Quiz7/8/9

The area under the normal curve to the right of μ equals

1/2

Put the following in order from narrowest to widest interval. Assume the sample size and sample proportion is the same for all four confidence intervals. ​(a) 99​% confidence interval ​(b) 92​% confidence interval ​(c) 80​% confidence interval ​(d) 95​% confidence interval

C B D A

Without doing any​ computation, put the following in order from least to​ greatest, assuming the population is normally distributed with μ=200 and σ=15. ​(a) ​P(190≤x≤210​) for a random sample of size n=10 ​(b) ​P(190≤x≤210​) for a random sample of size n=20 ​(c) ​P(190≤x≤210​)

C,A,B

The population proportion and sample proportion always have the same value.

FALSE

True or​ false: The​ chi-square distribution is symmetric.

False. The​ chi-square distribution is skewed to the right.

True or​ false: To construct a confidence interval about a population variance or standard​ deviation, either the population from which the sample is drawn must be​ normal, or the sample size must be large

False. To construct a confidence interval about a population variance or standard​ deviation, the sample must come from a normally distributed population.

For the shape of the distribution of the sample proportion to be approximately​ normal, it is required that ​np (1 -p ) greater than or equals​______.

For the shape of the distribution of the sample proportion to be approximately​ normal, it is required that ​np (1 -p ) greater than or equals 10.

One graph in the figure represents a normal distribution with mean μ=16 and standard deviation σ=1. The other graph represents a normal distribution with mean μ=6 and standard deviation σ=1. Determine which graph is which and explain how you know.

Graph A has a mean of μ=6 and graph B has a mean of μ=16 because a larger mean shifts the graph to the right

Explain what ​"95​% ​confidence" means in a 95​% confidence interval.

If 100 different confidence intervals are​ constructed, each based on a different sample of size n from the same​ population, then we expect 95 of the intervals to include the parameter and 5 to not include the parameter.

Explain what ​"95​% ​confidence" means in a 95​% confidence interval. What does ​"95​% ​confidence" mean in a 95​% confidence​ interval?

If 100 different confidence intervals are​ constructed, each based on a different sample of size n from the same​ population, then we expect 95 of the intervals to include the parameter and 5 to not include the parameter.

Fill in the blanks below to make a true statement.

In a binomial experiment with n trials and probability of success​ p, if np(1-p) >/10, the binomial random variable X is approximately normal with ux=np and ox=square root np(1-p)

In a survey conducted by the Gallup​ Organization, 1100 adult Americans were asked how many hours they worked in the previous week. Based on the​ results, a​ 95% confidence interval for the mean number of hours worked had a lower bound of 42.7 and an upper bound of 44.5. Provide two recommendations for decreasing the margin of error of the interval.

Increase the sample size. Decrease the confidence level

A trade magazine routinely checks the​ drive-through service times of​ fast-food restaurants. An 80​% confidence interval that results from examining 717 customers in one​ fast-food chain's​ drive-through has a lower bound of 170.7 seconds and an upper bound of 174.3 seconds. What does this​ mean?

One can be 80​% confident that the mean​ drive-through service time of this​ fast-food chain is between 170.7 seconds and 174.3 seconds.

The procedure for constructing a confidence interval about a mean is​ _______, which means minor departures from normality do not affect the accuracy of the interval.

ROBUST

The number of adult Americans from a random sample of n adults who support a bill proposing to extend daylight savings time is a binomial random variable. Assume that its probability will be approximated using the normal distribution. Describe the area under the normal curve that will be computed in order to determine the probability that more than 684 Americans support the bill.

Right of x= 684.5

Suppose the proportion of a population that has a certain characteristic is 0.45

Suppose the proportion of a population that has a certain characteristic is 0.45. The mean of the sampling distribution of p hat from this population is ri p hat = 0.45 .

Assume that the probability of the binomial random variable will be approximated using the normal distribution. Describe the area under the normal curve that will be computed. Find the probability that there are exactly 12 defective parts in a shipment.

The area between 11.5 and 12.5

Which of the following are properties of the normal​ curve?

The graph of a normal curve is symmetric. Your answer is correct. The area under the normal curve to the right of the mean is 0.5 The high point is located at the value of the mean.

