Statistics Chapter 8-10 Exam
Formulas:
Keyword: population proportion - p̂ = x/n, p̂ ± za/2√p̂(1-p̂)/n population mean - x ± ta/2 ∙ s/√n standard deviation - √(n-1)/x^2 a/2 ∙ s, √(n-1)/x^2 1-a/2 ∙ s σx = σ/√n, μx = μ, σp = √p(1-p)/n The test statistic for the hypothesis test on a population proportion z0=(p̂-p0)/[√p0(1-p0)/n] test statistic for hypothesis test on a population mean t0=(x-μ0)/(s/√n) sample standard deviations χ20 = [(n-1)s^2]/(σo^2)
Katrina wants to estimate the proportion of adults who read at least 10 books last year. To do so, she obtains a simple random sample of 100 adults and constructs a 95% confidence interval. Matthew also wants to estimate the proportion of adults who read at least 10 books last year. He obtains a simple random sample of 400 adults and constructs a 99% confidence interval. Assuming both Katrina and Matthew obtained the same point estimate, whose estimate will have the smaller margin of error? Justify your answer.
Matthew's estimate will have the smaller margin of error because the larger sample size more than compensates for the higher level of confidence.
For the following, indicate whether a confidence interval for a proportion or mean should be constructed to estimate the variable of interest. Justify your response. Does chewing your food for a longer period of time reduce one's caloric intake of food at dinner? A researcher requires a sample of 75 healthy males to chew their food twice as long as they normally do. The researcher then records the calorie consumption at dinner.
The confidence interval for a mean should be constructed because the variable of interest is an individual's reduction in caloric intake, which is a quantitative variable.
A group conducted a poll of 2067 likely voters just prior to an election. The results of the survey indicated that candidate A would receive 49% of the popular vote and candidate B would receive 45% of the popular vote. The margin of error was reported to be 5%. The group reported that the race was too close to call. Use the concept of a confidence interval to explain what this means.
The margin of error suggests candidate A may receive between 44% and 54% of the popular vote and candidate B may receive between 40% and 50% of the popular vote. Because the poll estimates overlap when accounting for margin of error, the poll cannot predict the winner.
Why is the sampling distribution of x approximately normal?
The sampling distribution of x is approximately normal because the sample size is large enough.
The standard deviation of the sampling distribution of x, denoted σx, is called the _____ _____ of the _____.
The standard deviation of the sampling distribution of x, denoted σx, is called the standard error of the mean.
State the conclusion based on the results of the test. The mean of the pressure required to open a certain valve is known to be μ=7.4 psi. Due to changes in the manufacturing process, the quality-control manager feels that the average pressure has changed. The null hypothesis was rejected.
There is sufficient evidence that the mean of the pressure required to open a certain valve has changed.
Is the statement below true or false? The distribution of the sample mean, x, will be normally distributed if the sample is obtained from a population that is normally distributed, regardless of the sample size.
True
Find the critical values χ21−α/2 and χ2α/2 for a 95% confidence level and a sample size of n=25.
Use chi-square calculator. χ2 1−α/2 = 12.401 χ2 α/2 = 39.364
Explain what "99% confidence" means in a 99% confidence interval.
What does "99% confidence" mean in a 99% confidence interval? A. The value of the parameter lies within 99% of a standard deviation of the estimate. B. The confidence interval includes 99% of all possible values for the parameter. C. If 100 different confidence intervals are constructed, each based on a different sample of size n from the same population, then we expect 99 of the intervals to include the parameter and 1 to not include the parameter. Your answer is correct. D. The probability that the value of the parameter lies between the lower and upper bounds of the interval is 99%. The probability that it does not is 1%.
The null and alternative hypotheses are given. Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed. What parameter is being tested? H0:p=0.5 H1:p≠0.5
What type of test is being conducted in this problem? Two-tailed test What parameter is being tested? Population proportion
Without doing any computation, put the following in order from least to greatest, assuming the population is normally distributed with μ=100 and σ=15. a) P(90≤x≤110) for a random sample of size n=20 (b) P(90≤x≤110) for a random sample of size n=50 (c) P(90≤x≤110)
c<a<b
Increasing the sample size:
decreases the probability because σx decreases as n increases.
