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STA296 EXAM 2

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ind the z* values based on a standard normal distribution for each of the following (use standard normal table). An 88% confidence interval for a correlation. (-1.645, 1.645) (-1.96, 1.96) (-1.185, 1.185) (-1.555, 1.555)

(-1.555, 1.555)

Find the z* values based on a standard normal distribution for each of the following (use standard normal table). A 94% confidence interval for a difference in proportions. (-1.96, 1.96) (-1.88, 1.88) (-1.59, 1.59) (-1.185, 1.185)

(-1.88, 1.88)

Match each p-value to the most appropriate conclusion.The evidence against the null and in favor of the alternative is very strong. 0.04 0.0001 0.65 0.07

0.0001

Suppose 1000 tests are run to test a null hypothesis using α=0.05. If the null hypothesis is true,about how many of these tests would you expect to show statistically significant results? 0 1000 5 cannot be determined from the information given 50

Determine the null and alternative hypotheses. Testing to see if there is evidence that the mean time spent studying per week for first-yearstudents is less than upperclass students.Let group 1 be the first year students and let group 2 be the upperclass students. H0: μ1= μ2 Ha: μ1≤ μ2 H0: μ1= μ2 Ha: μ1 < μ2 H0: μ1= μ2 Ha: μ1 ≥ μ2 H0: μ1= μ2 Ha: μ1 > μ2

H0: μ1= μ2 Ha: μ1 < μ2

The ______________ hypothesis contains the "=" sign.

Null

Examine the given statement, then identify whether the statement is a null hypothesis, an alternative hypothesis orneither. The percentage of viewers tuned to FOX News is equal to 85%. Alternative hypothesis Neither No answer text provided. Null hypothesis

Null hypothesis

The following is a set of hypotheses, some information from one or more samples, and a standarderror from a randomization distribution.Test H0 : μ=9 vs Ha : μ≠ 9 when the sample has n=72, x¯=9.07 and s=0.84 with SE=0.10.Find the value of the standardized z-test statistic and P-valueRound your answer to three decimal places. z= 0.700, P-value=0.484 z= -0.700, P-value=0.758 z= -0.700, P-value=0.242 z= 0.700, P-value=0.758

z= 0.700, P-value=0.484

Find the value of the standardized z-test statistic and the area below 202 for a N(160,25)distribution.Round your answer to three decimal places. z= 1.68, area = 0.046 z= -1.68, area = 0.046 z= -1.68, area = 0.954 z= 1.68, area = 0.954

z= 1.68, area = 0.954

In 2010, some researchers with the Pew Internet & American Life project interviewed a random sample of adults about theircell phone usage.One of the questions asked was whether the respondent had ever downloaded an application or ''app" totheir cell phone. The sample proportion who had, based on 1917 respondents who had cell phones, was p^=0.29. One suchdistribution, based on proportions from 5000 bootstrap samples, is shown in the figure below. The standard deviation ofthese proportions is 0.0102. Use this information to find a 90% confidence interval for the proportion of cell phone users (in2010) who have downloaded at least one app to their phone. For 90% confidence Level, find the z* values (-1.96, 1.96) (-1.28, 1.28) (-1.645, 1.645) (-1.18, 1.18)

(-1.645, 1.645)

The survey also asked participants for their level of education, and we wish to estimate the difference in the proportion to useonline dating between those with a college degree and those with a high school degree or less. The results are shown in thetwo-way table below. College High School TotalYes 157 70 227No 666 565 1231Total 823 635 1458 Let group 1 be the proportion of college graduates using online dating and group 2 be theproportion of high school graduates using online dating. For 95% confidence Level, find the z* values (-2.33, 2.33) (-2.575, 2.575) (-1.96, 1.96) (-1.645, 1.645)

(-1.96, 1.96)

Find the z* values based on a standard normal distribution for each of the following (use standard normal table). An 99% confidence interval for a correlation. (-2.575, 2.575) (-1.185, 1.185) (-1.645, 1.645) (-1.96, 1.96)

(-2.575, 2.575)

The survey also asked participants for their level of education, and we wish to estimate the difference in the proportion to useonline dating between those with a college degree and those with a high school degree or less. The results are shown in thetwo-way table below. College High School TotalYes 157 70 227No 666 565 1231Total 823 635 1458 Let group 1 be the proportion of college graduates using online dating and group 2 be theproportion of high school graduates using online dating. Use the fact that the standard error for the estimate is 0.019 to find a 95% confidence interval for thepopulation difference in proportions. Round your answers to three decimal places. (-0.118, -0.044) (0.044, 0.118) (-0.421, -0.347) (0.347, 0.421)

(0.044, 0.118)

In 2010, some researchers with the Pew Internet & American Life project interviewed a random sample of adults about theircell phone usage.One of the questions asked was whether the respondent had ever downloaded an application or ''app" totheir cell phone. The sample proportion who had, based on 1917 respondents who had cell phones, was p^=0.29. One suchdistribution, based on proportions from 5000 bootstrap samples, is shown in the figure below. The standard deviation ofthese proportions is 0.0102. Use this information to find a 90% confidence interval for the proportion of cell phone users (in2010) who have downloaded at least one app to their phone. The 90% confidence interval is (0.277, 0.303) (0.278, 0.302) (0.270, 0.310) (0.273, 0.307)

(0.273, 0.307)

Find the indicated confidence interval. Assume the standard error comes from a bootstrapdistribution that is approximately normally distributed.A 95% confidence interval for a mean μ if the sample has n=60 with x¯=62 and s=12, and thestandard error is SE=1.55.Round your answers to three decimal places. (58.962, 65.038) (59.450, 65.550) (38.480, 85.520) (42.260, 81.740)

(58.962, 65.038)

Which P-value provides the strongest evidence against the null hypothesis? 0.99 0.05 1 0.5 0.001

