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Q:
The F-test of the analysis of variance requires that the populations are normally distributed with equal variances.
Q:
If the p-value is sufficiently small for an ANOVA test, we can that all the population means are equal.
Q:
In an ANOVA test, the alternative hypothesis is that at least one population mean is different from the others.
Q:
The analysis of variance (ANOVA) technique analyzes the variance of the data to determine whether differences exist between the population means.
Q:
Side"by"side box"plots are typically a good way to begin the analysis when comparing two populations.
Q:
A test for independence is applied to a contingency table with 4 rows and 4 columns. The degrees of freedom for this chi-square test must equal 9.
Q:
Lilliefors test for normality compare two cumulative distribution functions (cdf's): the cdffrom a normal distribution and the cdfcorresponding to the given data (called the empirical cdf).
Q:
The Lilliefors test is used to test for normality.
Q:
The chi"square test for normality makes a comparison between the observed histogram and a histogram based on normality.
Q:
The test statistic employed to test is , which is Fdistributed with degrees of freedom.
Q:
When testing the equality of two population variances, the test statistic is the ratio of the population variances; namely .
Q:
In conducting hypothesis testing for difference between two means when samples are dependent (paired samples), the variable under consideration is ; the sample mean difference between the pairs.
Q:
In testing the difference between two population means using two independent samples, we use the pooled variance in estimating the standard error of the sampling distribution of the sample mean difference if the populations are normal with equal variances.
Q:
An example of a paired sample is the number of defective computer chips of a particular type from two different manufacturers.
Q:
Tests in which samples are not independent are referred to as matched pairs or paired samples.
Q:
A professor of statistics refutes the claim that the proportion of Republican voters in Michigan is at most 45%. To test the claim, the hypotheses:, , should be used.
Q:
The test statistic for a hypothesis test of a population proportion is the z-value.
Q:
If a null hypothesis about a population mean is rejected at the 0.025 level of significance, then it must also be rejected at the 0.01 level.
Q:
A low p"value provides evidence for accepting the null hypothesis and rejecting the alternative.
Q:
The p-value of a test is the probability of observing a test statistic at least as extreme as the one computed given that the null hypothesis is true.
Q:
The p-value of a test is the smallest level of significance at which the null hypothesis can be rejected.
Q:
Sample evidence is statistically significant at the level only if the p"value is larger than .
Q:
The significance level also determines the rejection region.
Q:
The analyst gets to choose the significance level . It is typically chosen to be 0.50, but it is occasionally chosen to be 0.05.
Q:
The power of a test is the probability of rejecting the null hypothesis when the alternative hypothesis is true.
Q:
A Type II error is committed when we incorrectly accept an alternative hypothesis that is false.
Q:
The probability of making a Type I error and the level of significance are the same.
Q:
Type I errors are usually considered more "costly" although this can lead to conservative decision making.
Q:
A Type I error probability is represented by; it is the probability of incorrectly rejecting a null hypothesis that is true.
Q:
The rejection region is the set of sample data that leads to the rejection of the alternative hypothesis.
Q:
A one-tailed alternative is one that is supported by evidence in either direction.
Q:
An alternative hypothesis can have the signs >, <, or ?.
Q:
An alternative or research hypothesis is usually the hypothesis a researcher wants to prove.
Q:
A null hypothesis is a statement about the value of a population parameter. It is usually the current thinking, or "status quo".
Q:
The idea of the chi"square test for independence is to:
a. compare the quantile-quantile (Q-Q) plot with what would be expected under independence
b. compare the actual counts in a contingency table with what would be expected under independence
c. compare the cumulative distribution with what would be expected under independence
d. None of these options
Q:
The test statistic in a hypothesis test for a population proportion is
a. t-value calculated from the sample
b. z-value calculated from the sample
c. F-value calculated from the sample
d. the sample proportion
Q:
Which of the following signs is possible in an alternative hypothesis?
a. =
b. =
c. =
d. ?
