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Indicate whether or not the appropriate hypothesis test would be for a difference in population means for the following scenario. If not, explain why not. Scenario 2: The report “Highest Paying Jobs for 2009–10 Bachelor’s Degree Graduates” (National Association of Colleges and Employers, February 2010) states that the mean yearly salary offer for students graduating with accounting degrees in 2010 is $48,722. A random sample of 50 accounting graduates at a large university resulted in a mean offer of $49,850 and a standard deviation of $3,300. You would like to determine if there is strong support for the claim that the mean salary offer for accounting graduates of this university is higher than the 2010 national average of $48,722.
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no
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Indicate whether or not the appropriate hypothesis test would be for a difference in population means for the following scenario. If not, explain why not. Scenario 3: Each person in a random sample of 228 male teenagers and a random sample of 306 female teenagers was asked how many hours he or she spent online in a typical week (Ipsos, January 25, 2006). The sample mean and standard deviation were 15.1 hours and 11.4 hours for males and 14.1 and 11.8 for females. You would like to determine if there is convincing evidence that the mean number of hours spent online in a typical week is greater for male teenagers than for female teenagers.
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yes
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The paper “Sodium content of Lunchtime Fast Food Purchases at Major U.S. Chains” (Archives of Internal Medicine [2010]: 732–734) reported that for a random sample of 850 meal purchases made at Burger King, the mean sodium content was 1,685 mg, and the standard deviation was 828 mg. For a random sample of 2,107 meal purchases made at McDonald’s, the mean sodium content was 1,477 mg, and the standard deviation was 812 mg. Based on these data, is it reasonable to conclude that there is a difference in mean sodium content for meal purchases at Burger King and meal purchases at McDonald’s? Use α = 0.05.
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H0: µb - µm = 0 versus Ha: µm - µb ≠ 0, t = 6.22, P ≈ 0.000, reject H0. There is convincing evidence that the mean sodium content is not the same for meal purchases at Burger King and McDonalds.
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For the following situation, state whether the parameter of interest is a mean or a proportion: In a survey, one hundred college students are asked how many hours per week they spend on the Internet.
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Mean. Each student reports a numerical value: a number of hours.
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For the following situation, state whether the parameter of interest is a mean or a proportion: In a survey, one hundred college students are asked: “What percentage of the time you spend on the Internet is part of your course work?”
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Mean. Each student reports a number (it is a percentage, and we can average over these percentages).
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For the following situation, state whether the parameter of interest is a mean or a proportion: In a survey, one hundred college students are asked whether or not they cited information from Wikipedia in their papers.
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Proportion. Each student reports Yes or No, so this is a categorical variable and we use a proportion.
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For the following situation, state whether the parameter of interest is a mean or a proportion: In a survey, one hundred college students are asked what percentage of their total weekly spending is on alcoholic beverages.
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Mean. Each student reports a number (a percentage).
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For the following situation, state whether the parameter of interest is a mean or a proportion: In a sample of one hundred recent college graduates, it is found that 85 percent expect to get a job within one year of their graduation date.
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Proportion. Each student reports whether or not he got a job, so this is a categorical variable and we use a proportion.
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For the following situation, state whether the parameter of interest is a mean or a proportion: A poll shows that 64% of Americans personally worry a great deal about federal spending and the budget deficit.
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Proportion. Each respondent reports whether or not they worry a great deal about federal spending and the budget deficit, so this is a categorical variable and we use a proportion.
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For the following situation, state whether the parameter of interest is a mean or a proportion: A survey reports that local TV news has shown a 17% increase in revenue within a two year period while newspaper revenues decreased by 6.4% during this time period.
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Mean. Each TV news program and newspaper report a number: revenue.
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For the following situation, state whether the parameter of interest is a mean or a proportion: In a survey, high school and college students are asked whether or not they use geolocation services on their smart phones.
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Proportion. Each student reports whether or not they use geolocation services on their smart phones, so this is a categorical variable and we use a proportion.
