Use the given information to find the minimum sample size required to estimate an unknown population mean μ. How many adults must be randomly selected to estimate the mean FICO (credit rating) score of working adults in a country? We want 90% confidence that the sample mean is within 3 points of the population mean, and the population standard deviation is 65. The minimum sample size required is adults. (Round up to the nearest whole number.)
A study of the ages of motorcyclists killed in crashes involves the random selection of 156 drivers with a mean of 36.17 years. Assuming that σ = 8.2 years, construct and interpret a 99% confidence interval estimate of the mean age of all motorcyclists killed in crashes. Click here to view a t distribution table. 1 Click here to view page 1 of the standard normal distribution table. 2 Click here to view page 2 of the standard normal distribution table. 3 What is the 99% confidence interval for the population mean µ? < µ < (Round to two decimal places as needed.) Notice that the confidence interval limits do not include ages below 20 years. What does this mean? A. The mean age of the sample will most likely not be less than 20 years old. B. Motorcyclists under the age of 20 never die in crashes. C. The mean age of the population will most likely not be less than 20 years old. D. The mean age of the population will never be less than 20 years old.
A random sample of the birth weights of 186 babies has a mean of 3103 g and a standard deviation of 676 g. Construct a 98% confidence interval estimate of the mean birth weight for all such babies. 4 Click the icon to view a table of critical t-values. What is the confidence interval estimate of the mean birth weight for all such babies? g < µ < g (Round to one decimal place as needed.)
Listed below are measured amounts of lead (in micrograms per cubic meter, or µg/m3) in the air. The EPA has established an air quality standard for lead of 1.5 µg/m3. The measurements shown below were recorded at a building on different days. Use the given values to construct a 95% confidence interval estimate of the mean amount of lead in the air. Is there anything about this data set suggesting that the confidence interval might not be very good? Click here to view a t distribution table. 5 Click here to view page 1 of the standard normal distribution table 6 Click here to view page 2 of the standard normal distribution table. 7 What is the confidence interval for the population mean µ? µg/m3 < µ < µg/m3 (Round to three decimal places as needed.) Is there anything about this data set suggesting that the confidence interval might not be very good? A. Yes, because the confidence interval does not contain 1.5µg/m3. B. Yes, the value of 5.40 appears to be an outlier. C. Yes, because the sample mean is greater than 1.5µg/m3. D. No, the data appears to be normally distributed.
Is the confidence interval 0.0455g < σ < 0.0602g equivalent to the expression (0.0455g, 0.0602g)? Is it equivalent to the expression 0.05285g ± 0.00735g? Choose the correct answer below. A. Yes; No B. No; No C. Yes; Yes D. No; Yes
Find the critical values χ2L and χ2R that correspond to the given confidence level and sample size. 90%; n = 29 χ2L = (Round to three decimal places as needed.) χ2R = (Round to three decimal places as needed.)
A simple random sample from a population with a normal distribution of 98 body temperatures has x¯ = 98.20◦F and s = 0.63◦F. Construct a 90% confidence interval estimate of the standard deviation of body temperature of all healthy humans. Is it safe to conclude that the population standard deviation is less than 1.20◦F ? 11 Click the icon to view the table of Chi-Square critical values. ◦F < σ < ◦F (Round to two decimal places as needed.) Is it safe to conclude that the population standard deviation is less than 1.20◦F ? A. This conclusion is not safe because 1.20◦F is in the confidence interval. B. This conclusion is safe because 1.20◦F is outside the confidence interval. C. This conclusion is not safe because 1.20◦F is outside the confidence interval. D. This conclusion is safe because 1.20◦F is in the confidence interval.
Twelve different video games showing substance use were observed and the duration of times of game play (in seconds) are listed below. The design of the study justifies the assumption that the sample can be treated as a simple random sample. Use the sample data to construct a 90% confidence interval estimate of σ, the standard deviation of the duration times of game play. Assume that this sample was obtained from a population with a normal distribution. 4,684 3,879 4,671 4,076 4,832 4,197 4,593 4,974 4,208 4,914 4,800 4,544 13 Click the icon to view the table of Chi-Square critical values. The confidence interval estimate is sec < σ < sec. (Round to one decimal place as needed.)