0.001

A study investigated whether dogs change their behavior depending on whether a person displays reliable or unreliablebehavior. Dogs were shown two containers, one empty and one containing a dog biscuit. An experimenter pointed to one ofthe two containers. If the experimenter pointed to the container with the treat on the first trial, 16 of 26 dogs followed theexperimenter's cue on the second trial. However, if the experimenter misled the dog on the first trial, only 7 of 26 dogsfollowed the cue on the second trial. Test to see if the proportion following the cue is different depending on whether theperson exhibited reliable or unreliable behavior. The standard error for the difference in proportions is 0.138. Use a 5%significance level.Let group 1 be the dogs to follow a cue from a person who is reliable and let group 2 be the dogs to follow a cue from aperson who is unreliable. What is the p-value? Round your answer to three decimal places. 0.012 0.006 0.994 0.345

0.012

Match each p-value to the most appropriate conclusion. The result is significant at the 5% level but not at the 1% level. 0.65 0.07 0.04 0.0001

0.04

Match each p-value to the most appropriate conclusion. The evidence against the null is significant only at the 10% level but not at the 5% and 1% level 0.07 0.65 0.04 0.0001

0.07

A study examines whether giving antibiotics in infancy increases the likelihood that the child will be overweight later inlife. Prescription records were examined to determine whether or not antibiotics were prescribed during the first year of achild's life, and each child was classified as overweight or not at age 12. The researchers compared the proportionoverweight in each group. The study concludes that: "Infants receiving antibiotics in the first year of life were more likelyto be overweight later in childhood compared with those who were unexposed (32.4% versus 18.2% at age 12,P-value=0.002)." using α = 0.05Let group 1 be the children who have been given antibiotics and let group 2 be the children who have not been givenantibiotics. What is the value of the sample statistic? 0.142 0.182 0.002 0.324

0.142

Use a table of areas to find the specified area under the standard normal curve. The area that lies between -1.10 and -0.36 0.2237 0.7763 0.4951 0.2239 -0.2237

0.2237

Suppose that a 95% confidence interval for μ is (54.8, 60.8). Which of the following is most likelythe p-value for the test of H0: μ =56 versus Ha: μ ≠ 56? 0.016 0.231 0.001 0.031

0.231

0.32

Match each p-value to the most appropriate conclusion. There is really no evidence supporting the alternative hypothesis 0.07 0.04 0.65 0.0001

0.65

It is of interest to test the hypotheses Ho: p = 0.8 versus Ha: p < 0.8. The sample outcome, based on n= 10 observations, is p^ = 0.7 , and the randomization statistic to be calculated is p^ , The p-value forthis test was found to be 0.322. If the test was performed correctly, where should the randomizationdistribution be centered? 0.8 0.322 10 0.050 0.7

0.8

Use a table of areas to find the specified area under the standard normal curve. The area that lies to the left of 1.13 0.8907 0.4354 0.8708 0.1292 0.8485

0.8708

Use a table of areas to find the specified area under the standard normal curve. The area that lies to the right of -1.82 0.9656 -0.0344 0.4828 0.4656 0.0344

0.9656

A hypothesis test is a "two-tailed" if the alternative hypothesis contains a _______ sign.

Examine the given statement, then identify whether the statement is a null hypothesis, an alternative hypothesis or neither. The mean income of workers who have majored in history is less than $25,000. Null hypothesis Alternative hypothesis Neither No answer text provided.

Alternative hypothesis

A study examines whether giving antibiotics in infancy increases the likelihood that the child will be overweight later inlife. Prescription records were examined to determine whether or not antibiotics were prescribed during the first year of achild's life, and each child was classified as overweight or not at age 12. The researchers compared the proportionoverweight in each group. The study concludes that: "Infants receiving antibiotics in the first year of life were more likelyto be overweight later in childhood compared with those who were unexposed (32.4% versus 18.2% at age 12,P-value=0.002)." using α = 0.05Let group 1 be the children who have been given antibiotics and let group 2 be the children who have not been givenantibiotics. Is the explanatory variable categorical or quantitative? Quantitative. Categorical No answer text provided. No answer text provided.

Categorical

A study examines whether giving antibiotics in infancy increases the likelihood that the child will be overweight later inlife. Prescription records were examined to determine whether or not antibiotics were prescribed during the first year of achild's life, and each child was classified as overweight or not at age 12. The researchers compared the proportionoverweight in each group. The study concludes that: "Infants receiving antibiotics in the first year of life were more likelyto be overweight later in childhood compared with those who were unexposed (32.4% versus 18.2% at age 12,P-value=0.002)." using α = 0.05Let group 1 be the children who have been given antibiotics and let group 2 be the children who have not been givenantibiotics. Is the response variable categorical or quantitative? Quantitative. Categorical No answer text provided. No answer text provided.

Categorical

For each of the following questions, indicate whether it is best assessed by using a confidence interval or a hypothesis test orwhether statistical inference is not relevant to answer it. What proportion of people using a public restroom wash their hands after going to the bathroom? Confidence interval Hypothesis test Inference not relevant No answer text provided.

Confidence interval

In 2010, some researchers with the Pew Internet & American Life project interviewed a random sample of adults about theircell phone usage.One of the questions asked was whether the respondent had ever downloaded an application or ''app" totheir cell phone. The sample proportion who had, based on 1917 respondents who had cell phones, was p^=0.29. One suchdistribution, based on proportions from 5000 bootstrap samples, is shown in the figure below. The standard deviation ofthese proportions is 0.0102. Use this information to find a 90% confidence interval for the proportion of cell phone users (in2010) who have downloaded at least one app to their phone. Is the following statement an appropriate interpretation of this interval? We are 90% sure that thetrue the proportion of cell phone users (in 2010) who have downloaded at least one app to theirphone. between *** and ***(The numbers are the answer from the question above) Incorrect No answer text provided. No answer text provided. Correct

Correct

The dataset Exercise Hours contains information on the amount of exercise (hours per week) for a sample of statisticsstudents. The mean amount of exercise was 9.4 hours for the 30 female students in the sample and 12.4 hours for the 20 malestudents. A randomization distribution of differences in means based on these data, under a null hypothesis of no differencein mean exercise time between females and males, is centered near zero and reasonably normally distributed. The standarderror for the difference in means, as estimated from the standard deviation of the randomization differences, is SE=2.38. Usethis information to test, at a 5% level, whether the data show that the mean exercise time for female statistics students is lessthan the mean exercise time of male statistics students. What is the conclusion? - Do not reject H0 and find evidence that females exercise less. - Reject H0 and find evidence that females do not exercise less. - Do not reject H0 and do not find evidence that females exercise less. - Reject H0 and find evidence that females exercise less.

