Practice Math Statistics Problems and Master the Concepts

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

Khaleesi Herbert 2021-09-15

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. $x^{4} + x - 3 = 0,$ (1,2)

Daniaal Sanchez 2021-08-16

Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters mu=71 and sigma^2=6.25 . What percentage of 25-year-old men are over 6 feet, 2 inches tall? What percentage of men in the 6-footer club are over 6 feet, 5 inches?

BenoguigoliB 2021-08-14

The pathogen Phytophthora capsici causes bell pepper plants to wilt and die. A research project was designed to study the effect of soil water content and the spread of the disease in fields of bell peppers. It is thought that too much water helps spread the disease. The fields were divided into rows and quadrants. The soil water content (percent of water by volume of soil) was determined for each plot. An important first step in such a research project is to give a statistical description of the data.
Soil Water Content for Bell Pepper Study
$\begin{array}{cc} 15 & 14 & 14 & 14 & 3 & 12 & 11 & 11 & 11 & 11 & 10 & 11 & 13 & 16 & 10 \\ 9 & 15 & 12 & 9 & 10 & 7 & 14 & 13 & 14 & 8 & 9 & 8 & 11 & 13 & 13 \\ 15 & 12 & 9 & 10 & 9 & 9 & 16 & 16 & 12 & 10 & 11 & 11 & 12 & 15 & 8 \\ 10 & 10 & 10 & 11 & 9 \end{array}$
(a) Make a box-and-whisker plot of the data. Find the interquartile range.

Nann 2021-08-14

Resource managers of forest game lands were concerned about the size of the deer and rabbit populations during the winter months in a particular forest. As an estimate of population size, they proposed using the average number of pellet groups for rabbits and deer per 30-foot-square plots. From an aerial photograph, the forest was divided into N= 10,000 30-foot-square grids. A simple random sample of 2 = 500 plots was taken, and the number of pellet groups was observed for rabbits and for deer. The results of this study are summarized in the accompanying table.
a. Estimate $μ_{1}, and μ_{2}$ , the average number of pellet groups for deer and rabbits, respectively, per 30-foot-square-plots. Place bounds on the errors of estimation.
b. Estimate the difference in the mean size of pellet groups per plot for the two animals, with an appropriate margin of error.
Deer
Sample mean = 2.30
Sample variance = 0.65
Rabbits
Sample mean = 4.52
Sample variance = 0.97

sagnuhh 2021-08-14

When σ is unknown and the sample size is $n \geq 30$ , 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 $n \geq 30$ , 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?

Isa Trevino 2021-08-14

You are interested in finding a $95 %$ 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.
$\begin{array}{cc} 9 & 6 & 10 & 15 & 19 & 6 & 23 & 26 & 19 & 16 & 11 & 25 & 16 & 11 \end{array}$
a) To compute the confidence interval use a t or z distribution.
b) With $95 %$ 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.

permaneceerc 2021-08-14

The two-way table summarizes information about eye color and gender in a random sample of 200 high school students.

Josalynn 2021-08-14

The two-way table gives the party affiliation of all members of the House of Representatives from eight different regions in the United States in July 2014 (there were 3 vacant seats at that time).

waigaK 2021-08-12

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 $90 %$ confidence interval estimate of the proportion of correct responses made by touch therapists.
Round to three decimal places as demanded-? $< p <$ ?

ediculeN 2021-08-12

Suppose that each student in a sample had been categorized with respect to political views, marijuana usage, and religious preference, with the categories of this latter factor being Protestant, Catholic, and other. The data could be displayed in three different two-way tables, one corresponding to each category of the third factor. With pijk = P(political category i, marijuana category j, and religious category k), the null hypothesis of independence of all three factors states that $p_{i j k} = p_{i} . . p_{j} . . p_{k}$
Let nijk denote the observed frequency in cell (i, j, k). Show how to estimate the expected cell counts assuming that H0 is true ( ${\hat{e}}_{i j k} = n {\hat{p}}_{i j k}$ , so the ${\hat{p}}_{i j k}$ ’s must be determined). Then use the general rule of thumb to determine the number of degrees of freedom for the chi-squared statistic.