A physician wants to develop criteria for determining whether a patient’s pulse rate is atypical, and she wants to determine whether there are significant differences between males and females. Use the sample pulse rates below. Male 96 64 72 84 64 68 72 84 72 76 Female 60 96 84 88 68 96 84 64 72 124 Click here to view a t distribution table. 8 Click here to view page 1 of the standard normal distribution table. 9 Click here to view page 2 of the standard normal distribution table. 10 a. Construct a 95% confidence interval estimate of the mean pulse rate for males. < µ < (Round to one decimal place as needed.) b. Construct a 95% confidence interval estimate of the mean pulse rate for females. < µ < (Round to one decimal place as needed.) c. Compare the preceding results. Can we conclude that the population means for males and females are different? A. Yes, because the two confidence intervals do not overlap, we can conclude that the two population means are different. B. Yes, the population mean for females appears to be greater than the population mean for males. C. No, because the two confidence intervals overlap, we cannot conclude that the two population means are different. D. Yes, the population mean for males appears to be greater than the population mean for females.
The body mass index (BMI) for a sample of men and a sample of women are given below. Assume the samples are simple random samples obtained from populations with normal distributions. Men 23.2 26.3 28.2 32.2 22.9 31.5 26.2 30.2 32.3 25.4 Women 19.4 17.4 19.9 28.4 18.5 18.7 20.8 30.3 18.3 17.6 12 Click the icon to view the table of Chi-Square critical values. a. Construct a 95% confidence interval estimate of the standard deviation of BMls for men. < σmen < (Round to two decimal places as needed.) b. Construct a 95% confidence interval estimate of the standard deviation of BMls for women. < σwomen < (Round to two decimal places as needed.) c. Compare and interpret the results. A. Since the intervals do not overlap, the populations appear to have different amounts of variation. B. Since the intervals do not overlap, the populations appear to have amounts of variation that are not substantially different. C. Since the intervals overlap, the populations appear to have different amounts of variation. D. Since the intervals overlap, the populations appear to have amounts of variation that are not substantially different.
In the study of population dynamics one of the most famous models for a growing but bounded population is the logistic equation dP/dt = P(a - bP), where a and b are positive constants. Although we will come back to this equation and solve it by an alternative method in Section 3.2, solve the DE this first time using the fact that it is a Bernoulli equation.
Examine the given statement, then express the null hypothesis H0 and the alternative hypothesis H1 in symbolic form. Less than 23% of Internet users pay bills online. Which of the following is the hypothesis test to be conducted? A. H0 : p < 0.23 H1 : p = 0.23 c. H0 : p > 0.23 H1 : p = 0.23 E. H0 : p 6= 0.23 H1 : p = 0.23 B. H0 : p = 0.23 H1 : p < 0.23 D. H0 : p = 0.23 H1 : p 6= 0.23 F. H0 : p = 0.23 H1 : p > 0.23
A mathematical model for the rate at which a drug disseminates into the bloodstream is given by dx dt = r - kx, where r and k are positive constants. The function x(t) describes the concentration of the drug in the bloodstream at time t. (a) Since the DE is autonomous, use the phase portrait concept of Section 2.1 to find the limiting value of x(t) as t → ∞. lim t→∞ x(t) = (b) Solve the DE subject to x(0) = 0. x(t) = Sketch the graph of x(t) and verify your prediction in part (a). At what time is the concentration one-half this limiting value?
The graph of f is given. Use the graph to compute the quantities asked for. (If an answer does not exist, enter DNE.) (a) limx→-16 f(x) (b) limx→-8+ f(x) (c) limx→-8- f(x) (d) limx→-8 f(x) (e) f(-8) (f) limx→+∞ f(x)
The graph of f is given. Use the graph to compute the quantities asked for. (If an answer does not exist, enter DNE.) (a) limx→-18 f(x) (b) limx→-9+ f(x) (c) limx→-9- f(x) (a) limx→-18 f(x) (b) limx→-9+ f(x) (c) limx→-9- f(x) (d) limx→-9 f(x) (e) f(-9) (f) limx→+∞ f(x)
Use the given information to find the P-value. The test statistic in a right tailed test is z = 1.88. P-value = (Round to four decimal places as needed.)
