Loretta, who turns eighty this year, has just learned about blood pressure problems in the elderly and is interested in how her blood pressure compares to those of her peers. Specifically, she is interested in her systolic blood pressure, which can be problematic among the elderly. She has uncovered an article in a scientific journal that reports that the mean systolic blood pressure measurement for women over seventy-five is 133.0 mmHg, with a standard deviation of 5.1 mmHg. Assume that the article reported correct information. Complete the following statements about the distribution of systolic blood pressure measurements for women over seventy-five. a) According to Chebyshev's theorem, at least ?36% 56% 75% 84% or 89% of the measurements lie between 122.8 mmHg and 143.2 mmHg. b) Acc

Question
Measurement
Loretta, who turns eighty this year, has just learned about blood pressure problems in the elderly and is interested in how her blood pressure compares to those of her peers. Specifically, she is interested in her systolic blood pressure, which can be problematic among the elderly. She has uncovered an article in a scientific journal that reports that the mean systolic blood pressure measurement for women over seventy-five is 133.0 mmHg, with a standard deviation of 5.1 mmHg.
Assume that the article reported correct information. Complete the following statements about the distribution of systolic blood pressure measurements for women over seventy-five.
a) According to Chebyshev's theorem, at least $$?36\% 56\% 75\% 84\%\ or\ 89\%$$ of the measurements lie between 122.8 mmHg and 143.2 mmHg.
b) According to Chebyshev's theorem, at least $$8/9 (about\ 89\%)$$ of the measurements lie between mmHg and mmHg. (Round your answer to 1 decimal place.)

2021-01-14
Step 1
Chebyshev’s inequality:
Chebyshev’s rule is appropriate for any distribution. That is, Chebyshev’s inequality applies to all distributions, regardless of shape. Moreover, it provides the minimum percentage of the observation that lies within k standard deviations of the mean. The Chebyshev’s rule states that, for any quantitative data set and any real number greater than k, at least $$(1-\frac{1}{k^{2}})$$ observations lie within k standard deviations to either side of the mean.
It is possible that very few measurements will fall within one standard deviation of the mean.
If $$k= 2$$, at least 3/4 of the measurements lie within 2 standard deviations to either side of the mean.
If $$k = 3$$, at least 8/9 of the measurements lie within 3 standard deviations to either side of the mean.
Generally, for any number k greater than 1, at least $$(1-\frac{1}{k^{2}})$$ of the measurements will fall within k standard deviations of the mean.
Step 2
It is given that the mean systolic blood pressure measurement for women over seventy-five is 133.0 mmHg, with a standard deviation of 5.1 mmHg.
That is, $$\mu=133$$ and standard deviation $$\sigma=5.1$$
One standard deviation below and above the mean is as follows:
$$(\mu−\sigma, \mu+\sigma)=(133−5.1,133+5.1)=(127.9,138.1)$$
Two standard deviation below and above the mean is as follows:
$$(\mu−2\sigma, \mu+2\sigma)=(133−2\times 5.1,133+2\times 5.1)=(122.8,143.2)$$
Three standard deviation below and above the mean is as follows:
$$(\mu−3\sigma, \mu+3\sigma)=(133−3times 5.1,133+3times 5.1)=(117.7,148.3)$$
a.
The measurements 122.8 mmHg and 143.2 mmHg are two standard deviations away from the mean. According to Chebyshev’s rule, about $$3/4(=75\%)$$ of the measurements lie within 2 standard deviations to either side of the mean.
Thus, according to Chebyshev's theorem, at least 75\% of the measurements lie between 122.8 mmHg and 143.2 mmHg.

Relevant Questions

A random sample of $$\displaystyle{n}_{{1}}={16}$$ communities in western Kansas gave the following information for people under 25 years of age.
$$\displaystyle{X}_{{1}}:$$ Rate of hay fever per 1000 population for people under 25
$$\begin{array}{|c|c|} \hline 97 & 91 & 121 & 129 & 94 & 123 & 112 &93\\ \hline 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88 \\ \hline \end{array}$$
A random sample of $$\displaystyle{n}_{{2}}={14}$$ regions in western Kansas gave the following information for people over 50 years old.
$$\displaystyle{X}_{{2}}:$$ Rate of hay fever per 1000 population for people over 50
$$\begin{array}{|c|c|} \hline 94 & 109 & 99 & 95 & 113 & 88 & 110\\ \hline 79 & 115 & 100 & 89 & 114 & 85 & 96\\ \hline \end{array}$$
(i) Use a calculator to calculate $$\displaystyle\overline{{x}}_{{1}},{s}_{{1}},\overline{{x}}_{{2}},{\quad\text{and}\quad}{s}_{{2}}.$$ (Round your answers to two decimal places.)