Do not reject H0 and do not find evidence that females exercise less.

A student in an introductory statistics course investigated if there is evidence that the proportion of milk chocolateM&M's that are green differs from the proportion of dark chocolate M&M's that are green. She purchased a bag of eachvariety, and her data are summarized in the following table. Green Not Green TotalMilk Chocolate 8 33 41Dark Chocolate 4 38 42Total 12 71 83 Use your p-value to make a decision about these hypotheses using α= 0.05 . Be sure to word yourdecision in the context of the problem. Include an assessment of the strength of your evidence. - Reject H0, this p-value provides evidence that the proportion of green candies differs for milk chocolate and dark chocolate M&M's. - Reject H0, this p-value provides no evidence that the proportion of green candies differs for milk chocolate and dark chocolate M&M's. Do not reject H0, this p-value provides no evidence that the proportion of green candies differs for milk chocolate and dark chocolate M&M's. - Do not reject H0, this p-value provides evidence that the proportion of green candies differs for milk chocolate and dark chocolate M&M's.

Do not reject H0, this p-value provides no evidence that the proportion of green candies differs for milk chocolate and dark chocolate M&M's.

Smaller sample sizes make it easier to achieve statistical significance if the alternative hypothesis istrue. No answer text provided. No answer text provided. True False

False

Determine the null and alternative hypotheses. The mean starting salary for students who have majored in statistics is $55,000. H0 : μ = 55,000 Ha : μ ≠ 55,000 H0 : μ = 55,000 Ha : μ < 55,000 None of these H0 : μ = 55,000 Ha : μ > 55,000 H0 : μ ≠ 55,000 Ha : μ = 55,000

H0 : μ = 55,000 Ha : μ ≠ 55,000

Which could be the null hypothesis for the true proportion of fireflies unable to produce light? H0: p ≈ 0.0012 H0: p = 0.0012 H0: p < 0.0012 H0: p > 0.0012 H0: p ≠0.0012

H0: p = 0.0012

In 2012 the Centers for Disease Control and Prevention reported that in a sample of 4,349 African Americans 31% wereVitamin D deficient. A 90% confidence interval based on this sample is (0.30, 0.32). It is believed that among the generalpopulation of Americans 8% suffer from Vitamin D deficiency.Define the appropriate parameter and state the appropriate hypotheses for testing the claim that,among African Americans, Vitamin D deficiency occurs at a rate other than 8%. H0: p = 0.08 Ha: p ≠ 0.08 H0: p > 0.08 Ha: p = 0.08 H0: p = 0.08 Ha: p > 0.08 H0: p ≠0.08 Ha: p = 0.08

H0: p = 0.08 Ha: p ≠ 0.08

In a survey of 1000 American adults conducted in April 2012, 43% reported having gone through an entire week withoutpaying for anything in cash. Test to see if this sample provides evidence that the proportion of all American adults going aweek without paying cash is greater than 40%. Use the fact that a randomization distribution is approximately normallydistributed with a standard error of SE=0.016. Show all details of the test and use a 5% significance level. State the null and alternative hypotheses. H0: p = 0.4 Ha: p > 0.4 H0: p<0.4 Ha: p = 0.4 H0: p = 0.4 Ha: p≠ 0.4 H0: p = 0.4 Ha: p < 0.4

H0: p = 0.4 Ha: p > 0.4

An article presented a method for estimating the body fat percentage of north Americanporcupines; the method was illustrated with a sample of n=25 porcupines. Based on this sample, a95% bootstrap confidence interval for the average body fat percentage of porcupines is 17.4% to25.8%. Which of the following null hypotheses would be rejected based on this confidenceinterval? H0: p = 18.6% H0: p = 26.6% H0: p = 22.9% H0: p = 20.0%

H0: p = 26.6%

A study examines whether giving antibiotics in infancy increases the likelihood that the child will be overweight later inlife. Prescription records were examined to determine whether or not antibiotics were prescribed during the first year of achild's life, and each child was classified as overweight or not at age 12. The researchers compared the proportionoverweight in each group. The study concludes that: "Infants receiving antibiotics in the first year of life were more likelyto be overweight later in childhood compared with those who were unexposed (32.4% versus 18.2% at age 12,P-value=0.002)." using α = 0.05Let group 1 be the children who have been given antibiotics and let group 2 be the children who have not been givenantibiotics. Define the appropriate parameter(s) and state the hypotheses for testing . H0: p1 = p2 Ha: p1 > p2 H0: p1 = p2 Ha: p1 < p2 H0: p1= p2 Ha: p1 ≠ p2 H0: p1 = p2 Ha: p1 ≥ p2

H0: p1 = p2 Ha: p1 > p2

A study investigated whether dogs change their behavior depending on whether a person displays reliable or unreliablebehavior. Dogs were shown two containers, one empty and one containing a dog biscuit. An experimenter pointed to one ofthe two containers. If the experimenter pointed to the container with the treat on the first trial, 16 of 26 dogs followed theexperimenter's cue on the second trial. However, if the experimenter misled the dog on the first trial, only 7 of 26 dogsfollowed the cue on the second trial. Test to see if the proportion following the cue is different depending on whether theperson exhibited reliable or unreliable behavior. The standard error for the difference in proportions is 0.138. Use a 5%significance level.Let group 1 be the dogs to follow a cue from a person who is reliable and let group 2 be the dogs to follow a cue from aperson who is unreliable. State the null and alternative hypotheses. H0: p1 < p2 Ha: p1 = p2 H0: p1 = p2 Ha: p1 > p2 H0: p1 = p2 Ha: p1≠ p2 H0: p1 = p2 Ha: p1 < p2

H0: p1 = p2 Ha: p1≠ p2

The survey also asked participants for their level of education, and we wish to estimate the difference in the proportion to useonline dating between those with a college degree and those with a high school degree or less. The results are shown in thetwo-way table below.College High School TotalYes 157 70 227No 666 565 1231Total 823 635 1458 State the null and alternative hypotheses. Let group 1 be the proportion of college graduates using online dating and group 2 be theproportion of high school graduates using online dating. H0: p1 = p2Ha: p1≠ p2 H0: p1 < p2 Ha: p1 = p2 H0: p1 = p2 Ha: p1 < p2 H0: p1 = p2 Ha: p1 > p2

H0: p1 = p2Ha: p1≠ p2

Which is the null hypothesis for testing that the average (μ) miles per gallon of a new SUV called the Aquarius is better than 25. H0: μ = 50 H0: μ = 25 H0: μ ≈ 25 H0: μ ≠ 25 None of these.

