Studies indicate that drinking water supplied by some old lead-lined city piping systems may contain harmful levels of lead. Based on data presented by Karalekas and colleagues,it appears that the distribution of lead content readings for individual water specimens has mean .033 mg/L and standard deviation .10 mg/L. Explain why it is obvious that the lead content readings are not normally distributed.

Question
Studies indicate that drinking water supplied by some old lead-lined city piping systems may contain harmful levels of lead. Based on data presented by Karalekas and colleagues,it appears that the distribution of lead content readings for individual water specimens has mean \(.033 mg/L\) and standard deviation \(.10 mg/L\). Explain why it is obvious that the lead content readings are not normally distributed.

Answers (1)

2021-03-03
The mean lead content reading in drinking water supplied by the old piping systems in the city involving lead-lined pipes is, \(\mu = 0.033mg/L\) and the standard deviation is \(\sigma = 0.10mg/L\).
In case the reading is zero, it would mean that the particular specimen of drinking water does not have any trace of lead in it. However, a negative reading is meaningless and impossible.
According to the empirical rule, if a random variable follows normal distribution, then about 68% of the observations must lie within the interval \((\mu +- \sigma)\), about 95% of the observations must lie within the interval \((\mu \pm 2\sigma)\), and approximately all the observations must lie within the interval \((\mu \pm 3\sigma)\).
The interval \((\mu \pm \sigma)\) is calculated below:
\((\mu \pm \sigma) = (0.033 — 0.10, 0.033 + 0.10)= (-0.067, 0.133)\).
Note that, the interval \((\mu \pm \sigma)\) for this data set contains negative values, which is not possible according to the given scenario. Evidently, the wider intervals \((\mu \pm 2\sigma)\ and\ (\mu \pm 3\sigma)\) will also contain negative values.
This suggests that the empirical rule does not hold for the given data set.
Hence, the readings on lead content do not have a normal distribution.
0