(ii) Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use $$\displaystyle\alpha={0.05}.$$
(a) What is the level of significance?
State the null and alternate hypotheses.
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}<\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}\ne\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}$$
(b) What sampling distribution will you use? What assumptions are you making?
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations,
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
The Student's t. We assume that both population distributions are approximately normal with known standard deviations,
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimalplaces.)
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimal places.)
(c) Find (or estimate) the P-value.
P-value $$\displaystyle>{0.250}$$
$$\displaystyle{0.125}<{P}-\text{value}<{0},{250}$$
$$\displaystyle{0},{050}<{P}-\text{value}<{0},{125}$$
$$\displaystyle{0},{025}<{P}-\text{value}<{0},{050}$$
$$\displaystyle{0},{005}<{P}-\text{value}<{0},{025}$$
P-value $$\displaystyle<{0.005}$$
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value
factor in determining the usefulness of an examination as a measure of demonstrated ability is the amount of spread that occurs in the grades. If the spread or variation of examination scores is very small, it usually means that the examination was either too hard or too easy. However, if the variance of scores is moderately large, then there is a definite difference in scores between "better," "average," and "poorer" students. A group of attorneys in a Midwest state has been given the task of making up this year's bar examination for the state. The examination has 500 total possible points, and from the history of past examinations, it is known that a standard deviation of around 60 points is desirable. Of course, too large or too small a standard deviation is not good. The attorneys want to test their examination to see how good it is. A preliminary version of the examination (with slight modifications to protect the integrity of the real examination) is given to a random sample of 20 newly graduated law students. Their scores give a sample standard deviation of 70 points. Using a 0.01 level of significance, test the claim that the population standard deviation for the new examination is 60 against the claim that the population standard deviation is different from 60.
(a) What is the level of significance?
State the null and alternate hypotheses.
$$H_{0}:\sigma=60,\ H_{1}:\sigma\ <\ 60H_{0}:\sigma\ >\ 60,\ H_{1}:\sigma=60H_{0}:\sigma=60,\ H_{1}:\sigma\ >\ 60H_{0}:\sigma=60,\ H_{1}:\sigma\ \neq\ 60$$
(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)
What are the degrees of freedom?
What assumptions are you making about the original distribution?
We assume a binomial population distribution.We assume a exponential population distribution. We assume a normal population distribution.We assume a uniform population distribution.
1. Find each of the requested values for a population with a mean of $$? = 40$$, and a standard deviation of $$? = 8$$ A. What is the z-score corresponding to $$X = 52?$$ B. What is the X value corresponding to $$z = - 0.50?$$ C. If all of the scores in the population are transformed into z-scores, what will be the values for the mean and standard deviation for the complete set of z-scores? D. What is the z-score corresponding to a sample mean of $$M=42$$ for a sample of $$n = 4$$ scores? E. What is the z-scores corresponding to a sample mean of $$M= 42$$ for a sample of $$n = 6$$ scores? 2. True or false: a. All normal distributions are symmetrical b. All normal distributions have a mean of 1.0 c. All normal distributions have a standard deviation of 1.0 d. The total area under the curve of all normal distributions is equal to 1 3. Interpret the location, direction, and distance (near or far) of the following zscores: $$a. -2.00 b. 1.25 c. 3.50 d. -0.34$$ 4. You are part of a trivia team and have tracked your team’s performance since you started playing, so you know that your scores are normally distributed with $$\mu = 78$$ and $$\sigma = 12$$. Recently, a new person joined the team, and you think the scores have gotten better. Use hypothesis testing to see if the average score has improved based on the following 8 weeks’ worth of score data: $$82, 74, 62, 68, 79, 94, 90, 81, 80$$. 5. You get hired as a server at a local restaurant, and the manager tells you that servers’ tips are $42 on average but vary about $$12 (\mu = 42, \sigma = 12)$$. You decide to track your tips to see if you make a different amount, but because this is your first job as a server, you don’t know if you will make more or less in tips. After working 16 shifts, you find that your average nightly amount is$44.50 from tips. Test for a difference between this value and the population mean at the $$\alpha = 0.05$$ level of significance.