H0: μ = 25

Determine the null and alternative hypotheses. An automobile manufacturer claims that its new sedan will average better than 25 miles per gallonin the city. Let μ represent the true average mileage of the new sedan. H0: μ = 25 Ha: μ > 25 H0: μ = 25 Ha: μ ≠ 25 H0: μ = 25 Ha: μ ≥ 25 H0: μ = 25 Ha: μ ≤ 25 H0: μ = 25 Ha: μ < 25

H0: μ = 25 Ha: μ > 25

H0: μ = 3.63 Ha: μ > 3.6

The dataset Exercise Hours contains information on the amount of exercise (hours per week) for a sample of statisticsstudents. The mean amount of exercise was 9.4 hours for the 30 female students in the sample and 12.4 hours for the 20 malestudents. A randomization distribution of differences in means based on these data, under a null hypothesis of no differencein mean exercise time between females and males, is centered near zero and reasonably normally distributed. The standarderror for the difference in means, as estimated from the standard deviation of the randomization differences, is SE=2.38. Usethis information to test, at a 5% level, whether the data show that the mean exercise time for female statistics students is lessthan the mean exercise time of male statistics students.State the null and alternative hypotheses. Let group 1 be the female participants and group 2 be the male participants. H0: μ1 = μ2 Ha: μ1 < μ2 H0: μ1 = μ2 Ha: μ1 > μ2 H0: μ1 < μ2 Ha: μ1 = μ2 H0: μ1 = μ2 Ha: μ1≠≠ μ2

H0: μ1 = μ2 Ha: μ1 < μ2

H0:Pm.c.=Pd.c.Ha:Pm.c.≠Pd.c

For each of the following questions, indicate whether it is best assessed by using a confidence interval or a hypothesis test orwhether statistical inference is not relevant to answer it. Do basketball players hit a higher proportion of free throws when they are playing at home thanwhen they are playing away? Hypothesis test Inference not relevant Confidence interval No answer text provided.

Hypothesis test

For each of the following questions, indicate whether it is best assessed by using a confidence interval or a hypothesis test orwhether statistical inference is not relevant to answer it. In 2010, what percent of the US Senate voted to confirm Elena Kagan as a member of the SupremeCourt? Inference not relevant No answer text provided. Hypothesis test Confidence interval

Inference not relevant

A pharmaceutical company is testing to see whether its new drug is significantly better than the existing drug on the market.It is more expensive than the existing drug. Which makes more sense to use, a relatively large significance level (such as α=0.10) or a relativelysmall significance level (such as α=0.01), for the company? No answer text provided. No answer text provided. Small Large

Large

In the situation below, indicate whether it makes more sense to use a relatively large significance level (such as α=0.10) or arelatively small significance level (such as α=0.01). Testing to see whether taking a vitamin supplement each day has significant health benefits. Thereare no (known) harmful side effects of the supplement. No answer text provided. No answer text provided. Small Large

Large

A study examines whether giving antibiotics in infancy increases the likelihood that the child will be overweight later inlife. Prescription records were examined to determine whether or not antibiotics were prescribed during the first year of achild's life, and each child was classified as overweight or not at age 12. The researchers compared the proportionoverweight in each group. The study concludes that: "Infants receiving antibiotics in the first year of life were more likelyto be overweight later in childhood compared with those who were unexposed (32.4% versus 18.2% at age 12,P-value=0.002)." using = 0.05Let group 1 be the children who have been given antibiotics and let group 2 be the children who have not been givenantibiotics. Can we conclude that whether or not children receive antibiotics in infancy causes the difference inproportion classified as overweight? No answer text provided. No No answer text provided. Yes

A study suggests that exposure to UV rays through the car window may increase the risk of skin cancer.The study reviewedthe records of all 1050 skin cancer patients referred to the St. Louis University Cancer Center in 2004. Of the 42 patients withmelanoma, the cancer occurred on the left side of the body in 31 patients and on the right side in the other 11. The authors hypothesize that skin cancers are more prevalent on the left because of the sunlightcoming in through car windows. (Windows protect against UVB rays but not UVA rays.) Do thedata in this study support a conclusion that more melanomas occur on the left side because ofincreased exposure to sunlight on that side for drivers? No answer text provided. No No answer text provided. Yes

The survey also asked participants for their level of education, and we wish to estimate the difference in the proportion to useonline dating between those with a college degree and those with a high school degree or less. The results are shown in thetwo-way table below. College High School TotalYes 157 70 227No 666 565 1231Total 823 635 1458 Let group 1 be the proportion of college graduates using online dating and group 2 be theproportion of high school graduates using online dating. Is it plausible that there is no difference between college graduates and high school graduates inhow likely they are to use online dating? Use the confidence interval from part above to answer. No yes No answer text provided. No answer text provided.