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For the following situation, state whether the parameter of interest is a mean or a proportion: In a survey, smart phone users are asked whether or not they use a web-based taxi service.
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Proportion. Each user reports whether or not they use a web-based taxi service, so this is a categorical variable and we use a proportion.
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For the following situation, state whether the parameter of interest is a mean or a proportion: In a survey, smart phone users are asked how many times they used a web-based taxi service over the last year.
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Mean. Each user reports a number: how many times they used a web-based taxi service over the last year.
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As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. What population is under consideration in the data set?
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The sample is from all computer chips manufactured at the factory during the week of pro- duction. We might be tempted to generalize the population to represent all weeks, but we should exercise caution here since the rate of defects may change over time.
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As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. What parameter is being estimated?
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The fraction of computer chips manufactured at the factory during the week of production that had defects.
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As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. What is the point estimate for the parameter?
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We estimate the parameter by computing the observed value in the data: p^ = 27/212 = 0.127
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As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. What is the name of the statistic we use to measure the uncertainty of the point estimate?
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We quantify this uncertainty using the standard error, which may be abbreviated as SE.
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As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. Compute the value of the standard error for this context.
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Compute the standard error using the SE formula and plugging in the point estimate p^ = 0:127 for p: SE = sqrt(0.127(1 – 0.127) / 212 = 0.023
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As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. The historical rate of defects is 10%. Should the engineer be surprised by the observed rate of defects during the current week?
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The standard error is the standard deviation of p^. A value of 0.10 would be about one standard error away from the observed value, which would not represent a very uncommon deviation. (Usually beyond about 2 standard errors is a good rule of thumb.) The engineer should not be surprised.
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Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt. What population is under consideration in the data set?
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The sample is from all adults in the United States, so US adults is the population under consideration.
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Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt. What parameter is being estimated?
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The fraction of US adults who could not cover a $400 expense without borrowing money or selling something.
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Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt. What is the point estimate for the parameter?
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We estimate the parameter by computing the observed value in the data: p^ = 322 / 765 = 0.421
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Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt. What is the name of the statistic we use to measure the uncertainty of the point estimate?
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We quantify this uncertainty using the standard error, which may be abbreviated as SE.
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Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt. Compute the value from part (d) for this context.
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We can compute the standard error using the SE formula and plugging in the point estimate p^ = 0:421 for p: SE = sqrt( 0.421(1 – 0.421) / 765) = 0.0179
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Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt. A cable news pundit thinks the value is actually 50%. Should she be surprised by the data?
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The standard error can be thought of as the standard deviation of ^p. A value of 0.50 would be over 4 standard errors from the observed value, which would represent a very uncommon observation. The news pundit should be surprised by the data.
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A nonprofit wants to understand the fraction of households that have elevated levels of lead in their drinking water. They expect at least 5% of homes will have elevated levels of lead, but not more than about 30%. They randomly sample 800 homes and work with the owners to retrieve water samples, and they compute the fraction of these homes with elevated lead levels. They repeat this 1,000 times and build a distribution of sample proportions. What is this distribution called?
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Sampling distribution.
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A nonprofit wants to understand the fraction of households that have elevated levels of lead in their drinking water. They expect at least 5% of homes will have elevated levels of lead, but not more than about 30%. They randomly sample 800 homes and work with the owners to retrieve water samples, and they compute the fraction of these homes with elevated lead levels. They repeat this 1,000 times and build a distribution of sample proportions. Would you expect the shape of this distribution to be symmetric, right skewed, or left skewed? Explain your reasoning.
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To know whether the distribution is skewed, we need to know the proportion. We've been told the proportion is likely above 5% and below 30%, and the success-failure condition would be satisfied for any of these values. If the population proportion is in this range, the sampling distribution will be symmetric.
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A nonprofit wants to understand the fraction of households that have elevated levels of lead in their drinking water. They expect at least 5% of homes will have elevated levels of lead, but not more than about 30%. They randomly sample 800 homes and work with the owners to retrieve water samples, and they compute the fraction of these homes with elevated lead levels. They repeat this 1,000 times and build a distribution of sample proportions. If the proportions are distributed around 8%, what is the variability of the distribution?