Exit polling is a popular technique used to determine the outcome of an election prior to results being tallied. Suppose a referendum to increase funding for education is on the ballot in a large town (voting population over 100,000). An exit poll of 400 voters finds that 204 voted for the referendum. How likely are the results of your sample if the population proportion of voters in the town in favor of the referendum is 0.49? Based on your result, comment on the dangers of using exit polling to call elections. How likely are the results of your sample if the population proportion of voters in the town in favor of the referendum is 0.49?
find μp and σp use normalcdf with p=0.54 and mean and standard deviation The probability that more than 204 people voted for the referendum is 0.2118. Comment on the dangers of using exit polling to call elections. Choose the correct answer below. The result is not unusual because the probability that p is equal to or more extreme than the sample proportion is greater than 5%. Thus, it is not unusual for a wrong call to be made in an election if exit polling alone is considered.
Clayton Kershaw of the Los Angeles Dodgers is one of the premier pitchers in baseball. His most popular pitch is a four-seam fastball. The accompanying data represent the pitch speed (in miles per hour) for a random sample of 15 of his four-seam fastball pitches. Complete parts (a) through (f).
formula: x ± ta/2 ∙ s/√n -------------------------------------------------------------- (a) Is "pitch speed" a quantitative or qualitative variable? Why is it important to know this when determining the type of confidence interval you may construct? The variable "pitch speed" is a quantitative variable. This is important to know because confidence intervals for a mean are constructed on quantitative data while confidence intervals for a proportion are constructed on qualitative data with two possible outcomes. (b) Draw a normal probability plot. Then determine whether "pitch speed" could come from a population that is normally distributed by using the correlation coefficient of the normal probability plot. (It's the linear one that has 90-100 on the x axis) Using the correlation coefficient of the normal probability plot, is it reasonable to conclude that the population is normally distributed? Since the absolute value of the correlation coefficient between the expected z-scores and the ordered observed data, 0.994, exceeds the critical value 0.514, it is reasonable to conclude that the data come from a population that is normally distributed. (for 0.514 use the table of critical values for n=15) (c) Draw a boxplot to verify the data set has no outliers. (It's the one that is centered) (d) Are the requirements for constructing a confidence interval for the mean pitch speed of Clayton Kershaw's four-seam fastball satisfied? An interval can be constructed because the data are approximately normal and there are no outliers. (e) Construct and interpret a 95% confidence interval for the mean pitch speed of Clayton Kershaw's four-seam fastball. 1-0.95= 0.05/2 = 0.025 invT(0.025, 14) = 2.145 94.8 +- 2.145 ∙ 0.692/√15 = 94.52, 95.29 (f) Do you believe that a 95% confidence interval for the mean pitch speed of four-seam fastballs for all major league pitchers' would be narrower or wider? Why? The interval for all pitchers in a league would be wider because the variability between pitchers is most likely greater than the variability between pitches for one pitcher.
In a random sample of 100 audited estate tax returns, it was determined that the mean amount of additional tax owed was $3468 with a standard deviation of $2558. Construct and interpret a 90% confidence interval for the mean additional amount of tax owed for estate tax returns.
formula: x ± ta/2 ∙ s/√n -------------------------------------------------------------- Find and interpret a 90% confidence interval for the mean additional amount of tax owed for estate tax returns. 1-0.90=0.10/2 = 0.05 invT(0.05,99) = 1.660 3468 +- 1660 ∙ 2558/√100 = 3043, 3893 One can be 90% confident that the mean additional tax owed is between $3043 and $3893.
The following data represent the pH of rain for a random sample of 12 rain dates in a particular region. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. The sample standard deviation is s=0.327. Construct and interpret a 95% confidence interval for the standard deviation pH of rainwater in this region. 4.48, 5.23, 5.11, 4.86, 4.77, 4.85, 5.68, 4.81, 5.12, 4.85, 4.82, 4.50
formula: √(n-1)/x^2 a/2 ∙ s, √(n-1)/x^2 1-a/2 ∙ s -------------------------------------------------------------- Use chi-square calculator or given table 21.920, 3.816 √(12-1)/21.920 ∙ 0.327 = 0.232, √(12-1)/3.816 ∙ 0.327 = 0.556 There is 95% confidence that the population standard deviation is between 0.232 and 0.556.