Is statistical inference intuitive to babies? In other words, are babies able to generalize from sample to population? In this study,1 8-month-old infants watched someone draw a sample of five balls from an opaque box. Each sample consisted of four balls of one color (red or white) and one ball of the other color. After observing the sample, the side of the box was lifted so the infants could see all of the balls inside (the population). Some boxes had an “expected” population, with balls in the same color proportions as the sample, while other boxes had an “unexpected” population, with balls in the opposite color proportion from the sample. Babies looked at the unexpected populations for an average of 9.9 seconds (sd = 4.5 seconds) and the expected populations for an average of 7.5 seconds (sd = 4.2 seconds). The sample size in each group was 20, and you may assume the data in each group are reasonably normally distributed. Is this convincing evidence that babies look longer at the unexpected population, suggesting that they make inferences about the population from the sample? Let group 1 and group 2 be the time spent looking at the unexpected and expected populations, respectively. A) Calculate the relevant sample statistic. Enter the exact answer. Sample statistic: _____ B) Calculate the t-statistic. Round your answer to two decimal places. t-statistic = ___________ C) Find the p-value. Round your answer to three decimal places. p-value =
Case: Dr. Jung’s Diamonds Selection
With Christmas coming, Dr. Jung became interested in buying diamonds for his wife. After perusing the Web, he learned about the “4Cs” of diamonds: cut, color, clarity, and carat. He knew his wife wanted round-cut earrings mounted in white gold settings, so he immediately narrowed his focus to evaluating color, clarity, and carat for that style earring.
After a bit of searching, Dr. Jung located a number of earring sets that he would consider purchasing. But he knew the pricing of diamonds varied considerably. To assist in his decision making, Dr. Jung decided to use regression analysis to develop a model to predict the retail price of different sets of round-cut earrings based on their color, clarity, and carat scores. He assembled the data in the file Diamonds.xls for this purpose. Use this data to answer the following questions for Dr. Jung.
1) Prepare scatter plots showing the relationship between the earring prices (Y) and each of the potential independent variables. What sort of relationship does each plot suggest?
2) Let X1, X2, and X3 represent diamond color, clarity, and carats, respectively. If Dr. Jung wanted to build a linear regression model to estimate earring prices using these variables, which variables would you recommend that he use? Why?
3) Suppose Dr. Jung decides to use clarity (X2) and carats (X3) as independent variables in a regression model to predict earring prices. What is the estimated regression equation? What is the value of the R2 and adjusted-R2 statistics?
4) Use the regression equation identified in the previous question to create estimated prices for each of the earring sets in Dr. Jung’s sample. Which sets of earrings appear to be overpriced and which appear to be bargains? Based on this analysis, which set of earrings would you suggest that Dr. Jung purchase?
5) Dr. Jung now remembers that it sometimes helps to perform a square root transformation on the dependent variable in a regression problem. Modify your spreadsheet to include a new dependent variable that is the square root on the earring prices (use Excel’s SQRT( ) function). If Dr. Jung wanted to build a linear regression model to estimate the square root of earring prices using the same independent variables as before, which variables would you recommend that he use? Why?
1
6) Suppose Dr. Jung decides to use clarity (X2) and carats (X3) as independent variables in a regression model to predict the square root of the earring prices. What is the estimated regression equation? What is the value of the R2 and adjusted-R2 statistics?
7) Use the regression equation identified in the previous question to create estimated prices for each of the earring sets in Dr. Jung’s sample. (Remember, your model estimates the square root of the earring prices. So you must actually square the model’s estimates to convert them to price estimates.) Which sets of earring appears to be overpriced and which appear to be bargains? Based on this analysis, which set of earrings would you suggest that Dr. Jung purchase?
8) Dr. Jung now also remembers that it sometimes helps to include interaction terms in a regression model—where you create a new independent variable as the product of two of the original variables. Modify your spreadsheet to include three new independent variables, X4, X5, and X6, representing interaction terms where: X4 = X1 × X2, X5 = X1 × X3, and X6 = X2 × X3. There are now six potential independent variables. If Dr. Jung wanted to build a linear regression model to estimate the square root of earring prices using the same independent variables as before, which variables would you recommend that he use? Why?
9) Suppose Dr. Jung decides to use color (X1), carats (X3) and the interaction terms X4 (color * clarity) and X5 (color * carats) as independent variables in a regression model to predict the square root of the earring prices. What is the estimated regression equation? What is the value of the R2 and adjusted-R2 statistics?
10) Use the regression equation identified in the previous question to create estimated prices for each of the earring sets in Dr. Jung’s sample. (Remember, your model estimates the square root of the earring prices. So you must square the model’s estimates to convert them to actual price estimates.) Which sets of earrings appear to be overpriced and which appear to be bargains? Based on this analysis, which set of earrings would you suggest that Dr. Jung purchase?