e. All of these options
Q:
Which of the following statements are true regarding the chi " square goodness-of-fit test for normality?
a. The test doesdepend on which and how many categories we use for the histogram.
b. The test is notvery effective unless the sample size is large, say, at least 80 or 100.
c. The test tends to be toosensitive if the sample size is really large.
d. None of these options
e. All of these options
Q:
The power of a test is the probability that we
a. reject the null hypothesis when the alternative hypothesis is false
b. reject the null hypothesis when the alternative hypothesis is true
c. accept the null hypothesis when the alternative hypothesis is false
d. accept the null hypothesis when the alternative hypothesis is true
Q:
A type I error occurs when the:
a. null hypothesis is incorrectly accepted when it is false
b. null hypothesis is incorrectly rejected when it is true
c. sample mean differs from the population mean
d. test is biased
Q:
If a teacher is trying to prove that new method of teaching economics is more effective than traditional one, he/she will conduct a:
a. one-tailed test
b. two-tailed test
c. point estimate of the population parameter
d. confidence interval
Q:
Which pair of the following tests is used to test for normality?a. A t-test and an ANOVA testb. An Empirical cumulative distribution function test and an F-testc. A Chi-Square test and a Lilliefors testd. A Quantile-Quantile plot and a p-value test
Q:
The alternative hypothesis is also known as the:
a. elective hypothesis
b. optional hypothesis
c. research hypothesis
d. null hypothesis
Q:
The p"value of a sample is the probability of seeing a sample with
a. at most as much evidence in favor of the null hypothesis as the sample actually observed.
b. at most as much evidence in favor of the alternative hypothesis as the sample actually observed.
c. at least as much evidence in favor of the null hypothesis as the sample actually observed.
d. at least as much evidence in favor of the alternative hypothesis as the sample actually observed.
Q:
One-way ANOVA is often used in situations where:
a. there are two populations
b. randomly selected populations
c. randomized experiments with a single population
d. none of these options
Q:
The null and alternative hypotheses divide all possibilities into:
a. two sets that overlap
b. two non-overlapping sets
c. two sets that may or may not overlap
d. as many sets as necessary to cover all possibilities
Q:
A null hypothesis can only be rejected at the 5% significance level if and only if:
a. a 95% confidence interval includes the hypothesized value of the parameter
b. a 95% confidence interval does not include the hypothesized value of the parameter
c. the null hypothesis is biased
d. the null hypotheses includes sampling error
Q:
One-way ANOVA is used when analyzing the:
a. difference between more than two population means
b. results of a two-tailed test
c. results from a large sample
d. difference between two population variances
Q:
A p-value is considered "convincing" if it is:
a. less than 0.01
b. between 0.01 and 0.05
c. between 0.05 and 0.10
d. greater than 0.10
Q:
The hypothesis that an analyst is trying to prove is called the:
a. elective hypothesis
b. alternative hypothesis
c. optional hypothesis
d. null hypothesis
Q:
The ANOVA test is based on which assumptions?
a. The samples are independent and randomly selected from the populations
b. The populations are normally distributed
c. All population variances are equal
d. All of these options
e. None of these options
Q:
Smaller p-values indicate more evidence in support of the:
a. null hypothesis
b. alternative hypothesis
c. quality of the researcher
d. None of these options
Q:
Which of the following values is not typically used for?
a. 0.01
b. 0.05
c. 0.10
d. 0.50
Q:
In statistical analysis, the burden of proof lies traditionally with the :
a. alternative hypothesis
b. null hypothesis
c. analyst
d. facts presented to the statistical analyst
Q:
NARRBEGIN: SA_122_123A defensive driving training company is interested in evaluating the relative effectiveness of its two main modes of training; online and traditional classroom. The company has collected a random sample of 300 customers in a particular area, with the following results: OnlineClassroomPass143128Fail1712Total160140NARREND(A) Construct a 95% confidence interval for the difference between the proportions of online and classroom customers who pass the final exam.(B) Interpret the confidence interval obtained in (A).