Determine μx and σx from the given parameters of the population and sample size. μ=89, σ=32, n=64
formulas: σx = σ/√n, μx = μ -------------------------------------------------------------- 32/√64 = 4, μ=89 μx = 89, σx = 4
Sleep apnea is a disorder in which you have one or more pauses in breathing or shallow breaths while you sleep. In a cross-sectional study of 320 adults who suffer from sleep apnea, it was found that 224 had gum disease. Note: In the general population, about 17.5% of individuals have gum disease. Complete parts (a) through (c) below.
formulas: p̂ = x/n, p̂ ± za/2√p̂(1-p̂)/n -------------------------------------------------------------- (a) What does it mean for this study to be cross-sectional? The data were obtained at a specific point in time and the study was an observational study. (b) What is the variable of interest in this study? Is it qualitative or quantitative? Explain. Whether a person with sleep apnea has gum disease or not Determine whether the variable in question is qualitative or quantitative. Choose the correct answer below. The variable is qualitative because it classifies the individuals in the study. (c) Estimate the proportion of individuals who suffer from sleep apnea who have gum disease with 99% confidence. Interpret your result. Select the correct choice below and fill in the answer boxes to complete your choice. There is 99% confidence that the population proportion of individuals who suffer from sleep apnea who have gum disease is between 0.634 and 0.766
A simple random sample of size n=400 individuals who are currently employed is asked if they work at home at least once per week. Of the 400 employed individuals surveyed, 30 responded that they did work at home at least once per week. Construct a 99% confidence interval for the population proportion of employed individuals who work at home at least once per week.
formulas: p̂ = x/n, p̂ ± za/2√p̂(1-p̂)/n -------------------------------------------------------------- p=x/n = 0.075, 1 - 0.99 = 0.01 0.01/2 = 0.005 invnorm(-0.005, 0, 1)=2.575 0.075+-2.575√0.075(1-0.075)/400= 0.041, 0.109 The lower bound is 0.041 The upper bound is 0.109
To test H0: μ=107 versus H1: μ≠107 a simple random sample of size n=35 is obtained. Complete parts a through e below.
formulas: t0= (x-μ0)/(s/√n) -------------------------------------------------------------- (a) Does the population have to be normally distributed to test this hypothesis? Why? No, because n≥30 (b) If x=104.0 and s=5.7, compute the test statistic. t0=(104.0-107)/(5.7/√35) t0=−3.11. (c) Draw a t-distribution with the area that represents the P-value shaded. Choose the correct graph below. (its the graph that is shaded evenly on each side) (d) Approximate the P-value. Choose the correct answer below. tcdf(-3.11, 9999, 34)=0.998 1-0.998=0.002 P-value=0.002 0.002<P-value<0.005 Interpret the P-value. Choose the correct answer below. If 1000 random samples of size n=35 are obtained, about 4 samples are expected to result in a mean as extreme or more extreme than the one observed if μ=107. (e) If the researcher decides to test this hypothesis at the α=0.01 level of significance, will the researcher reject the null hypothesis? Yes
A college entrance exam company determined that a score of 20 on the mathematics portion of the exam suggests that a student is ready for college-level mathematics. To achieve this goal, the company recommends that students take a core curriculum of math courses in high school. Suppose a random sample of 200 students who completed this core set of courses results in a mean math score of 20.4 on the college entrance exam with a standard deviation of 3.6. Do these results suggest that students who complete the core curriculum are ready for college-level mathematics? That is, are they scoring above 20 on the mathematics portion of the exam? Complete parts a) through d) below.
formulas: t0= (x-μ0)/(s/√n) -------------------------------------------------------------- (a) State the appropriate null and alternative hypotheses. Fill in the correct answers below. The appropriate null and alternative hypotheses are H0: μ=20 versus H1: μ>20. b) Verify that the requirements to perform the test using the t-distribution are satisfied. Select all that apply. A. The sample size is larger than 30. Your answer is correct. B. A boxplot of the sample data shows no outliers. C. The students' test scores were independent of one another. Your answer is correct. D. The students were randomly sampled. Your answer is correct. E. The sample data come from a population that is approximately normal. F. None of the requirements are satisfied. (c) Use the P-value approach at the α=0.10 level of significance to test the hypotheses in part (a). Identify the test statistic. t0= (x-μ0)/(s/√n) t0= 20.4-20)/(3.6/√200) t0=1.57 Approximate the P-value. tcdf(1.57, 9999, 199) P-value = 0.059 The P-value is in the range 0.05<P-value<0.10. (d) Write a conclusion based on the results. Reject the null hypothesis and claim that there is sufficient evidence to conclude that the population mean is greater than 20.