Classify the conclusion of the significance test as a Type I error, a Type II error, or No error. A manufacturer claims that the mean amount of juice in its 16 ounce bottles is 16.1 ounces. Aconsumer advocacy group wants to perform a significance test to determine whether the meanamount is actually less than this. The hypotheses are:H0: μ = 16.1 ouncesHa: μ < 16.1 ouncesSuppose that the results of the sample lead to rejection of the null hypothesis. Classify thatconclusion as a Type I error, a Type II error, or a correct decision, if in fact the mean amount ofjuice, μ, is less than 16.1 ounces. Type I error No answer text provided. Type II error No error

No error

A study examines whether giving antibiotics in infancy increases the likelihood that the child will be overweight later inlife. Prescription records were examined to determine whether or not antibiotics were prescribed during the first year of achild's life, and each child was classified as overweight or not at age 12. The researchers compared the proportionoverweight in each group. The study concludes that: "Infants receiving antibiotics in the first year of life were more likelyto be overweight later in childhood compared with those who were unexposed (32.4% versus 18.2% at age 12,P-value=0.002)." using α = 0.05Let group 1 be the children who have been given antibiotics and let group 2 be the children who have not been givenantibiotics. Is this an experiment or an observational study? - Experiment - Observational study

Observational study

A study examines whether giving antibiotics in infancy increases the likelihood that the child will be overweight later inlife. Prescription records were examined to determine whether or not antibiotics were prescribed during the first year of achild's life, and each child was classified as overweight or not at age 12. The researchers compared the proportionoverweight in each group. The study concludes that: "Infants receiving antibiotics in the first year of life were more likelyto be overweight later in childhood compared with those who were unexposed (32.4% versus 18.2% at age 12,P-value=0.002)." using = 0.05Let group 1 be the children who have been given antibiotics and let group 2 be the children who have not been givenantibiotics. Use the p-value to make a decision about these hypotheses using α = 0.05 No answer text provided. Reject H0. No answer text provided. Do not reject H0.

Reject H0

In a survey of 1000 American adults conducted in April 2012, 43% reported having gone through an entire week withoutpaying for anything in cash. Test to see if this sample provides evidence that the proportion of all American adults going aweek without paying cash is greater than 40%. Use the fact that a randomization distribution is approximately normallydistributed with a standard error of SE=0.016. Show all details of the test and use a 5% significance level. What is the conclusion? - Reject H0 and find evidence that the proportion is not greater than 40%. - Do not reject H0 and find evidence that the proportion is greater than 40%. - Reject H0 and find evidence that the proportion is greater than 40%. - Do not reject H0 and do not find evidence that the proportion is greater than 40%.

Reject H0 and find evidence that the proportion is greater than 40%.

A study suggests that exposure to UV rays through the car window may increase the risk of skin cancer.The study reviewedthe records of all 1050 skin cancer patients referred to the St. Louis University Cancer Center in 2004. Of the 42 patients withmelanoma, the cancer occurred on the left side of the body in 31 patients and on the right side in the other 11. A randomization distribution gives the p-value as 0.003 for testing the hypotheses given .What is the conclusion of the test in the context of this study? - Do not reject H0, there is no evidence that melanomas are more likely on the left side. - Do not reject H0, there is evidence that melanomas are more likely on the left side. 2 - Reject H0, there is evidence that melanomas are more likely on the left side. - Reject H0, there is evidence that melanomas are not more likely on the left side.

Reject H0, there is evidence that melanomas are more likely on the left side.

A study investigated whether dogs change their behavior depending on whether a person displays reliable or unreliablebehavior. Dogs were shown two containers, one empty and one containing a dog biscuit. An experimenter pointed to one ofthe two containers. If the experimenter pointed to the container with the treat on the first trial, 16 of 26 dogs followed theexperimenter's cue on the second trial. However, if the experimenter misled the dog on the first trial, only 7 of 26 dogsfollowed the cue on the second trial. Test to see if the proportion following the cue is different depending on whether theperson exhibited reliable or unreliable behavior. The standard error for the difference in proportions is 0.138. Use a 5%significance level.Let group 1 be the dogs to follow a cue from a person who is reliable and let group 2 be the dogs to follow a cue from aperson who is unreliable. What is the conclusion? Reject H0. Do not reject H0. Reject Ha. Do not reject Ha.

Reject H0.

A study suggests that exposure to UV rays through the car window may increase the risk of skin cancer.The study reviewedthe records of all 1050 skin cancer patients referred to the St. Louis University Cancer Center in 2004. Of the 42 patients withmelanoma, the cancer occurred on the left side of the body in 31 patients and on the right side in the other 11. A bootstrap 95% confidence interval for the proportion of melanomas occurring on the left is 0.579to 0.861. Use the confidence interval to predict the results of the hypothesis test . Reject H0. No answer text provided. No answer text provided. Do not reject H0.

Reject H0.

As of August 8, 2012, the national average price for a gallon of regular unleaded gasoline was $3.63. The prices for asample of n = 10 gas stations in the state of Illinois are provided: $3.759 $3.919 $3.859 $4.099 $4.299 $4.309 $3.999$4.099 $3.749 $3.659 Use your p-value to make a decision about these hypotheses using α = 0.05. Be sure to word yourdecision in the context of the problem. Include an assessment of the strength of your evidence. - Reject H0, This p-value provides weak evidence that the average gas price in Illinois is greater than 3.63 (the national average). - Reject H0. This p-value provides very strong evidence that the average gas price in Illinois is greater than 3.63 (the national average). - Do not reject H0, we cannot conclude that the average gas price in Illinois is greater than 3.63 (the national average). - Reject H0, This p-value provides mild evidence that the average gas price in Illinois is greater than 3.63 (the national average). - Do not reject H0,This p-value provides no evidence that the average gas price in Illinois is greater than 3.63 (the national average).

Reject H0. This p-value provides very strong evidence that the average gas price in Illinois is greater than 3.63 (the national average).

A pharmaceutical company is testing to see whether its new drug is significantly better than the existing drug on the market.It is more expensive than the existing drug. Which makes more sense to use, a relatively large significance level (such as α=0.10) or a relativelysmall significance level (such as α=0.01), for the consumers? No answer text provided. No answer text provided. Large Small

Small

In the situation below, indicate whether it makes more sense to use a relatively large significance level (such as α=0.10) or arelatively small significance level (such as α=0.01). Testing a new drug with potentially dangerous side effects to see if it is significantly better than thedrug currently in use. If it is found to be more effective, it will be prescribed to millions of people No answer text provided. Large No answer text provided. Small

Small

Which of the following statements is false? - The P-value is the probability, when the null hypothesis is true, of obtaining a sample as extreme as (or more extreme than) the observed sample. - The P-value is between 0 and 1. - The P-value assumes Ha is true. - The smaller the P-value, the stronger the evidence is against H0.

The P-value assumes Ha is true.