Q:
NARRBEGIN: SA_119_121
A large regional department store is evaluating the effectiveness of its credit card program, which costs it approximately $1m per year to administer. The store believes that for the credit card program to be worthwhile, the administrative costs should be no more than 10% of the total of the average annual account balances. Rather than reviewing each of the 15,000 individual accounts, the store's analysts randomly selected a sample of 500 average annual balances from the frame. The sample mean and sample standard deviation were $215.75 and $55.90, respectively.
NARREND
(A) Construct a 95% confidence interval for the mean of the average annual credit account balances.
(B) Interpret the 95% confidence interval constructed in (A).
(C) Use the confidence interval constructed for (A) to help the store evaluate its criteria for whether or not the credit card program is worthwhile.
Q:
NARRBEGIN: SA_117_118The widths of 100 elevator rails have been measured. The sample mean and standard deviation of the elevator rails are 2.05 inches and 0.01 inch.NARREND(A) Construct a 95% confidence interval for the average width of an elevator rail. Do we need to assume that the width of elevator rails follows a normal distribution?(B) How large a sample of elevator rails would we have to measure to ensure that we could estimate, with 95% confidence, the average diameter of an elevator rail within 0.01 inch?
Q:
NARRBEGIN: SA_108_112The following values have been calculated using the TDIST and TINV functions in Excel. These values come from a t- distribution with 15 degrees of freedom.These values represent the probability to the right of the given positive values.Valuet -probability1.000.16661.200.12441.400.0909These values represent the positive t- value for a given probability in both tails (sum of both tails).Probabilityt -value0.201.34060.101.75310.052.1315NARRENDIn constructing confidence interval estimate for the difference between the means of two populations, where the unknown population variances are assumed not to be equal, summary statistics computed from two independent samples are as follows: , , , , , and . Construct 90% confidence interval for .
Q:
NARRBEGIN: SA_108_112The following values have been calculated using the TDIST and TINV functions in Excel. These values come from a t- distribution with 15 degrees of freedom.These values represent the probability to the right of the given positive values.Valuet -probability1.000.16661.200.12441.400.0909These values represent the positive t- value for a given probability in both tails (sum of both tails).Probabilityt -value0.201.34060.101.75310.052.1315NARRENDIn past years, approximately 25% of all U.S. families purchased potato chips at least once a month. We are interested in determining the fraction of all U.S. families that currently purchase potato chips at least once a month. How many families must we survey if we want to be 99% sure that our estimate of the fraction of U.S. families currently purchasing potato chips at least once a month is accurate within 2%?
Q:
NARRBEGIN: SA_108_112The following values have been calculated using the TDIST and TINV functions in Excel. These values come from a t- distribution with 15 degrees of freedom.These values represent the probability to the right of the given positive values.Valuet -probability1.000.16661.200.12441.400.0909These values represent the positive t- value for a given probability in both tails (sum of both tails).Probabilityt -value0.201.34060.101.75310.052.1315NARRENDYou are trying to estimate the average amount a family spends on food during a year. In the past the standard deviation of the amount a family has spent on food during a year has been approximately $800. If you want to be 95% sure that you estimated average family food expenditures within $50, how many families do you need to survey?
Q:
NARRBEGIN: SA_108_112The following values have been calculated using the TDIST and TINV functions in Excel. These values come from a t- distribution with 15 degrees of freedom.These values represent the probability to the right of the given positive values.Valuet -probability1.000.16661.200.12441.400.0909These values represent the positive t- value for a given probability in both tails (sum of both tails).Probabilityt -value0.201.34060.101.75310.052.1315NARRENDWhat would be the t-values where 0.95 of the values would fall within this interval?
Q:
NARRBEGIN: SA_108_112The following values have been calculated using the TDIST and TINV functions in Excel. These values come from a t- distribution with 15 degrees of freedom.These values represent the probability to the right of the given positive values.