Ramsey 2021-08-12

Water flows through a water hose at a rate of $Q_{1} = 680 c \frac{m^{3}}{s}$ , the diameter of the hose is $d_{1} = 2.2 c m$ . A nozzle is attached to the water hose. The water leaves the nozzle at a velocity of $v_{2} = 9.2 \frac{m}{s}$ .
1) Enter an expression for the cross-sectional area of the hose, $A_{1}$ , in terms of its diameter, $d_{1}$
2) Calculate the numerical value of $A_{1},$ in square centimeters.
3) Enter an expression for the speed of the water in the hose, $v_{1}$ , in terms of the volume floe rate $Q_{1}$ and cross-sectional area $A_{1}$
4) Calculate the speed of the water in the hose, $v_{1}$ in meters per second.
5) Enter an expression for the cross-sectional area of the nozzle, $A_{2}$ , in terms of $v_{1}, v_{2}$ and $A_{1}$
6) Calculate the cross-sectional area of the nozzle, $A_{2}$ in square centimeters.

Line 2021-08-11

State whether the inl'esrigarion in question is an observational study or a designed experiment. Justify your answer in each case. According to a study published in the Journal of the American Public Health Association, left-handed people do not die at an earlier age than right-handed people, contrary to the conclusion of a highly publicized report done 2 years earlier. The investigation involved a 6-year study of 3800 people in East Boston older than age 65. Researchers at Harvard University and the National Institute of Aging found that the "lefties" and "righties" died at exactly the same rate. "There was no difference, period," said Dr. J. Guralnik, an epidemiologist at the institute and one of the coauthors of the report.

Nannie Mack 2021-08-10

Suppose that you want to perform a hypothesis test based on independent random samples to compare the means of two populations. For each part, decide whether you would use the pooled t-test, the nonpooled t-test, the Mann– Whitney test, or none of these tests if preliminary data analyses of the samples suggest that the two distributions of the variable under consideration are a. normal but do not have the same shape. b. not normal but have the same shape. c. not normal and do not have the same shape. both sample sizes are large.

UkusakazaL 2021-08-10

Use the confidence interval to find the margin of error and the sample mean.
(0.142, 0.300)
The margin of error is ?.
The sample mean is ?.

UkusakazaL 2021-08-10

The high price of medicines is a source of major expense for those seniors in the United States who have to pay for these medicines themselves. A random sample of 2000 seniors who pay for their medicines showed that they spent an average of $$ 4600$ last year on medicines with a standard deviation of $$ 800$ . a) Make a $98 %$ confidence interval for the corresponding population mean. b) Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss allpossible alternatives. Which alternative is the best?

UkusakazaL 2021-08-10

A firm employs 189 junior accountants. In a random sample of 50 of these, the mean number of hours overtime billed in a particular week was 9.7, and the sample standard deviation was 6.2 hours.
a. Find a $95 %$ confidence interval for the mean number of hours overtime billed per junior accountant in this firm that week.
b. Find a $99 %$ confidence interval for the total number of hours overtime billed by junior accountants in the firm during the week of interest.

UkusakazaL 2021-08-10

To monitor complex chemical processes, chemical engineers will consider key process indicators, which may be just yield but most often depend on several quantities. Before trying to improve a process, 9 measurements were made on a key performance indicator. 123, 106, 114, 128, 113, 109, 120, 102, 111. Assume that the key performance indicator has a normal distribution. (a) Find the sample variance s 2 of the sample. (b) Find a $95 %$ confidence interval for $σ 2$ . (c) Find a $98 %$ confidence interval for $σ$ .

UkusakazaL 2021-08-10

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 $96 %$ 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 $888.83 \leq μ \leq 899.56$ , choose 889 and 900.
$\begin{array}{cc} 775 & 773 & 807 & 797 & 791 & 769 & 789 & 768 & 805 & 763 & 771 & 792 \end{array}$

UkusakazaL 2021-08-10

Out of 200 people sampled, 116 preferred Candidate A. Based on this, estimate what proportion of the voting population prefers Candidate A with $99 %$ confidence.
Out of 500 people sampled, 455 had kids.
The best point estimate for pp is.
The margin of error for a $99 %$ confidence interval is.

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