A credit score is used by credit agencies (such as mortgage companies and banks) to assess the credit worthiness of individuals. Values range from 300 to 850, with a credit score over 700 considered to be a quality credit risk. According to a survey, the mean credit score is 710.3. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 48 high-income individuals and found the sample mean credit score to be 722.6 with a standard deviation of 81.1. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the α=0.05 level of significance.
formulas: t0= (x-μ0)/(s/√n) -------------------------------------------------------------- State the null and alternative hypotheses. Fill in the correct answers below. H0:μ=710.3 H1:μ>710.3 Identify the t-statistic. t0= (722.6-710.3)/(81.1/√48) t0=1.05 Approximate the P-value. tcdf(1.05, 9999, 47)=0.150 The P-value is in the range P-value>0.10. Make a conclusion regarding the hypothesis. Do not reject the null hypothesis. There is not sufficient evidence to conclude that the mean credit score of high-income individuals is greater than 710.3
Twenty years ago, 54% of parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. A recent survey found that 238 of 700 parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. Do parents feel differently today than they did twenty years ago? Use the α=0.05 level of significance.
formulas: z0=(p̂-p0)/[√p0(1-p0)/n], p̂=x/n -------------------------------------------------------------- 700(0.54)(1-0.54)= 173.9 Because np0(1−p0) = 173.9>10, the sample size is less than 5% of the population size, and the sample can be reasonably assumed to be random, the requirements for testing the hypothesis are satisfied. What are the null and alternative hypotheses? H0:p=0.54 versus H1: p≠0.54 Find the test statistic. p̂=238/700 = 0.34 z0=(0.34-0.54)/[√0.54(1-0.54)/700], z0= -10.62 Find the P-value. normalcdf(-1E99, -10.62, 0, 1)=0 P-value=0 Determine the conclusion for this hypothesis test. Choose the correct answer below. Since P-value<α, reject the null hypothesis and conclude that there is sufficient evidence that parents feel differently today.
Test the hypothesis using the P-value approach. Be sure to verify the requirements of the test. H0:p=0.5 versus H1: p>0.5 n=250; x=140; α=0.01
formulas: z0=(p̂-p0)/[√p0(1-p0)/n], p̂=x/n -------------------------------------------------------------- Calculate the test statistic, z0. p̂=140/250=0.56 z0=(0.56-0.5)/[√0.5(1-0.5)/250] z0=1.90 Identify the P-value. normalcdf(-1E99, 1.90, 0, 1)= 0.971 1-0.971 P-value=0.029 Choose the correct result of the hypothesis test for the P-value approach below. Do not reject the null hypothesis; because the P-value is greater than α.
To test H0: σ=2.3 versus H1: σ>2.3, a random sample of size n=18 is obtained from a population that is known to be normally distributed. Complete parts (a) through (d).
formulas: χ20 = [(n-1)s^2]/(σo^2) -------------------------------------------------------------- (a) If the sample standard deviation is determined to be s=2.4, compute the test statistic. χ20 = [(18-1)2.4^2]/(2.3^2) χ20=18.510 (b) If the researcher decides to test this hypothesis at the α=0.01 level of significance, determine the critical value. use chi-square calculator df=17, 0.01 χ2 0.01=33.409 (c) Draw a chi-square distribution and depict the critical region. (It's the graph with the shaded region to the very right) (d) Will the researcher reject the null hypothesis? Do not reject H0 because χ2 0 < χ2 0.01.
To test H0: σ=1.7 versus H1: σ>1.7, a random sample of size n=20 is obtained from a population that is known to be normally distributed. (a) If the sample standard deviation is determined to be s=1.9, compute the test statistic. (b) If the researcher decides to test this hypothesis at the α=0.10 level of significance, use technology to determine the P-value. (c) Will the researcher reject the null hypothesis?
formulas: χ20 = [(n-1)s^2]/(σo^2) -------------------------------------------------------------- (a) The test statistic is χ20 = [(20-1)1.9^2]/(1.7^2) χ20=23.73 (b) x^2cdf(23.73, 9999, 19)= 0.207 The P-value is 0.207. (c) Since the P-value is greater than the level of significance, the researcher will not reject the null hypothesis H0: σ=1.7.
Suppose a professional golfing association requires that the standard deviation of the diameter of a golf ball be less than 0.004 inch. Determine whether these randomly selected golf balls conform to this requirement at the α=0.10 level of significance. Assume that the population is normally distributed. 1.682 1.677 1.679 1.682 1.676 1.677 1.681 1.677 1.684 1.675 1.679 1.679
formulas: χ20 = [(n-1)s^2]/(σo^2) -------------------------------------------------------------- What are the correct hypotheses for this test? H0:σ=0.004 versus H1:σ<0.004 Find the sample standard deviation. (create list on ti-84) s=0.00276 Use s to calculate the value of the test statistic. χ20 = [(12-1)0.00276^2]/(0.004^2) χ20=5.24 Identify the critical value(s) for this test. use chi-square calculator df=11, 0.10 χ2 0.10=5.578 The critical value is 5.578 What is the correct conclusion at the α=0.10 level of significance? Since the test statistic is less than the critical value, reject the null hypothesis. There is sufficient evidence to conclude that these golf balls conform to the requirement at the 0.10 level of significance.