Q:
NARRBEGIN: SA_108_112The following values have been calculated using the TDIST and TINV functions in Excel. These values come from a t- distribution with 15 degrees of freedom.These values represent the probability to the right of the given positive values.Valuet -probability1.000.16661.200.12441.400.0909These values represent the positive t- value for a given probability in both tails (sum of both tails).Probabilityt -value0.201.34060.101.75310.052.1315NARRENDWhat would be the t-values where 0.10 of the values are in both tails (sum of both tails)?
Q:
NARRBEGIN: SA_108_112The following values have been calculated using the TDIST and TINV functions in Excel. These values come from a t- distribution with 15 degrees of freedom.These values represent the probability to the right of the given positive values.Valuet -probability1.000.16661.200.12441.400.0909These values represent the positive t- value for a given probability in both tails (sum of both tails).Probabilityt -value0.201.34060.101.75310.052.1315NARRENDWhat would be the t-value where 0.05 of the values are in the upper tail?
Q:
NARRBEGIN: SA_108_112The following values have been calculated using the TDIST and TINV functions in Excel. These values come from a t- distribution with 15 degrees of freedom.These values represent the probability to the right of the given positive values.Valuet -probability1.000.16661.200.12441.400.0909These values represent the positive t- value for a given probability in both tails (sum of both tails).Probabilityt -value0.201.34060.101.75310.052.1315NARRENDWhat is the probability of a t-value between "1.40 and +1.40?
Q:
NARRBEGIN: SA_108_112The following values have been calculated using the TDIST and TINV functions in Excel. These values come from a t- distribution with 15 degrees of freedom.These values represent the probability to the right of the given positive values.Valuet -probability1.000.16661.200.12441.400.0909These values represent the positive t- value for a given probability in both tails (sum of both tails).Probabilityt -value0.201.34060.101.75310.052.1315NARRENDWhat is the probability of a t-value larger than 1.20?
Q:
NARRBEGIN: SA_108_112The following values have been calculated using the TDIST and TINV functions in Excel. These values come from a t- distribution with 15 degrees of freedom.These values represent the probability to the right of the given positive values.Valuet -probability1.000.16661.200.12441.400.0909These values represent the positive t- value for a given probability in both tails (sum of both tails).Probabilityt -value0.201.34060.101.75310.052.1315NARRENDWhat is the probability of a t-value smaller than 1.00?
Q:
NARRBEGIN: SA_101_107The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the feasibility of providing a dental insurance plan. A random sample of 12 employees in 2004 reveals the following family dental expenses (in dollars): 115, 370, 250, 93, 540, 225, 177, 425, 318, 182, 275, and 228. Use StatTools for your calculations.NARREND(A) Construct a 90% confidence interval estimate of the mean family dental expenses for all employees of this corporation.(B) What assumption about the population distribution must be made to answer (A)?(C) Interpret the 90% confidence interval constructed in (A).(D) Suppose you used a 95% confidence interval in (A). What would be your answer?(E) Suppose the fourth value were 593 instead of 93. What would be your answer to (A)? What effect does this change have on the confidence interval?(F) Construct a 90% confidence interval estimate for the standard deviation of family dental expenses for all employees of this corporation.(G) Interpret the 90% confidence interval constructed in (E).
Q:
NARRBEGIN: SA_97_100The average annual household income levels of citizens of selected U.S. cities are shown below.CityHouseholdCityHouseholdCityHouseholdIndexIncomeIndexIncomeIndexIncome1$54,30021$53,50041$61,5002$61,80022$45,60042$53,0003$61,40023$70,10043$51,0004$50,80024$108,70044$55,6005$56,20025$46,40045$51,6006$48,30026$56,70046$57,2007$61,60027$59,10047$54,3008$63,20028$46,30048$51,5009$55,20029$52,90049$53,50010$58,00030$56,30050$61,80011$77,60031$67,30051$44,80012$47,60032$63,80052$57,40013$62,70033$70,60053$48,10014$46,20034$49,80054$52,70015$64,30035$51,30055$57,40016$56,00036$56,60056$65,50017$53,40037$49,60057$59,60018$56,80038$67,40058$62,00019$51,20039$53,70059$49,70020$59,00040$48,70060$54,400NARREND(A) Use Excel to obtain a simple random sample of size 10 from this frame.(B) Using the sample generated in (A), construct a 95% confidence interval for the mean average annual household income level of citizens in the selected U.S. cities. Assume that the population consists of all average annual household income levels in the given frame.(C) Interpret the 95% confidence interval constructed in (B).(D) Does the 95% confidence interval contain the actual population mean? If not, explain why not. What proportion of many similarly constructed confidence intervals should include the true population mean value?