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}&{H}{o}{u}{s}{e}{w}{\quad\text{or}\quad}{k}{H}{o}{u}{r}{s}\backslash{h}{l}\in{e}{G}{e}{n}{d}{e}{r}&{S}{a}\mp\le\ {S}{i}{z}{e}&{M}{e}{a}{n}&{S}{\tan{{d}}}{a}{r}{d}\ {D}{e}{v}{i}{a}{t}{i}{o}{n}\backslash{h}{l}\in{e}{W}{o}{m}{e}{n}&{473473}&{33.133}{.1}&{14.214}{.2}\backslash{h}{l}\in{e}{M}{e}{n}&{488488}&{18.618}{.6}&{15.715}{.7}\backslash{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ a. Based on this​ study, calculate how many more hours per​ week, on the​ average, women spend on housework than men. b. Find the standard error for comparing the means. What factor causes the standard error to be small compared to the sample standard deviations for the two​ groups? The cause the standard error to be small compared to the sample standard deviations for the two groups. c. Calculate the​ 95% confidence interval comparing the population means for women Interpret the result including the relevance of 0 being within the interval or not. The​ 95% confidence interval for ​$$\displaystyle{\left(\mu_{{W}}-\mu_{{M}}​\right)}$$ is: (Round to two decimal places as​ needed.) The values in the​ 95% confidence interval are less than 0, are greater than 0, include 0, which implies that the population mean for women could be the same as is less than is greater than the population mean for men. d. State the assumptions upon which the interval in part c is based. Upon which assumptions below is the interval​ based? Select all that apply. A.The standard deviations of the two populations are approximately equal. B.The population distribution for each group is approximately normal. C.The samples from the two groups are independent. D.The samples from the two groups are random.
You may need to use the appropriate appendix table or technology to answer this question.
Money reports that the average annual cost of the first year of owning and caring for a large dog in 2017 is $1,448. The Irish Red and White Setter Association of America has requested a study to estimate the annual first-year cost for owners of this breed. A sample of 50 will be used. Based on past studies, the population standard deviation is assumed known with $$\displaystyle\sigma=\{230}.$$ $$\begin{matrix} 1,902 & 2,042 & 1,936 & 1,817 & 1,504 & 1,572 & 1,532 & 1,907 & 1,882 & 2,153 \\ 1,945 & 1,335 & 2,006 & 1,516 & 1,839 & 1,739 & 1,456 & 1,958 & 1,934 & 2,094 \\ 1,739 & 1,434 & 1,667 & 1,679 & 1,736 & 1,670 & 1,770 & 2,052 & 1,379 & 1,939\\ 1,854 & 1,913 & 2,163 & 1,737 & 1,888 & 1,737 & 2,230 & 2,131 & 1,813 & 2,118\\ 1,978 & 2,166 & 1,482 & 1,700 & 1,679 & 2,060 & 1,683 & 1,850 & 2,232 & 2,294 \end{matrix}$$ (a) What is the margin of error for a $$95\%$$ confidence interval of the mean cost in dollars of the first year of owning and caring for this breed? (Round your answer to nearest cent.) (b) The DATAfile Setters contains data collected from fifty owners of Irish Setters on the cost of the first year of owning and caring for their dogs. Use this data set to compute the sample mean. Using this sample, what is the $$95\%$$ confidence interval for the mean cost in dollars of the first year of owning and caring for an Irish Red and White Setter? (Round your answers to nearest cent.)$_______ to \$________
1. A researcher is interested in finding a 98% confidence interval for the mean number of times per day that college students text. The study included 144 students who averaged 44.7 texts per day. The standard deviation was 16.5 texts. a. To compute the confidence interval use a ? z t distribution. b. With 98% confidence the population mean number of texts per day is between and texts. c. If many groups of 144 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population number of texts per day and about percent will not contain the true population mean number of texts per day. 2. You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you believe the population standard deviation is approximately $$\displaystyle\sigma={40.4}$$ dollars. You would like to be 90% confident that your estimate is within 1.5 dollar(s) of average spending on the birthday parties. How many parents do you have to sample? n = 3. You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately $$\displaystyle\sigma={57.5}$$. You would like to be 95% confident that your estimate is within 0.1 of the true population mean. How large of a sample size is required?
Juan makes a measurement in a chemistry laboratory and records the result in his lab report. The stardard deviation of lab measurements made by students is $$\sigma=10$$ milligrams. Juan repeats the measurement 3 times and records the mean xbar of his 3 measurements.
A researcher is interested in finding a $$90\%$$ confidence interval for the mean number minutes students are concentrating on their professor during a one hour statistics lecture. The study included 117 students who averaged 40.9 minutes concentrating on their professor during the hour lecture. The standard deviation was 11.8 minutes. Round answers to 3 decimal places where possible.
With $$90\%$$ confidence the population mean minutes of concentration is between ____ and ____ minutes.