The graph of f is given. Use the graph to compute the quantities asked for. (If an answer does not exist, enter DNE.) (a) limx→6 f(x) (b) limx→0+ f(x) (a) limx→6 f(x) (b) limx→0+ f(x) (c) limx→0- f(x) (d) limx→0 f(x) (e) f(0) (f) limx→-∞ f(x)
Find the value of the test statistic z using z = pˆ-p / √pq/n . The claim is that the proportion of adults who smoked a cigarette in the past week is less than 0.25, and the sample statistics include n = 1217 subjects with 292 saying that they smoked a cigarette in the past week.
The graph of f is given. Use the graph to compute the quantities asked for. (If an answer does not exist, enter DNE.) (a) limx→0 f(x) (b) limx→4 f(x) (c) limx→-∞ f(x) (d) limx→+∞ f(x)
The graph of f is given. Use the graph to compute the quantities asked for. (If an answer does not exist, enter DNE.) (a) limx→8 f(x) (b) limx→-8 f(x)
The graph of f is given. Use the graph to compute the quantities asked for, if a = 2. (If an answer does not exist, enter DNE.) (a) limx→4 f(x) (b) limx→2+ f(x) (c) limx→2- f(x) (d) limx→2 f(x) (e) f(2)
Use the given information to find the P-value. With H1 : p ≠ 1/5, the test statistic is z = -1.09. P-value = (Round to four decimal places as needed.)
The graph of f is given. Use the graph to compute the quantities asked for, if a = 1. HINT [See Examples 4 - 5.] (If you need to use ∞ or -∞, enter INFINITY or - INFINITY, respectively. If an answer does not exist, enter DNE.) (a) limx→-1 f(x) (b) limx→0+ f(x) (c) limx→0- f(x) (d) limx→0 f(x) (e) f(0) (f) limx→+∞ f(x)
The graph of f is given. Use the graph to compute the quantities asked for, if a = 17. [See Example 4.] (If an answer does not exist, enter DNE.) (a) limx→-17 f(x) (b) limx→0+ f(x) (c) limx→0- f(x) (d) limx→0 f(x) (e) f(0) (f) limx→-∞ f(x)
The graph of f is given. Use the graph to compute the quantities asked for. (If an answer does not exist, enter DNE.) (a) limx→-1 f(x) (b) limx→0+ f(x) (c) limx→0- f(x) (d) limx→0 f(x) (e) f(0) (f) f(-1)
State the final conclusion in simple nontechnical terms. Original claim: The proportion of male golfers is less than 0.6. Initial conclusion: Fail to reject the null hypothesis. Which of the following is the correct conclusion? A. There is sufficient evidence to support the claim that the proportion of male golfers is less than 0.6. B. There is not sufficient evidence to support the claim that the proportion of male golfers is less than 0.6.
The graph of f is given. Use the graph to compute the quantities asked for. (If an answer does not exist, enter DNE.) (a) limx→0- f(x) (b) limx→3+ f(x) (c) limx→0 f(x) (d) limx→3 f(x) (e) r(0) (f) f(3)
The graph shows a rough representation of the (aggregate) market depth of the stocks comprising the S&P 500 on the day of the U.S. stock market crash at 2:45 pm on May 6, 2010 (the "Flash Crash"; t is the time of the day in hours, and m(t) is the market depth in millions of shares). (a) Compute the following. (If a limit does not exist, enter DNE, and say why.) lim t→14.75- m(t) = The limit does exist. The limit does not exist because it diverges. The limit does not exist because the limit from the left does not match the limit from the right. lim t→14.75+ m(t) = The limit does exist. The limit does not exist because it diverges. The limit does not exist because the limit from the left does not match the limit from the right. lim t→14.75 m(t) = The limit does exist. The limit does not exist because it diverges. The limit does not exist because the limit from the left does not match the limit from the right. m(14.75) = (b) What do the answers to part (a) tell you about the market depth? Just prior to 2:45 pm, the $&P market depth was approaching million shares, but then fell suddenly to million shares at 2:45 exactly, then immediately following 2:45 it began to recover at values close to million shares.