A jar of peanuts is supposed to have 24 ounces of peanuts. The filling machine inevitably experiences fluctuations in filling, so a quality-control manager randomly samples 12 jars of peanuts from the storage facility and measures their contents. She obtains the accompanying data. Complete parts (a) through (d) below.
formulas: √(n-1)/x^2 a/2 ∙ s, √(n-1)/x^2 1-a/2 ∙ s -------------------------------------------------------------- (a) Verify that the data are normally distributed by constructed a normal probability plot. Choose the correct normal probability plot below. (It's the graph that has 23.2-24.6 on the x-axis and is increasing and roughly linear) Are the data normally distributed? Yes, (because it is roughly linear) (b) Determine the sample standard deviation. Make a list with the data on the TI-84 s=0.363 (c) Construct a 90% confidence interval for the population standard deviation of the number of ounces of peanuts. 1-0.90= 0.10/2 = 0.05 Use Chi-Square Calculator 19.675, 4.575 √(12-1)/19.675 ∙ 0.363 = 0.272, √(12-1)/4.575 ∙ 0.363 = 0.564 There is 90% confidence that the population standard deviation is between 0.272 and 0.564 (d) The quality control manager wants the machine to have a population standard deviation below 0.20 ounce. Does the confidence interval validate this desire? C. No—the lower bound of the confidence interval is greater than 0.20.
Since this result is unusual:
it is reasonable to conclude that the population mean is higher than 2.
reject
p value is less than a (level of significance)
the smaller the probability:
the more unusual the result is
μx = 700, σx = 40 If the sample size is n=9 what is the standard deviation of the population from which the sample was drawn?
σx = σ/√n σx = 40/√9 = 13.33333333 13.33333333 x 9 = 120
do not reject
χ2 0 < χ2 0.01.
Determine the critical values for these tests of a population standard deviation. (a) A right-tailed test with 16 degrees of freedom at the α=0.05 level of significance (b) A left-tailed test for a sample of size n=31 at the α=0.1 level of significance (c) A two-tailed test for a sample of size n=24 at the α=0.1 level of significance
(a) Use given table row 16, column 0.05 = 26.296 The critical value for this right-tailed test is 26.296 (b) Use given table row 30, column 0.90 = 20.599 The critical value for this left-tailed test is 20.599 (c) Use given table row 23, column 0.95 and 0.05 = 13.091, 35.172. The critical values for this two-tailed test are 13.091, 35.172.
Complete parts (a) through (c) below. (a) Determine the critical value(s) for a right-tailed test of a population mean at the α=0.10 level of significance with 10 degrees of freedom. (b) Determine the critical value(s) for a left-tailed test of a population mean at the α=0.10 level of significance based on a sample size of n=20. (c) Determine the critical value(s) for a two-tailed test of a population mean at the α=0.05 level of significance based on a sample size of n=17.
(a) Use the given table Row 10, column 0.10 tcrit = + 1.372 (b) Use the given table Row 19, column 0.10 tcrit = − 1.328 (c) Use the given table Row 16, column 0.025 tcrit= ± 2.120
A researcher wishes to estimate the percentage of adults who support abolishing the penny. What size sample should be obtained if he wishes the estimate to be within 2 percentage points with 99 % confidence if: (a) He uses a previous estimate of 25%? (b) He does not use any prior estimates?
(a) 1-0.99 = 0.01 0.01/2 = 0.005 invnorm(0.005, 0, 1) = 2.575 E = 2% E = 0.02 p = 25% p = 0.25 n = p(1-p)[(z a/2)/E]^2 n = 0.25(1-0.25)(2.575/0.02)^2 n = 3109 (b) n = 0.25 [(z a/2)/E]^2 n = 0.25 (2.575/0.02)^2 n = 4145
For students who first enrolled in two-year public institutions in a recent semester, the proportion who earned a bachelor's degree within six years was 0.398. The president of a certain junior college believes that the proportion of students who enroll in her institution have a higher completion rate. (a) State the null and alternative hypotheses in words. (b) State the null and alternative hypotheses symbolically. (c) Explain what it would mean to make a Type I error. (d) Explain what it would mean to make a Type II error.