Relevant Questions

asked 2021-01-17
A new thermostat has been engineered for the frozen food cases in large supermarkets. Both the old and new thermostats hold temperatures at an average of \(25^{\circ}F\). However, it is hoped that the new thermostat might be more dependable in the sense that it will hold temperatures closer to \(25^{\circ}F\). One frozen food case was equipped with the new thermostat, and a random sample of 21 temperature readings gave a sample variance of 5.1. Another similar frozen food case was equipped with the old thermostat, and a random sample of 19 temperature readings gave a sample variance of 12.8. Test the claim that the population variance of the old thermostat temperature readings is larger than that for the new thermostat. Use a \(5\%\) level of significance. How could your test conclusion relate to the question regarding the dependability of the temperature readings? (Let population 1 refer to data from the old thermostat.)
(a) What is the level of significance?
State the null and alternate hypotheses.
\(H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}>?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}\neq?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}?_{2}^{2},H1:?_{1}^{2}=?_{2}^{2}\)
(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)
What are the degrees of freedom?
\(df_{N} = ?\)
\(df_{D} = ?\)
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings. Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.
asked 2021-02-11
Several models have been proposed to explain the diversification of life during geological periods. According to Benton (1997), The diversification of marine families in the past 600 million years (Myr) appears to have followed two or three logistic curves, with equilibrium levels that lasted for up to 200 Myr. In contrast, continental organisms clearly show an exponential pattern of diversification, and although it is not clear whether the empirical diversification patterns are real or are artifacts of a poor fossil record, the latter explanation seems unlikely. In this problem, we will investigate three models fordiversification. They are analogous to models for populationgrowth, however, the quantities involved have a differentinterpretation. We denote by N(t) the diversification function,which counts the number of taxa as a function of time, and by rthe intrinsic rate of diversification.
(a) (Exponential Model) This model is described by \(\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{e}}}{N}\ {\left({8.86}\right)}.\) Solve (8.86) with the initial condition N(0) at time 0, and show that \(\displaystyle{r}_{{{e}}}\) can be estimated from \(\displaystyle{r}_{{{e}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ {\left({8.87}\right)}\)
(b) (Logistic Growth) This model is described by \(\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{l}}}{N}\ {\left({1}\ -\ {\frac{{{N}}}{{{K}}}}\right)}\ {\left({8.88}\right)}\) where K is the equilibrium value. Solve (8.88) with the initial condition N(0) at time 0, and show that \(\displaystyle{r}_{{{l}}}\) can be estimated from \(\displaystyle{r}_{{{l}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ +\ {\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}\ {\left({8.89}\right)}\) for \(\displaystyle{N}{\left({t}\right)}\ {<}\ {K}.\)
(c) Assume that \(\displaystyle{N}{\left({0}\right)}={1}\) and \(\displaystyle{N}{\left({10}\right)}={1000}.\) Estimate \(\displaystyle{r}_{{{e}}}\) and \(\displaystyle{r}_{{{l}}}\) for both \(\displaystyle{K}={1001}\) and \(\displaystyle{K}={10000}.\)
(d) Use your answer in (c) to explain the following quote from Stanley (1979): There must be a general tendency for calculated values of \(\displaystyle{\left[{r}\right]}\) to represent underestimates of exponential rates,because some radiation will have followed distinctly sigmoid paths during the interval evaluated.
(e) Explain why the exponential model is a good approximation to the logistic model when \(\displaystyle\frac{{N}}{{K}}\) is small compared with 1.
asked 2020-11-08
Replacement of paint on highways and streets represents a large investment of funds by state and local governments each year. A new, cheaper brand of paint is tested for durability after one month’s time by reflectometer readings. For the new brand to be acceptable, it must have a mean reflectometer reading greater than 19.6. The sample data, based on 35 randomly selected readings, show \(x =19.8\ and\ s=1.5\). Do the sample data provide sufficient evidence to conclude that the new brand is acceptable? Conduct hypothesis test using \(a=.05\). Use the traditional approach and the p-value approach to hypothesis testing! Show all of the steps of the hypothesis test for each approach.
asked 2020-10-23
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.
asked 2021-02-06
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 $________
asked 2020-11-08
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{matrix} 15 & 14 & 14 & 14 & 13 & 12 & 11 & 11 & 11 & 11 & 10 & 11 & 13 & 16 \\ 9 & 15 & 12 & 9 & 10 & 7 & 14 & 13 & 14 & 8 & 9 & 8 & 11 & 13 \\ 15 & 12 & 9 & 10 & 9 & 9 & 16 & 16 & 12 & 10 & 11 & 11 & 12 & 15 \\ 10 & 10 & 10 & 11 & 9 \end{matrix} If you have a statistical calculator or computer, use it to find the actual sample mean and sample standard deviation.
asked 2021-02-25
Iron is very important for babies' growth. A common belief is that breastfeeding will help the baby to get more iron than formula feeding. To justify the belief, a study followed 2 groups of babies from born to 6 months. With one group babies are breast fed, and the other group are formula fed without iron supplements. Data below shows iron levels of those two groups of babies. \(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}{G}{r}{o}{u}{p}&{S}{a}\mp\le\ {s}{i}{z}{e}&{m}{e}{a}{n}&{S}{\tan{{d}}}{a}{r}{d}\ {d}{e}{v}{i}{a}{t}{i}{o}{n}\backslash{h}{l}\in{e}{B}{r}{e}\ast-{f}{e}{d}&{23}&{13.3}&{1.7}\backslash{h}{l}\in{e}{F}{\quad\text{or}\quad}\mu{l}{a}-{f}{e}{d}&{23}&{12.4}&{1.8}\backslash{h}{l}\in{e}{D}{I}{F}{F}={B}{r}{e}\ast-{F}{\quad\text{or}\quad}\mu{l}{a}&{23}&{0.9}&{1.4}\backslash{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\) (1) There are two groups we need to compare for the study: Breast-Fed and Formula- Fed. Are those two groups dependent or independent? Based on your answer, what inference procedure should we apply for this research? (2) Please perform the inference you decided in (1), and make sure to follow the 5-step procedure for any hypothesis test. (3) Based on your conclusion in (2), what kind of error could you make? Explain the type of error using the context words for this research
asked 2021-02-23
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?
asked 2020-11-12
(a) The company's production equipment produces metal discs weighing 200 g. It should be noted that the weight of the discs corresponds to the normal distribution. To check machine consistency, 20 discs are randomly selected with an average weight of 205 g and a standard deviation of 7 g. What is the 99% confidence interval for the average weight of the selected discs?
(b) A company launched a new model of golf ball. It claimed that the driving distance is at least 300m. A sample of 20 balls yields a sample mean of 295m and sample standard deviation of 8m. It is assumed that the driving distance is normally distributed.
Conduct an appropriate hypothesis testing at the 0.05 level of significance. Is there any evidence that the average travel distance stated by the company is true?
asked 2021-02-25
Give a full and correct answer Why is it important that a sample be random and representative when conducting hypothesis testing? Representative Sample vs. Random Sample: An Overview Economists and researchers seek to reduce sampling bias to near negligible levels when employing statistical analysis. Three basic characteristics in a sample reduce the chances of sampling bias and allow economists to make more confident inferences about a general population from the results obtained from the sample analysis or study: * Such samples must be representative of the chosen population studied. * They must be randomly chosen, meaning that each member of the larger population has an equal chance of being chosen. * They must be large enough so as not to skew the results. The optimal size of the sample group depends on the precise degree of confidence required for making an inference. Representative sampling and random sampling are two techniques used to help ensure data is free of bias. These sampling techniques are not mutually exclusive and, in fact, they are often used in tandem to reduce the degree of sampling error in an analysis and allow for greater confidence in making statistical inferences from the sample in regard to the larger group. Representative Sample A representative sample is a group or set chosen from a larger statistical population or group of factors or instances that adequately replicates the larger group according to whatever characteristic or quality is under study. A representative sample parallels key variables and characteristics of the large society under examination. Some examples include sex, age, education level, socioeconomic status (SES), or marital status. A larger sample size reduced sampling error and increases the likelihood that the sample accurately reflects the target population. Random Sample A random sample is a group or set chosen from a larger population or group of factors of instances in a random manner that allows for each member of the larger group to have an equal chance of being chosen. A random sample is meant to be an unbiased representation of the larger population. It is considered a fair way to select a sample from a larger population since every member of the population has an equal chance of getting selected. Special Considerations: People collecting samples need to ensure that bias is minimized. Representative sampling is one of the key methods of achieving this because such samples replicate as closely as possible elements of the larger population under study. This alone, however, is not enough to make the sampling bias negligible. Combining the random sampling technique with the representative sampling method reduces bias further because no specific member of the representative population has a greater chance of selection into the sample than any other. Summarize this article in 250 words.
...