Before the September 11, 2001 attacks, an airline’s stock was trading at around $36 per share. Immediately after the attacks, the share price dropped by $14.40. Let U(t) be this cost at time t, and take t = 11 to represent September 11, 2001 . What does the given information tell you about lim t→11 U(t) ? lim t→11 U(t) = $36.00 lim t→11 U(t) = $14.40 lim t→11 U(t) = $25.20 lim t→11 U(t) does not exist
In 1997, a survey of 980 households showed that 174 of them use e-mail. Use those sample results to test the claim that more than 15% of households use email. Use a 0.05 significance level. Use this information to answer the following questions. a. Which of the following is the hypothesis test to be conducted? A. H0 : p = 0.15 H1 : p 6= 0.15 c. H0 : p > 0.15 H1 : p = 0.15 E. H0 : p < 0.15 H1 : p = 0.15 B. H0 : p = 0.15 H1 : p < 0.15 D. H0 : p = 0.15 H1 : p > 0.15 F. H0 : p 6= 0.15 H1 : p = 0.15 b. What is the test statistic? z = (Round to two decimal places as needed.) c. What is the P-value? P-value = (Round to three decimal places as needed.) d. What is the conclusion? There is not sufficient evidence to support the claim that more than 15% of households use e-mail. There is sufficient evidence to support the claim that more than 15% of households use e-mail. e. Is the conclusion valid today? Why or why not? A. Yes, the conclusion is valid today because the requirements to perform the test are satisfied. B. No, the conclusion is not valid today because the population characteristics of the use of e-mail are changing rapidly. C. You can make no decisions about the validity of the conclusion today.
Annual U.S. imports from a certain country in the years 1996 through 2003 can be approximated by I(t) = t^2 + 3.4t + 47 (1 ≤ t ≤ 9) billion dollars, where t represents time in years since 1995. Annual U.S. exports to this country in the same years can be approximated by E(t) = 0.4t^2 - 1.5t + 12 (0 ≤ t ≤ 10) billion dollars. Assuming the trends shown in the above models continue indefinitely, numerically estimate the following. (If an answer does not exist, enter DNE.) lim t→+∞ E(t) and lim t→+∞ E(t) I(t) lim t→+∞ E(t) = lim t→+∞ E(t) I(t) = 0 Interpret your answers. In the long term, U.S. imports from the country will rise without bound and be 0.4 times U.S. exports to the country. In the long term, U.S. imports from the country will fall without bound and be 0.4 times U.S. exports to the country. In the long term, U.S. exports to the country will rise without bound and be 0.4 times U.S. imports from the country. In the long term, U.S. exports to the country will fall without bound and be 0.4 times U.S. imports from the country. Comment on the results. In the real world, imports and exports cannot rise without bound. Thus, the given models should not be extrapolated far into the future. In the real world, imports and exports can fall without bound. Thus, the given models can be extrapolated far into the future. In the real world, imports and exports do not change, they always stay fixed. Thus, the given models should not be extrapolated far into the future. In the real world, imports and exports can rise without bound. Thus, the given models can be extrapolated far into the future.