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We use the standard error to describe the variability: SE = sqrt( 0.08(1 – 0.08) / 800) = 0.0096
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A nonprofit wants to understand the fraction of households that have elevated levels of lead in their drinking water. They expect at least 5% of homes will have elevated levels of lead, but not more than about 30%. They randomly sample 800 homes and work with the owners to retrieve water samples, and they compute the fraction of these homes with elevated lead levels. They repeat this 1,000 times and build a distribution of sample proportions. Suppose the researchers’ budget is reduced, and they are only able to collect 250 observations per sample, but they can still collect 1,000 samples. They build a new distribution of sample proportions. How will the variability of this new distribution compare to the variability of the distribution when each sample contained 800 observations?
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The distribution will tend to be more variable when we have fewer observations per sample.
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Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class project, students randomly sample 40 students and check if those students made the list. They repeat this 1,000 times and build a distribution of sample proportions. What is this distribution called?
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Sampling distribution
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Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class project, students randomly sample 40 students and check if those students made the list. They repeat this 1,000 times and build a distribution of sample proportions. Would you expect the shape of this distribution to be symmetric, right skewed, or left skewed? Explain your reasoning.
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Since the proportion is p = 0.16 and n = 40, the success-failure condition is not satis_ed, with the expected number of successes being just 40 x 0.16 = 6.4. When we have too few expected successes, the sampling distribution of p^ is right-skewed.
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Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class project, students randomly sample 40 students and check if those students made the list. They repeat this 1,000 times and build a distribution of sample proportions. Calculate the variability of this distribution.
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The variability can still be calculated, even if we cannot model ^p using a normal distribution: SE = sqrt( 0.16(1 – 0.16) / 40) = 0.058
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Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class project, students randomly sample 40 students and check if those students made the list. They repeat this 1,000 times and build a distribution of sample proportions. Suppose the students decide to sample again, this time collecting 90 students per sample, and they again collect 1,000 samples. They build a new distribution of sample proportions. How will the variability of this new distribution compare to the variability of the distribution when each sample contained 40 observations?
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When there are more observations in the sample, the point estimate tends to be less variable. This means the distribution will tend to be less variable when we have more observations per sample. Beyond the required answer: the success-failure condition would be satisfied with this larger sample, so the distribution would also be symmetric in this scenario.
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In 2013, the Pew Research Foundation reported that “45% of U.S. adults report that they live with one or more chronic conditions” However, this value was based on a sample, so it may not be a perfect estimate for the population parameter of interest on its own. The study reported a standard error of about 1.2%, and a normal model may reasonably be used in this setting. Create a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions. Also interpret the confidence interval in the context of the study.
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Recall that the general formula is point estimate ±_z*x SE. First, identify the three different values. The point estimate is 45%, z* = 1.96 for a 95% confidence level, and SE = 1.2%. Then, plug the values into the formula: 45% ± 1:96 x 1.2% = (42.6%; 47.4%) We are 95% confident that the proportion of US adults who live with one or more chronic conditions is between 42.6% and 47.4%.
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A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter.12. The standard error for this estimate was 2.4%, and a normal distribution may be used to model the sample proportion. Construct a 99% confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter, and interpret the confidence interval in context.
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Recall that the general formula is point estimate ± z* x SE. First, identify the three different values. The point estimate is 45%, z* = 2.58 for a 99% confidence level, and SE = 2.4%. Then, plug the values into the formula: 52% ± 2:58 x 2.4% = (45.8%; 58.2%) We are 99% confident that 45.8% to 58.2% of U.S. adult Twitter users get some news on Twitter.