(a) State the null hypothesis in words. Choose the correct answer below. Among students who enroll at the certain junior college, the completion rate is 0.398. State the alternative hypothesis in words. Choose the correct answer below. Among students who enroll at the certain junior college, the completion rate is greater than 0.398. (b) State the hypotheses symbolically H0: p = 0.398 H1: p > 0.398 (c) What would it mean to make a Type I error? The president rejects the hypothesis that the proportion of students who earn a bachelor's degree within six years is equal to 0.398 when, in fact, the proportion is equal to 0.398. (d) What would it mean to make a Type II error? The president fails to reject the hypothesis that the proportion of students who earn a bachelor's degree within six years is equal to 0.398 when, in fact, the proportion is greater than 0.398
Suppose a simple random sample of size n=64 is obtained from a population that is skewed right with μ=70 and σ=16. (a) Describe the sampling distribution of x. (b) What is P x>72.4? (c) What is P x≤65.4? (d) What is P 68.1<x<73.8?
(a) The distribution is approximately normal. Find the mean and standard deviation of the sampling distribution of x. μx = 70 σx = 2 (b) (72.4 - 70)/2 = 1.2 normalcdf(-1E99, 1.2, 0, 1) = 0.8849 1-0.8849 - 0.1151 P x>72.4 = 0.1151 (c) (65.4 - 70)/2 = -2.3 normalcdf(-1E99, -2.3, 0, 1) = 0.0107 P x≤65.4 = 0.0107 (d) (73.8 - 70)/2 = 1.9 normalcdf(-1E99, 1.9, 0, 1) = 0.9713 (68.1 - 70)/2 = -0.95 normalcdf(-1E99, -0.95, 0, 1) = 0.1711 0.9713 - 0.1711 = 0.8002 P 68.1<x<73.8 = 0.8002
In a survey of 2065 adults in a certain country conducted during a period of economic uncertainty, 53% thought that wages paid to workers in industry were too low. The margin of error was 9 percentage points with 90% confidence. For parts (a) through (d) below, which represent a reasonable interpretation of the survey results? For those that are not reasonable, explain the flaw.
(a) We are 90% confident 53% of adults in the country during the period of economic uncertainty felt wages paid to workers in industry were too low. Is the interpretation reasonable? A. The interpretation is reasonable. B. The interpretation is flawed. The interpretation provides no interval about the population proportion. Your answer is correct. C. The interpretation is flawed. The interpretation suggests that this interval sets the standard for all the other intervals, which is not true. D. The interpretation is flawed. The interpretation indicates that the level of confidence is varying. (b) We are 81% to 99% confident 53% of adults in the country during the period of economic uncertainty felt wages paid to workers in industry were too low. Is the interpretation reasonable? A. The interpretation is reasonable. B. The interpretation is flawed. The interpretation indicates that the level of confidence is varying. Your answer is correct. C. The interpretation is flawed. The interpretation provides no interval about the population proportion. D. The interpretation is flawed. The interpretation suggests that this interval sets the standard for all the other intervals, which is not true. (c) We are 90% confident that the interval from 0.44 to 0.62 contains the true proportion of adults in the country during the period of economic uncertainty who believed wages paid to workers in industry were too low. Is the interpretation reasonable? A. The interpretation is reasonable. Your answer is correct. B. The interpretation is flawed. The interpretation suggests that this interval sets the standard for all the other intervals, which is not true. C. The interpretation is flawed. The interpretation indicates that the level of confidence is varying. D. The interpretation is flawed. The interpretation provides no interval about the population proportion. (d) In 90% of samples of adults in the country during the period of economic uncertainty, the proportion who believed wages paid to workers in industry were too low is between 0.44 and 0.62. Is the interpretation reasonable? A. The interpretation is reasonable. B. The interpretation is flawed. The interpretation indicates that the level of confidence is varying. C. The interpretation is flawed. The interpretation suggests that this interval sets the standard for all the other intervals, which is not true. Your answer is correct. D. The interpretation is flawed. The interpretation provides no interval about the population proportion.
majority =
0.5
True or False: The population proportion and sample proportion always have the same value.
False
For the shape of the distribution of the sample proportion to be approximately normal, it is required that np(1−p)≥______.
For the shape of the distribution of the sample proportion to be approximately normal, it is required that np(1−p)≥1010.