Which of the following would be an appropriate alternative hypothesis? The sample mean is greater than 3.4. The sample mean is equal to 3.4. The population mean is equal to 3.4. The sample mean is not equal to 3.4. The population mean is greater than 3.4.

The population mean is greater than 3.4.

Which of the following would be an appropriate null hypothesis? The population proportion is not equal to 0.41. The sample proportion is equal to 0.41. The population proportion is equal to 0.41. The sample proportion is less than 0.41. The population proportion is less than 0.41.

The population proportion is equal to 0.41.

Which of the following would be an appropriate null hypothesis? The population proportion is not equal to 0.41. The sample proportion is equal to 0.41. The population proportion is equal to 0.41.t The sample proportion is less than 0.41. The population proportion is less than 0.41.

The population proportion is equal to 0.41.

Does consuming beer attract mosquitoes? An experiment was done in Africa to test possible ways to reduce the spread ofmalaria by mosquitoes. In the experiment, 43 volunteers were randomly assigned to consume either a liter of beer or aliter of water, and the attractiveness to mosquitoes of each volunteer was measured. The experiment was designed to testwhether beer consumption increases mosquito attraction. The report states that "Beer consumption, as opposed to waterconsumption, significantly increased the activation(P<0.001)". How strong is the evidence for the result? - There is no evidence. - There is mild evidence. - There is no evidence at all. Results are not significant. - There is strong evidence.

There is strong evidence.

By replicating a study and finding significant results again, we can be more confident that theresults are indeed significant. No answer text provided. No answer text provided. False True

True

Confidence intervals are for estimating the value of a parameter, while hypothesis tests testwhether the data provide evidence against a specific hypothesized value. No answer text provided. No answer text provided. False True

True

We are determining the value of a parameter when we know the entire population. Inference is notrelevant in this situation No answer text provided. True No answer text provided.

True

when we want to answer specific question about a population parameter (use sample data). Thehypothesis test is most appropriate False No answer text provided. True No answer text provided.

True

when we want to estimate a population parameter (use sample data). A confidence interval is mostappropriate. No answer text provided. True No answer text provided. False

True

The level of significance, α, is the probability of making a Type I error Correct decision Type β error Type II error

Type I erro

In the past, the mean lifetime for a certain type of flashlight battery has been 9.5 hours. Themanufacturer has introduced a change in the production method and wants to perform asignificance test to determine whether the mean lifetime has increased as a result. The hypotheses are:H0: μ = 9.5 hoursHa: μ > 9.5 hoursSuppose that the results of the sample lead to rejection of the null hypothesis. Classify that conclusion as a Type I error, a Type II error, or a correct decision, if in fact the mean running time has not increased. Type I error No error No answer text provided. Type II error

Type I error

For the given significance test, explain the meaning of a Type I error, a Type II error, or a correct decision as specified. At one school, the average amount of time tenth-graders spend watching television each week is21.6 hours. The principal introduces a campaign to encourage the students to watch less television.One year later, the principal performs a significance test using α = 0.05 to determine whether theaverage amount of time spent watching television per week has decreased. The hypotheses are:H0: μ = 21.6 hoursHa: μ < 21.6 hoursIf the P-value = 0.04 and a decision error is made, what type of error is it? Explain. - Type I error. We conclude that the average amount of time spent watching television each week is less than 21.6 hours when it in fact is not. - Type I error. We conclude that the average amount of time spent watching television each week is 21.6 hours when it is in fact less. - Type II error. We conclude that the average amount of time spent watching television each week is 21.6 hours when it is in fact less. - Type II error. We conclude that the average amount of time spent watching television each week is less than 21.6 hours when it in fact is not.

Type I error. We conclude that the average amount of time spent watching television each week is less than 21.6 hours when it in fact is not.

In the past, the mean lifetime for a certain type of flashlight battery has been 9.6 hours. Themanufacturer has introduced a change in the production method and wants to perform asignificance test to determine whether the mean lifetime has increased as a result. The hypotheses are:H0: μ = 9.6 hoursHa: μ > 9.6 hoursSuppose that the results of the sample lead to nonrejection of the null hypothesis. Classify that conclusion as a Type I error, a Type II error, or a correct decision, if in fact the mean running time has increased. Type II error No error No answer text provided. Type I error

Type II error

For the given significance test, explain the meaning of a Type I error, a Type II error, or a correct decision as specified. A health insurer has determined that the "reasonable and customary" fee for a certain medicalprocedure is $1200. They suspect that the average fee charged by one particular clinic for thisprocedure is higher than $1200. The insurer performs a significance test to determine whether theirsuspicion is correct using α = 0.05. The hypotheses are:H0: μ = $1200Ha: μ > $1200If the P-value is 0.09 and a decision error is made, what type of error is it? Explain. - Type I error. We conclude that the average fee charged for the procedure is not higher than $1200 when it actually is higher. - Type I error. We conclude that the average fee charged for the procedure is higher than $1200 when it actually is not higher. - Type II error. We conclude that the average fee charged for the procedure is not higher than $1200 when it actually is higher. - Type II error. We conclude that the average fee charged for the procedure is higher than $1200 when it actually is not higher.

Type II error. We conclude that the average fee charged for the procedure is not higher than $1200 when it actually is higher.

In the situation below,describe what it means in that context to make a Type I and Type II error. Testing a new drug withpotentially dangerous side effects to see if it is significantly better than the drug currently in use. If it is found to be moreeffective, it will be prescribed to millions of people. Making a Type II error means: - We find evidence that the new drug is more effective but it is really not any better. - We do not find any difference between the drugs. - none of these - We find evidence that the new drug is more effective. - We do not find enough evidence that the new drug is more effective but it really is more effective.

We do not find enough evidence that the new drug is more effective but it really is more

n the situation below,describe what it means in that context to make a Type I and Type II error. Testing a new drug withpotentially dangerous side effects to see if it is significantly better than the drug currently in use. If it is found to be moreeffective, it will be prescribed to millions of people. Making a Type I error means - We find evidence that the new drug is more effective but it is really not any better. - We do not find any difference between the drugs. - none of these - We find evidence that the new drug is more effective. - We do not find enough evidence that the new drug is more effective but it really is more effective.