What happens to the graph of the normal curve as the mean​ increases? What happens to the graph of the normal curve as the standard deviation​ decreases?

The graph of the normal curve slides right The graph of the normal curve compresses and becomes steeper.

A study was conducted that resulted in the following relative frequency histogram. Determine whether or not the histogram indicates that a normal distribution could be used as a model for the variable.

The histogram is not bell-shaped, so a normal distribution could not be used as a model for the variable.

The data from a simple random sample with 25observations was used to construct the plots given below. The normal probability plot that was constructed has a correlation coefficient of 0.942.Judge whether a​ t-interval could be constructed using the data in the sample.

The normal probability plot does not suggest the data could come from a normal population because 0.942less than<. 959.959 and the boxplot shows ​outliers, so a​ t-interval could not be constructed.

There are two college entrance exams that are often taken by​ students, the ACT and the SAT. The composite score on the ACT is approximately normally distributed with mean 21.1 and standard deviation 5.1.The composite score on the SAT is approximately normally distributed with mean 1026and standard deviation 210.Suppose you scored 25on the ACT and 1179on the SAT. Which exam did you score better​ on? Justify your reasoning using the normal model.

The score on the ACT is​ better, because the percentile for the the ACT score is higher.

Explain why the​ t-distribution has less spread as the number of degrees of freedom increases.

The​ t-distribution has less spread as the degrees of freedom increase​ because, as n​ increases, s becomes closer to σ by the law of large numbers.

Explain why the​ t-distribution has less spread as the number of degrees of freedom increases.

The​ t-distribution has less spread as the degrees of freedom increase​ because, as n​ increases, s becomes closer to σ by the law of large numbers.

A group conducted a poll of 2076 likely voters just prior to an election. The results of the survey indicated that candidate A would receive 48​% of the popular vote and candidate B would receive 44​%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.

What does it mean to say the race was too close to​ call? A. The margin of error suggests candidate A may receive between 43​% and 53​% of the popular vote and candidate B may receive between 39​% and 49​% of the popular vote. Because the poll estimates overlap when accounting for margin of​ error, the poll cannot predict the winner.

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.

Whose estimate will have the smaller margin of error and​ why? -​Matthew's estimate will have the smaller margin of error because the larger sample size more than compensates for the higher level of confidence.

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.

Whose estimate will have the smaller margin of error and​ why? Matthew's estimate will have the smaller margin of error because the larger sample size more than compensates for the higher level of confidence

A survey of 2313 adults in a certain large country aged 18 and older conducted by a reputable polling organization found that 425 have donated blood in the past two years. Complete parts​ (a) through​ (c) below. a) Obtain a point estimate for the population proportion of adults in the country aged 18 and older who have donated blood in the past two years. p hat= b) Verify that the requirements for constructing a confidence interval about p are satisfied. ----a simple random​ sample, the value of ----- is -----​, which is ------10, and the ----- equal to​ 5% of the -----

a) 0.184 b) CAN BE ASSUMED TO BE n p hat(1-p hat) 346.909 GREATER THAN OR EQUAL TO SAMPLE SIZE CAN BE ASSUMED TO BE POPULATION SIZE c) 90%.... 0.170 and 0.197

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)

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. --One can be 95% confident that the mean pitch speed of Kershaws...... 1-0.95= 0.05/2 = 0.025invT(0.025, 14) = 2.14594.8 +- 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 survey of 2085 adults in a certain country conducted during a period of economic​ uncertainty, 57​% thought that wages paid to workers in industry were too low. The margin of error was 5 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 57​% 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? ​(b) We are 85​% to 95​% confident 57​% 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? c) We are 90​% confident that the interval from 0.52 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? (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 0.52 and 0.62. Is the interpretation​ reasonable?

a). The interpretation is flawed. The interpretation provides no interval about the population proportion. b.) The interpretation is flawed. The interpretation indicates that the level of confidence is varying. c.) The interpretation is reasonable. d.) The interpretation is flawed. The interpretation suggests that this interval sets the standard for all the other​ intervals, which is not true.

