American automobiles produced in 2012 and classified as “large” had a mean fuel economy of 19.6 miles per gallon with a standard deviation of 3.36 miles per gallon. A particular model on this list was rated at 23 miles per gallon, giving it a z-score of about 1.01. Which statement is true based on this information? A) Because the standard deviation is small compared to the mean, a Normal model is appropriate and we can say that about 84.4% of “large” automobiles have a fuel economy of 23 miles per gallon or less. B) Because a z-score was calculated, it is appropriate to use a Normal model to say that about 84.4% of “large” automobiles have a fuel economy of 23 miles per gallon or less. C) Because 23 miles per gallon is greater than the mean of 19.6 miles per gallon, the distribution is ske

Question
Modeling
asked 2021-01-17
American automobiles produced in 2012 and classified as “large” had a mean fuel economy of 19.6 miles per gallon with a standard deviation of 3.36 miles per gallon. A particular model on this list was rated at 23 miles per gallon, giving it a z-score of about 1.01. Which statement is true based on this information? A) Because the standard deviation is small compared to the mean, a Normal model is appropriate and we can say that about 84.4% of “large” automobiles have a fuel economy of 23 miles per gallon or less. B) Because a z-score was calculated, it is appropriate to use a Normal model to say that about 84.4% of “large” automobiles have a fuel economy of 23 miles per gallon or less. C) Because 23 miles per gallon is greater than the mean of 19.6 miles per gallon, the distribution is skewed to the right. This means the z-score cannot be used to calculate a proportion.D) Because no information was given about the shape of the distribution, it is not appropriate to use the z-score to calculate the proportion of automobiles with a fuel economy of 23 miles per gallon or less. E) Because no information was given about the shape of the distribution, it is not appropriate to calculate a z-score, so the z-score has no meaning in this situation.

Answers (1)

2021-01-18
Step 1 Normal distribution is used in various real life applications. It is a two-parameter distribution which denotes its mean and variance correspondingly. One of the properties is that the linear combination of number of random variables also has normal distribution which helps a lot in modeling. Step 2 The statement A) is true based on this information which is "Because the standard deviation is small compared to the mean, a Normal model is appropriate and we can say that about 84.4% of “large” automobiles have a fuel economy of 23 miles per gallon or less."
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Relevant Questions