In a study of 420,094 cell phone users, 183 subjects developed brain cancer. Test the claim that cell phone users develop brain cancer at a rate that is different from the rate of 0.0340% for people who do not use cell phones. Because this issue has such great importance, use a 0.005 significance level. Use this information to answer the following questions. a. Which of the following is the hypothesis test to be conducted? A. H0 : p = 0.00034 H1 : p > 0.00034 C. H0 : p < 0.00034 H1 : p = 0.00034 E. H0 : p > 0.00034 H1 : p = 0.00034 B. H0 : p = 0.00034 H1 : p < 0.00034 D. H0 : p = 0.00034 H1 : p 6= 0.00034 F. H0 : p 6= 0.00034 H1 : p = 0.00034 b. What is the test statistic? z = (Round to two decimal places as needed.) c. What is the P-value? P-value = (Round to four decimal places as needed.) d. What is the conclusion? A. There is not sufficient evidence to support the claim that cell phone users develop brain cancer at a rate that is different from the rate of 0.0340% for people who do not use cell phones. B. There is sufficient evidence to support the claim that cell phone users develop brain cancer at a rate that is different from the rate of 0.0340% for people who do not use cell phones. e. Should cell phone users be concerned about brain cancer? A. No, the study does not suggest that a cell phone user is at a higher risk for brain cancer than a non cell phone user. B. Yes, the study suggests that a cell phone user is at a higher risk for brain cancer than a non cell phone user.
When testing gas pumps for accuracy, fuel-quality enforcement specialists tested pumps and found that 1311 of them were not pumping accurately (within 3.3 oz when 5gal is pumped), and 5653 pumps were accurate. Use a 0.01 significance level to test the claim of an industry representative that less than 20% of the pumps are inaccurate. Use the P-value method and use the normal distribution as an approximation to the binomial distribution. Identify the null hypothesis and alternative hypothesis. A. H0 : p > 0.2 H1 : p = 0.2 c. H0 : p = 0.2 H1 : p > 0.2 E. H0 : p < 0.2 H1 : p = 0.2 B. H0 : p = 0.2 H1 : p 6= 0.2 D. H0 : p = 0.2 H1 : p < 0.2 H0 : p 6= 0.2 H1 : p = 0.2 F. H0 : p 6= 0.2 H1 : p = 0.2 The test statistic is z = (Round to four decimal places as needed.) The P-value is (Round to four decimal places as needed.) Because the P-value is (1) the significance level, (2) the null hypothesis. There is (3) evidence support the claim that less than 20% of the pumps are inaccurate. (1) greater than (2) fail to reject (3) sufficient less than reject insufficient
In a survey of 145 senior executives, 51% said that the most common job interview mistake is to have little or no knowledge of the company. Use a 0.01 significance level to test the claim that in the population of all senior executives, 40% say that the most common job interview mistake is to have little or no knowledge of the company. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution. Identify the null and alternative hypotheses. Choose the correct answer below. A. H0 : p = 0.4 H1 : p 6= 0.4 C. H0 : p = 0.6 H1 : p 6= 0.6 E. H0 : p = 0.6 H1 : p < 0.6 B. H0 : p = 0.6 H1 : p > 0.6 D. H0 : p = 0.4 H1 : p < 0.4 F. H0 : p = 0.4 H1 : p > 0.4 The test statistic is z = (Round to two decimal places as needed.) The P-value is. (Round to four decimal places as needed.) Identify the conclusion about the null hypothesis and the final conclusion that addresses the original claim. (1) H0. There (2) sufficient evidence to warrant rejection of the claim that among senior executives, 40% say that the most common job interview mistake is to have little or no knowledge of the company. (1) Fail to reject (2) is Reject is not
When 3011 adults were surveyed in a poll, 76% said that they use the Internet. Is it okay for a newspaper reporter to write that "3 / 4 of all adults use the Internet"? Why or why not? Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution. Identify the null and alternative hypotheses. Choose the correct answer below. A. H0 : p = 0.75 H1 : p < 0.75 C. H0 : p = 0.25 H1 : p < 0.25 E. H0 : p = 0.25 H1 : p 6= 0.25 B. H0 : p = 0.75 H1 : p 6= 0.75 D. H0 : p = 0.25 H1 : p > 0.25 F. H0 : p = 0.75 H1 : p > 0.75 The test statistic is z = . (Round to two decimal places as needed.) The P-value is (Round to four decimal places as needed.) Identify the conclusion about the null hypothesis and the final conclusion that addresses the original claim. (Assume a 0.05 significance level.) (1) H0. There (2) sufficient evidence to warrant rejection of the claim that 3/4 of all adults use the Internet. The reporter should write that 3/4 of all adults use the Internet. Reject (2) is not Fail to reject is