Researchers wanted to determine if having a television​ (TV) in the bedroom is associated with obesity. The researchers administered a questionnaire to 350 twelve-year-old adolescents. After analyzing the​ results, the researchers determined that the body mass index of the adolescents who had a TV in their bedroom was significantly higher than that of the adolescents who did not have a TV in their bedroom. Complete parts​ (a) through​ (e). (a) Why is this an observational​ study? What type of observational study is​ this? Why is this an observational​ study? What type of observational study is​ this? ​(b) What is the response variable in the​ study? -What is the explanatory​ variable? (c) Can you think of any lurking variables that may affect the results of the​ study? d) In the​ report, the researchers​ stated, "These results remain significant after adjustment for socioeconomic​ status." What does this​ mean? (e) Does a television in the bedroom cause a higher body mass​ index? Explain

a. This is an observational study because the researchers observe the behavior of the individuals in the study without trying to influence an explanatory variable of the study. cross-sectional b.) The response variable is the body mass index of the adolescents. -The explanatory variable is whether the adolescent has a TV in the bedroom or not. c) Yes. For​ example, possible lurking variables might be eating habits and the amount of exercise per week. d.) The researchers made an effort to avoid confounding by accounting for potential lurking variables. No, a television in the bedroom and obesity are associated because the body mass index of the adolescents who had a TV in their bedroom was significantly higher than that of the adolescents who did not have a TV in their bedroom.

The​ _______ represents the expected proportion of intervals that will contain the parameter if a large number of different samples of size n is obtained. It is denoted​ _______

level of confidence (1-infinity) times 100%

Two​ researchers, Jaime and​ Mariya, are each constructing confidence intervals for the proportion of a population who is​ left-handed. They find the point estimate is 0.13.Each independently constructed a confidence interval based on the point​ estimate, but​ Jaime's interval has a lower bound of 0.074and an upper bound of 0.186​,while​ Mariya's interval has a lower bound of 0.068and an upper bound of 0.139. Which interval is​ wrong? Why?

mariya​'s interval is wrong because it is not centered on the point estimate.

Suppose a simple random sample of size n is drawn from a large population with mean μ and standard deviation σ.The sampling distribution of x has mean μx= and standard deviation σx=

μ o/sq.root n

According to a​ study, the proportion of people who are satisfied with the way things are going in their lives is 0.80. Suppose that a random sample of 100 people is obtained. Complete parts​ (a) through​ (e) below.

​(a) Suppose the random sample of 100 people is​ asked, "Are you satisfied with the way things are going in your​ life?" Is the response to this question qualitative or​ quantitative? Explain. -The response is qualitative because the responses can be classified based on the characteristic of being satisfied or not. The sample proportion p is a random variable because the value of p varies from sample to sample. The variability is due to the fact that different people feel differently regarding their satisfaction.

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 192 had gum disease.​ Note: In the general​ population, about​ 17.5% of individuals have gum disease. Complete parts​ (a) through​ (c) below.

​(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. Identify the variable of interest. Choose the correct answer below. -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. -Qualitative—the variable classifies the individuals in the study. ​(c) Estimate the proportion of individuals who suffer from sleep apnea who have gum disease with 95​% confidence. Interpret your result. Select the correct choice below and fill in the answer boxes to complete your choice =192/320=.6 LB= 0.6+1.96 times sp.rt (0.6(1-.16)/320) UB=0.6-1.96 times sp.rt (0.6(1-.16)/320) -There is 95​% CONFIDENCE that the population proportion of individuals who suffer from sleep apnea who have gum disease is between .546 and .654

A television sports commentator wants to estimate the proportion of citizens who​ "follow professional​ football." Complete parts​ (a) through​ (c)

​(a) What sample size should be obtained if he wants to be within 4 percentage points with 95​% confidence if he uses an estimate of 52​% obtained from a​ poll? -the samp size is 600 (1.96/0.04)^2 0.52(0.48) = 599.2896 round up 600 (b) What sample size should be obtained if he wants to be within 4 percentage points with 95​% confidence if he does not use any prior​ estimates? -The sample size is 601 (1.96/0.04)^2 0.5 0.5 = 600.25 round up 601 ​(c) Why are the results from parts​ (a) and​ (b) so​ close? -The results are close because 0.52(1−0.52)=0.2496 is very close to 0.25.