\%}}$ confidence interval for the population proportion. 
Interpret your results. Choose the correct answer below. 
A) With ${\displaystyle 95\mathrm{\%}}$ confidence, it can be said that the sample proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval 
B) With ${\displaystyle 99\mathrm{\%}}$ confidence, it can be said that the population proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval. 
C) The endpoints of the given confidence interval show that adults pay bills online ${\displaystyle 99\mathrm{\%}}$ of the time.

A standard normal table, also called the unit normal table or Z table, may be a mathematical table for the values of ${\displaystyle \varphi }$, which are the values of the cumulative distribution function of the traditional distribution.
$\begin{array}{|cc|}\hline \text{confidence intervals}& Z\\ 90& 1.645\\ 95& 1.96\\ 96& 2.05\\ 97& 2.17\\ 98& 2.326\\ 99& 2.576\\ \hline\end{array}$
a) How do calculate Z-score using the confidence interval?

A set of data consists of 45 observations between 0 and 29. Which size would you advise for the class period?

A random sample of 820 households from a city, Stats City, were surveyed. From the statistical technology, we find a ${\displaystyle 95\mathrm{\%}}$ confidence interval for the mean number of people in the households was ${\displaystyle 1.85<\mu <3.25}$
What is the true meaning of the interval presented in the problem about Stats City?

You are interested in finding a ${\displaystyle 95\mathrm{\%}}$ confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 14 randomly selected physical therapy patients. Round answers to 3 decimal places where possible.
$\begin{array}{|cccccccccccccc|}\hline 9& 6& 10& 15& 19& 6& 23& 26& 19& 16& 11& 25& 16& 11\\ \hline\end{array}$
a) To compute the confidence interval use a t or z distribution.
b) With ${\displaystyle 95\mathrm{\%}}$ confidence the population mean number of visits per physical therapy patient is between "?" and "?" visits.
c) If many groups of 14 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About "?" percent of these confidence intervals will contain the true population mean number of visits per patient and about "?" percent will not contain the true population mean number of visits per patient.

When σ is unknown and the sample size is ${\displaystyle n\ge 30}$, there are tow methods for computing confidence intervals for μμ. Method 1: Use the Student's t distribution with d.f. = n - 1. This is the method used in the text. It is widely employed in statistical studies. The majority of statistical software programs also employ this technique. Method 2: When ${\displaystyle n\ge 30}$, use the sample standard deviation s as an estimate for σσ, and then use the standard normal distribution. This method is based on the fact that for large samples, s is a fairly good approximation for σσ. Also, for large n, the critical values for the Student's t distribution approach those of the standard normal distribution. Consider a random sample of size n = 31, with sample mean x¯=45.2 and sample standard deviation s = 5.3. (c) Compare intervals for the two methods. Would you say that confidence intervals based on a Student's t distribution tend to be longer than intervals based on the conventional normal distribution, making them more conservative?

Anna flipped a coin to choose either her right hand or her left hand, and also she asked the therapists to identify the named hand by placing their hand just under Anna's hand without seeing it and without touching it. Among 358 trials, the touch therapists were correct 172 times. Complete parts​ (a) through​ (d).
a) Given that Anna used a coin toss to elect either her right hand or her left hand, what proportion of correct responses would be anticipated if the touch therapists made arbitrary suppositions?( Type an integer or a numeric. Do not round.)
b) Using Anna's sample results, what's the stylish point estimate of the therapists' success rate?( Round to three decimal places as demanded.)
c) Using Anna's sample results, construct a ${\displaystyle 90\mathrm{\%}}$ confidence interval estimate of the proportion of correct responses made by touch therapists.
Round to three decimal places as demanded-?${\displaystyle <p<}$?

A local firm manufactures LED products that have a lifespan that is approximately normally distributed with a std. dev. of 30 hours. If a sample of 30 LED products has an average lifespan of 780 hours, find a ${\displaystyle 96\mathrm{\%}}$ confidence interval for the population mean of all LED products produced by this firm.
Choose 2 answers in nearest unit (ones) or in whole number.
Example, if your answer is ${\displaystyle 888.83\le \mu \le 899.56}$, choose 889 and 900.
$\begin{array}{|cccccccccccc|}\hline 775& 773& 807& 797& 791& 769& 789& 768& 805& 763& 771& 792\\ \hline\end{array}$