Q:
NARRBEGIN: SA_95_96An automobile dealer wants to estimate the proportion of customers who still own the cars they purchased six years ago. A random sample of 200 customers selected from the automobile dealer's records indicates that 88 still own cars that were purchased six years earlier.NARREND(A) Construct a 95% confidence interval estimate of the population proportion of all customers who still own the cars they purchased six years ago(B) How can the result in (A) be used by the automobile dealer to study satisfaction with cars purchased at the dealership?
Q:
NARRBEGIN: SA_91_94Q-Mart is interested in comparing customer who used its own charge card with those who use other types of credit cards. Q-Mart would like to know if customers who use the Q-Mart card spend more money per visit, on average, than customers who use some other type of credit card. They have collected information on a random sample of 38 charge customers as shown below. On average, the person using a Q-Mart card spends $192.81 per visit and customers using another type of card spend $104.47 per visit.Summary statistics for two samples FemaleMaleSample sizes2522Sample means102.23 86.46Sample standard deviations93.39359.695 Confidence interval for difference between means Sample mean difference 15.77 Pooled standard deviation79.466 Std error of difference 23.23 NARREND(A) Using a t - value of 2.0281, calculate a 95% confidence interval for the difference between the average Q-Mart charge and the average charge on another type of credit card.(B) What are the degrees of freedom for the t - multiple in this calculation? Explain how you would calculate the degrees of freedom in this case.(C) What is the assumption in this case that allows you to use the pooled standard deviation for this confidence interval?(D) Would you conclude that there is a significant difference between the two types of customers in this case? Explain.
Q:
NARRBEGIN: SA_88_90
Senior management of a consulting services firm is concerned about a growing decline in the firm's weekly number of billable hours. The firm expects each professional employee to spend at least 40 hours per week on work. In an effort to understand this problem better, management would like to estimate the standard deviation of the number of hours their employees spend on work-related activities in a typical week. Rather than reviewing the records of all the firm's full-time employees, the management randomly selected a sample of size 50 from the available frame. The sample mean and sample standard deviations were 48.5 and 7.5 hours, respectively.
NARREND
(A) Construct a 99% confidence interval for the standard deviation of the number of hours this firm's employees spend on work-related activities in a typical week.
(B) Interpret the 99% confidence interval constructed in (A).
(C) Given the target range of 40 to 60 hours of work per week, should senior management be concerned about the number of hours their employees are currently devoting to work? Explain why or why not.
Q:
NARRBEGIN: SA_83_87
THE FOLLOWING ITEMS REQUIRE THE USE OF EXCEL:
NARREND
(A) Compute has a t-distribution with 15 degrees of freedom.
(B) Compute has a t-distribution with 150 degrees of freedom.
(C) How do you explain the difference between the results obtained in (A) and (B)?
(D) Compute where Z is a standard normal random variable.
(E) Compare the results of (D) to the results obtained in (A) and (B). How do you explain the difference in these probabilities?