We find evidence that the new drug is more effective but it is really not any better.

A study examines whether giving antibiotics in infancy increases the likelihood that the child will be overweight later in life. Prescription records were examined to determine whether or not antibiotics were prescribed during the first year of achild's life, and each child was classified as overweight or not at age 12. The researchers compared the proportionoverweight in each group. The study concludes that: "Infants receiving antibiotics in the first year of life were more likelyto be overweight later in childhood compared with those who were unexposed (32.4% versus 18.2% at age 12,P-value=0.002)." using α = 0.05Let group 1 be the children who have been given antibiotics and let group 2 be the children who have not been givenantibiotics. What is the explanatory variable? - No answer text provided. - Whether or not the child was classified as overweight at age 12. - No answer text provided. - Whether or not infants received antibiotics during the first year of life.

Whether or not infants received antibiotics during the first year of life.

A study examines whether giving antibiotics in infancy increases the likelihood that the child will be overweight later inlife. Prescription records were examined to determine whether or not antibiotics were prescribed during the first year of achild's life, and each child was classified as overweight or not at age 12. The researchers compared the proportionoverweight in each group. The study concludes that: "Infants receiving antibiotics in the first year of life were more likelyto be overweight later in childhood compared with those who were unexposed (32.4% versus 18.2% at age 12,P-value=0.002)." using α = 0.05Let group 1 be the children who have been given antibiotics and let group 2 be the children who have not been givenantibiotics. What is the response variable? - No answer text provided. - No answer text provided. - Whether or not the child was classified as overweight at age 12. - Whether or not infants received antibiotics during the first year of life

Whether or not the child was classified as overweight at age 12.

Does consuming beer attract mosquitoes? An experiment was done in Africa to test possible ways to reduce the spread ofmalaria by mosquitoes. In the experiment, 43 volunteers were randomly assigned to consume either a liter of beer or aliter of water, and the attractiveness to mosquitoes of each volunteer was measured. The experiment was designed to testwhether beer consumption increases mosquito attraction. The report states that "Beer consumption, as opposed to waterconsumption, significantly increased the activation(P<0.001)". Based on these results, it is reasonable to conclude that consuming beer causes an increase inmosquito attraction? No No answer text provided. Yes No answer text provided.

Yes

n 2012 the Centers for Disease Control and Prevention reported that in a sample of 4,349 African Americans 31% wereVitamin D deficient. A 90% confidence interval based on this sample is (0.30, 0.32). It is believed that among the generalpopulation of Americans 8% suffer from Vitamin D deficiency. Does this confidence interval provide evidence that among African Americans Vitamin Ddeficiency occurs at a rate other than 8%? What significance level is being used to make thisdecision? Yes. at the 10% significance level No. at the 10% significance level No. at the 90% significance level Yes. at the 90% significance level

Yes. at the 10% significance level

For a given level of significance, increasing the sample size will ____________________ theprobability of committing a Type II error if the alternative hypothesis is true. not affect always increase sometimes decrease sometimes increase decrease

decrease

For a given level of significance, increasing the sample size will ____________________ theprobability of committing a Type I error if the alternative hypothesis is true. sometimes decrease always decrease not affect sometimes increase always increase

not affect

A Type II error occurs by - not rejecting the null hypothesis when the null hypothesis is false. - rejecting the null hypothesis when the null hypothesis is true. - rejecting the null hypothesis when the null hypothesis is false. - not rejecting the null hypothesis when the null hypothesis is true.

not rejecting the null hypothesis when the null hypothesis is false.

A study suggests that exposure to UV rays through the car window may increase the risk of skin cancer.The study reviewedthe records of all 1050 skin cancer patients referred to the St. Louis University Cancer Center in 2004. Of the 42 patients withmelanoma, the cancer occurred on the left side of the body in 31 patients and on the right side in the other 11. Suppose the question of interest is whether melanomas are more likely to occur on the left side thanon the right.State the null and alternative hypotheses. - H0: p = 0.5 Ha: p > 0.5 - H0: p = 0.5 Ha: p < 0.5 - H0: p = 0.5 Ha: p ≠0.5 - H0: p = 0.5 Ha: p≥ 0.5

p = 0.5 Ha: p > 0.5

In a survey of 1000 American adults conducted in April 2012, 43% reported having gone through an entire week withoutpaying for anything in cash. Test to see if this sample provides evidence that the proportion of all American adults going aweek without paying cash is greater than 40%. Use the fact that a randomization distribution is approximately normallydistributed with a standard error of SE=0.016. Show all details of the test and use a 5% significance level. What is the p-value? Round your answer to three decimal places. p-value = 0.702 p-value = 0.970 p-value = 0.298 p-value = 0.030

p-value = 0.030

The dataset Exercise Hours contains information on the amount of exercise (hours per week) for a sample of statisticsstudents. The mean amount of exercise was 9.4 hours for the 30 female students in the sample and 12.4 hours for the 20 malestudents. A randomization distribution of differences in means based on these data, under a null hypothesis of no differencein mean exercise time between females and males, is centered near zero and reasonably normally distributed. The standarderror for the difference in means, as estimated from the standard deviation of the randomization differences, is SE=2.38. Usethis information to test, at a 5% level, whether the data show that the mean exercise time for female statistics students is lessthan the mean exercise time of male statistics students. What is the p-value?Round your answer to three decimal places. p-value=0.104 p-value=0.896 p-value=0.215 p-value=0.785

p-value=0.104

A study examines whether giving antibiotics in infancy increases the likelihood that the child will be overweight later inlife. Prescription records were examined to determine whether or not antibiotics were prescribed during the first year of achild's life, and each child was classified as overweight or not at age 12. The researchers compared the proportionoverweight in each group. The study concludes that: "Infants receiving antibiotics in the first year of life were more likelyto be overweight later in childhood compared with those who were unexposed (32.4% versus 18.2% at age 12,P-value=0.002)." using α = 0.05Let group 1 be the children who have been given antibiotics and let group 2 be the children who have not been givenantibiotics. Give the notation for the sample statistic. p1 - p2 μ1 - μ2 x1¯−x2¯ p1^−p2^

p1^−p2^

he survey also asked participants for their level of education, and we wish to estimate the difference in the proportion to useonline dating between those with a college degree and those with a high school degree or less. The results are shown in thetwo-way table below. College High School TotalYes 157 70 227No 666 565 1231Total 823 635 1458 Let group 1 be the proportion of college graduates using online dating and group 2 be theproportion of high school graduates using online dating. What is the sample statistic? p1^−p2^= 0.384 p1^−p2^= 0.081 p1^−p2^= -0.081 p1^−p2^= -0.384

p1^−p2^= 0.081

p^=0.738

p^m.c−p^d.c

p^m.c−p^d.c=0.10

A Type I error occurs by - not rejecting the null hypothesis when the null hypothesis is true. - rejecting the null hypothesis when the null hypothesis is false. - rejecting the null hypothesis when the null hypothesis is true. - not rejecting the null hypothesis when the null hypothesis is false.

rejecting the null hypothesis when the null hypothesis is true.