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 2021-01-27
\(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}&{H}{o}{u}{s}{e}{w}{\quad\text{or}\quad}{k}{H}{o}{u}{r}{s}\backslash{h}{l}\in{e}{G}{e}{n}{d}{e}{r}&{S}{a}\mp\le\ {S}{i}{z}{e}&{M}{e}{a}{n}&{S}{\tan{{d}}}{a}{r}{d}\ {D}{e}{v}{i}{a}{t}{i}{o}{n}\backslash{h}{l}\in{e}{W}{o}{m}{e}{n}&{473473}&{33.133}{.1}&{14.214}{.2}\backslash{h}{l}\in{e}{M}{e}{n}&{488488}&{18.618}{.6}&{15.715}{.7}\backslash{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\) 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.
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 2021-01-28
Indicate true or false for the following statements. If false, specify what change will make the statement true.
a) In the two-sample t test, the number of degrees of freedom for the test statistic increases as sample sizes increase.
b) When the means of two independent samples are used to to compare two population means, we are dealing with dependent (paired) samples.
c) The \(\displaystyle{x}^{{{2}}}\) distribution is used for making inferences about two population variances.
d) The standard normal (z) score may be used for inferences concerning population proportions.
e) The F distribution is symmetric and has a mean of 0.
f) The pooled variance estimate is used when comparing means of two populations using independent samples.
g) It is not necessary to have equal sample sizes for the paired t test.
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 2020-10-23
The table below shows the number of people for three different race groups who were shot by police that were either armed or unarmed. These values are very close to the exact numbers. They have been changed slightly for each student to get a unique problem.
Suspect was Armed:
Black - 543
White - 1176
Hispanic - 378
Total - 2097
Suspect was unarmed:
Black - 60
White - 67
Hispanic - 38
Total - 165
Total:
Black - 603
White - 1243
Hispanic - 416
Total - 2262
Give your answer as a decimal to at least three decimal places.
a) What percent are Black?
b) What percent are Unarmed?
c) In order for two variables to be Independent of each other, the P \((A and B) = P(A) \cdot P(B) P(A and B) = P(A) \cdot P(B).\)
This just means that the percentage of times that both things happen equals the individual percentages multiplied together (Only if they are Independent of each other).
Therefore, if a person's race is independent of whether they were killed being unarmed then the percentage of black people that are killed while being unarmed should equal the percentage of blacks times the percentage of Unarmed. Let's check this. Multiply your answer to part a (percentage of blacks) by your answer to part b (percentage of unarmed).
Remember, the previous answer is only correct if the variables are Independent.
d) Now let's get the real percent that are Black and Unarmed by using the table?
If answer c is "significantly different" than answer d, then that means that there could be a different percentage of unarmed people being shot based on race. We will check this out later in the course.
Let's compare the percentage of unarmed shot for each race.
e) What percent are White and Unarmed?
f) What percent are Hispanic and Unarmed?
If you compare answers d, e and f it shows the highest percentage of unarmed people being shot is most likely white.
Why is that?
This is because there are more white people in the United States than any other race and therefore there are likely to be more white people in the table. Since there are more white people in the table, there most likely would be more white and unarmed people shot by police than any other race. This pulls the percentage of white and unarmed up. In addition, there most likely would be more white and armed shot by police. All the percentages for white people would be higher, because there are more white people. For example, the table contains very few Hispanic people, and the percentage of people in the table that were Hispanic and unarmed is the lowest percentage.
Think of it this way. If you went to a college that was 90% female and 10% male, then females would most likely have the highest percentage of A grades. They would also most likely have the highest percentage of B, C, D and F grades
The correct way to compare is "conditional probability". Conditional probability is getting the probability of something happening, given we are dealing with just the people in a particular group.
g) What percent of blacks shot and killed by police were unarmed?
h) What percent of whites shot and killed by police were unarmed?
i) What percent of Hispanics shot and killed by police were unarmed?
You can see by the answers to part g and h, that the percentage of blacks that were unarmed and killed by police is approximately twice that of whites that were unarmed and killed by police.
j) Why do you believe this is happening?
Do a search on the internet for reasons why blacks are more likely to be killed by police. Read a few articles on the topic. Write your response using the articles as references. Give the websites used in your response. Your answer should be several sentences long with at least one website listed. This part of this problem will be graded after the due date.
asked 2020-12-07
Would you rather spend more federal taxes on art? Of a random sample of \(n_{1} = 86\) politically conservative voters, \(r_{1} = 18\) responded yes. Another random sample of \(n_{2} = 85\) politically moderate voters showed that \(r_{2} = 21\) responded yes. Does this information indicate that the population proportion of conservative voters inclined to spend more federal tax money on funding the arts is less than the proportion of moderate voters so inclined? Use \(\alpha = 0.05.\) (a) State the null and alternate hypotheses. \(H_0:p_{1} = p_{2}, H_{1}:p_{1} > p_2\)
\(H_0:p_{1} = p_{2}, H_{1}:p_{1} < p_2\)
\(H_0:p_{1} = p_{2}, H_{1}:p_{1} \neq p_2\)
\(H_{0}:p_{1} < p_{2}, H_{1}:p_{1} = p_{2}\) (b) What sampling distribution will you use? What assumptions are you making? The Student's t. The number of trials is sufficiently large. The standard normal. The number of trials is sufficiently large.The standard normal. We assume the population distributions are approximately normal. The Student's t. We assume the population distributions are approximately normal. (c)What is the value of the sample test statistic? (Test the difference \(p_{1} - p_{2}\). Do not use rounded values. Round your final answer to two decimal places.) (d) Find (or estimate) the P-value. (Round your answer to four decimal places.) (e) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level alpha? At the \(\alpha = 0.05\) level, we reject the null hypothesis and conclude the data are statistically significant. At the \(\alpha = 0.05\) level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the \(\alpha = 0.05\) level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the \(\alpha = 0.05\) level, we reject the null hypothesis and conclude the data are not statistically significant. (f) Interpret your conclusion in the context of the application. Reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters.
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 2021-01-31
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.
asked 2021-03-07
The article “Modeling Arterial Signal Optimization with Enhanced Cell Transmission Formulations presents a new method for timing traffic signals in heavily traveled intersections. The effectiveness of the new method was evaluated in a simulation study. In 50 simulations, the mean improvement in traffic flow in a particular intersection was 654.1 vehicles per hour, with a standard deviation of 311.7 vehicles per hour.
a) Find a \(\displaystyle{95}\%\) confidence interval for the improvement in traffic flow due to the new system.
b) Find a \(\displaystyle{98}\%\) confidence interval for the improvement in traffic flow due to the new system.
c) A traffic engineer states that the mean improvement is between 581.6 and 726.6 vehicles per hour. With what level of confidence can this statement be made?
d) Approximately what sample size is needed so that a \(\displaystyle{95}\%\)
confidence interval will specify the mean to within \(\displaystyle\pm\ {50}\) vehicles per hour?
e) Approximately what sample size is needed so that a \(\displaystyle{98}\%\) confidence
interval will specify the mean to within \(\displaystyle\pm\ {50}\) vehicles per hour?
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