Q:
NARRBEGIN: SA_81_82A real estate agent has collected a random sample of 40 houses that were recently sold in Grand Rapids, Michigan. She is interested in comparing the appraised value and recent selling price (in thousands of dollars) of the houses in this particular market. The values of these two variables for each of the 40 randomly selected houses are shown below.HouseValuePriceHouseValuePrice1140.93140.2421136.57135.352132.42129.8922130.44121.543118.30121.1423118.13132.984122.14111.2324130.98147.535149.82145.1425131.33128.496128.91139.0126141.10141.937134.61129.3427117.87123.558121.99113.6128160.58162.039150.50141.0529151.10157.3910142.87152.9030120.15114.5511155.55157.7931133.17139.5412128.50135.5732140.16149.9213143.36151.9933124.56122.0814119.65120.5334127.97136.5115122.57118.6435101.93109.4116145.27149.5136131.47127.2917149.73146.8637121.27120.4518147.70143.8838143.55151.9619117.53118.5239136.89132.5420140.13146.0740106.11114.33NARREND(A) Use the sample data to generate a 95% confidence interval for the mean difference between the appraised values and selling prices of the houses sold in Grand Rapids.(B) Interpret the constructed confidence interval fin (A) for the real estate agent.
Q:
NARRBEGIN: SA_79_80
Auditors of Independent Bank are interested in comparing the reported value of all 1775 customer savings account balances with their own findings regarding the actual value of such assets. Rather than reviewing the records of each savings account at the bank, the auditors randomly selected a sample of 100 savings account balances from the frame. The sample mean and sample standard deviations were $505.75 and 360.95, respectively.
NARREND
(A) Construct a 90% confidence interval for the total value of all savings account balances within this bank. Assume that the population consists of all savings account balances in the frame.
(B) Interpret the 90% confidence interval constructed in (A).
Q:
NARRBEGIN: SA_75_78The percent defective for parts produced by a manufacturing process is targeted at 4%. The process is monitored daily by taking samples of sizes n = 160 units. Suppose that today's sample contains 14 defectives.NARREND(A) Determine a 95% confidence interval for the proportion defective for the process today.(B) Based on your answer to (A), is it still reasonable to think the overall proportion defective produced by today's process is actually the targeted 4%? Explain your reasoning.(C) The confidence interval in (A) is based on the assumption of a large sample size. Is this sample size sufficiently large in this example? Explain how you arrived at your answer.(D) How many units would have to be sampled to be 95% confident that you can estimate the fraction of defective parts within 2% (using the information from today's sample)?
Q:
NARRBEGIN: SA_73_74A market research consultant hired by Coke Classic Company is interested in estimating the difference between the proportions of female and male customers who favor Coke Classic over Pepsi Cola in Chicago. A random sample of 200 consumers from the market under investigation showed the following frequency distribution. MaleFemale Coke7238 110Pepsi583290 13070 200NARREND(A) Construct a 95% confidence interval for the difference between the proportions of male and female customers who prefer Coke Classic over Pepsi Cola.(B) Interpret the constructed confidence interval.
Q:
NARRBEGIN: SA_71_72A department store is interested in the average balance that is carried on its store's credit card. A sample of 40 accounts reveals an average balance of $1,250 and a standard deviation of $350.NARREND(A) Find a 95% confidence interval for the mean account balance on this store's credit card (the t-multiple with 39 degrees of freedom is 2.0227).(B) What sample size would be needed to ensure that we could estimate the true mean account balance and have only 5 chances in 100 of being off by more than $100?
Q:
NARRBEGIN: SA_69_70
A sample of 40 country CD recordings of Willie Nelson has been examined. The average playing time of these recordings is 51.3 minutes, and the standard deviation is 5.8 minutes.
NARREND
(A) Construct a 95% confidence interval for the mean playing time of all Willie Nelson recordings.
(B) Interpret the confidence interval you constructed in (A).
Q:
NARRBEGIN: SA_67_68A sample of 9 production managers with over 15 years of experience has an average salary of $71,000 and a sample standard deviation of $18,000.NARREND(A) You can be 95% confident that the mean salary for all production managers with at least 15 years of experience is between what two numbers (the t-multiple with 8 degrees of freedom is 2.306)? What assumption are you making about the distribution of salaries?(B) What sample size would be needed to ensure that we could estimate the true mean salary of all production managers with more than 15 years of experience and have only 5 chances in 100 of being off by more than $4200?