As of August 8, 2012, the national average price for a gallon of regular unleaded gasoline was $3.63. The prices for asample of n = 10 gas stations in the state of Illinois are provided: $3.759 $3.919 $3.859 $4.099 $4.299 $4.309 $3.999$4.099 $3.749 $3.659 Indicated weather the test is a left-tail test, a right-tail test, or a two-tailed test. right-tail test None of these two-tailed test left-tail test

right-tail test

right-tailed

The p-value is - the probability that the null hypothesis is true. - the probability that the alternative hypothesis is true. - the probability, when the null hypothesis is true, of obtaining a sample as extreme as (or more extreme than) the observed sample. - the probability, when the alternative hypothesis is true, of obtaining a sample as extreme as (or more extreme than) the observed sample.

the probability, when the null hypothesis is true, of obtaining a sample as extreme as (or more extreme than) the observed sample.

two-tailed test

x¯

x¯= 3.975

A study investigated whether dogs change their behavior depending on whether a person displays reliable or unreliablebehavior. Dogs were shown two containers, one empty and one containing a dog biscuit. An experimenter pointed to one ofthe two containers. If the experimenter pointed to the container with the treat on the first trial, 16 of 26 dogs followed theexperimenter's cue on the second trial. However, if the experimenter misled the dog on the first trial, only 7 of 26 dogsfollowed the cue on the second trial. Test to see if the proportion following the cue is different depending on whether theperson exhibited reliable or unreliable behavior. The standard error for the difference in proportions is 0.138. Use a 5%significance level.Let group 1 be the dogs to follow a cue from a person who is reliable and let group 2 be the dogs to follow a cue from aperson who is unreliable. Do we find evidence that the proportion of dogs following a cue is different depending on whetherthe person exhibited reliable or unreliable behavior? No answer text provided. No answer text provided. No yes

yes

Does consuming beer attract mosquitoes? An experiment was done in Africa to test possible ways to reduce the spread ofmalaria by mosquitoes. In the experiment, 43 volunteers were randomly assigned to consume either a liter of beer or aliter of water, and the attractiveness to mosquitoes of each volunteer was measured. The experiment was designed to testwhether beer consumption increases mosquito attraction. The report states that "Beer consumption, as opposed to waterconsumption, significantly increased the activation(P<0.001)".Is this convincing evidence that consuming beer is associated with higher mosquito attraction? No answer text provided. No yes

yes

The dataset Exercise Hours contains information on the amount of exercise (hours per week) for a sample of statisticsstudents. The mean amount of exercise was 9.4 hours for the 30 female students in the sample and 12.4 hours for the 20 malestudents. A randomization distribution of differences in means based on these data, under a null hypothesis of no differencein mean exercise time between females and males, is centered near zero and reasonably normally distributed. The standarderror for the difference in means, as estimated from the standard deviation of the randomization differences, is SE=2.38. Usethis information to test, at a 5% level, whether the data show that the mean exercise time for female statistics students is lessthan the mean exercise time of male statistics students. What is the test statistic? z = -0.79 z = 0.79 z = -1.26 z = 1.26

z = -1.26

In a survey of 1000 American adults conducted in April 2012, 43% reported having gone through an entire week withoutpaying for anything in cash. Test to see if this sample provides evidence that the proportion of all American adults going aweek without paying cash is greater than 40%. Use the fact that a randomization distribution is approximately normallydistributed with a standard error of SE=0.016. Show all details of the test and use a 5% significance level. What is the test statistic? Round your answer to two decimal places. z = -0.53 z = 1.88 z = 0.53 z = -1.88

z = 1.88

A study investigated whether dogs change their behavior depending on whether a person displays reliable or unreliablebehavior. Dogs were shown two containers, one empty and one containing a dog biscuit. An experimenter pointed to one ofthe two containers. If the experimenter pointed to the container with the treat on the first trial, 16 of 26 dogs followed theexperimenter's cue on the second trial. However, if the experimenter misled the dog on the first trial, only 7 of 26 dogsfollowed the cue on the second trial. Test to see if the proportion following the cue is different depending on whether theperson exhibited reliable or unreliable behavior. The standard error for the difference in proportions is 0.138. Use a 5%significance level.Let group 1 be the dogs to follow a cue from a person who is reliable and let group 2 be the dogs to follow a cue from aperson who is unreliable. What is the test statistic? Round your answer to two decimal places. z = 0.40 z = 2.51 z = -0.40 z = -2.51

z = 2.51

Find the value of the standardized z-test statistic and the area above 140 for a N(160,25)distribution.Round your answer to three decimal places. z= 0.8, area = 0.212 z= -0.8, area = 0.212 z= -0.8, area = 0.788 z= 0.8, area = 0.788

z= -0.8, area = 0.788

The following is a set of hypotheses, some information from one or more samples, and a standarderror from a randomization distribution.Test H0 : p1=p2 vs Ha : p1<p2 when the samples have n1=150 with p1^=0.19, and n2=90 withp2^=0.23. The standard error of p1^−p2^from the randomization distribution is 0.05.Find the value of the standardized z-test statistic and p-valueRound your answer to three decimal places. z= -0.8, p-value=0.212 z= 0.8, p-value=0.788 z= -0.8, p-value=0.788 z= 0.8, p-value=0.212

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