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In 2013, the Pew Research Foundation reported that “45% of U.S. adults report that they live with one or more chronic conditions”, and the standard error for this estimate is 1.2%. Identify the following statement as true or false. Provide an explanation to justify your answer: If we repeated this study 1,000 times and constructed a 95% confidence interval for each study, then approximately 950 of those confidence intervals would contain the true fraction of U.S. adults who suffer from chronic illnesses.
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True. Notice that the description focuses on the true population value.
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In 2013, the Pew Research Foundation reported that “45% of U.S. adults report that they live with one or more chronic conditions”, and the standard error for this estimate is 1.2%. Identify the following statement as true or false. Provide an explanation to justify your answer: The poll provides statistically significant evidence (at the α = 0.05 level) that the percentage of U.S. adults who suffer from chronic illnesses is below 50%.
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True. If we examine the 95% confidence interval computed in Exercise ??, we can see that 50% is not included in this interval. This means that in a hypothesis test, we would reject the null hypothesis that the proportion is 0.5.
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In 2013, the Pew Research Foundation reported that “45% of U.S. adults report that they live with one or more chronic conditions”, and the standard error for this estimate is 1.2%. Identify the following statement as true or false. Provide an explanation to justify your answer: Since the standard error is 1.2%, only 1.2% of people in the study communicated uncertainty about their answer.
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False. The standard error describes the uncertainty in the overall estimate from natural fluctuations due to randomness, not the uncertainty corresponding to individuals' responses.
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A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter, and the standard error for this estimate was 2.4%. Identify each of the following statements as true or false. Provide an explanation to justify each of your answers. The data provide statistically significant evidence that more than half of U.S. adult Twitter users get some news through Twitter. Use a significance level of α = 0.01.
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False. 50% is included in the 99% confidence interval, hence a null hypothesis of p = 0.50 would not be rejected at this level.
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A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter, and the standard error for this estimate was 2.4%. Identify each of the following statements as true or false. Provide an explanation to justify each of your answers. Since the standard error is 2.4%, we can conclude that 97.6% of all U.S. adult Twitter users were included in the study.
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False. The standard error measures the variability of the sample proportion, and is unrelated to the proportion of the population included in the study.
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A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter, and the standard error for this estimate was 2.4%. Identify each of the following statements as true or false. Provide an explanation to justify each of your answers. If we want to reduce the standard error of the estimate, we should collect less data.
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False. We need to increase the sample size to decrease the standard error.
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A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter, and the standard error for this estimate was 2.4%. Identify each of the following statements as true or false. Provide an explanation to justify each of your answers. If we construct a 90% confidence interval for the percentage of U.S. adults Twitter users who get some news through Twitter, this confidence interval will be wider than a corresponding 99% confidence interval.
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False. As the confidence level decreases so does the margin of error, and hence the width of the confidence interval.
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A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a simple random sample of 64 patients and determines the time (in minutes) between when they checked in to the ER until they were first seen by a doctor. A 95% confidence interval based on this sample is (128 minutes, 147 minutes), which is based on the normal model for the mean. Determine whether the following statement is true or false, and explain your reasoning: We are 95% confident that the average waiting time of these 64 emergency room patients is between 128 and 147 minutes.
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False, even if the population distribution is not normal, with a large enough sample size we can assume that the sampling distribution is nearly Normal and calculate a confidence interval. It would however be ideal to see a histogram or Normal probability plot of the population distribution to assess the level of skewness. If the distribution is very strongly skewed, the sample size of 64 may be insufficient for the sample mean to be approximately normally distributed.
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A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a simple random sample of 64 patients and determines the time (in minutes) between when they checked in to the ER until they were first seen by a doctor. A 95% confidence interval based on this sample is (128 minutes, 147 minutes), which is based on the normal model for the mean. Determine whether the following statement is true or false, and explain your reasoning: We are 95% confident that the average waiting time of all patients at this hospital’s emergency room is between 128 and 147 minutes.
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True, this is the correct interpretation of the confidence interval.
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A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a simple random sample of 64 patients and determines the time (in minutes) between when they checked in to the ER until they were first seen by a doctor. A 95% confidence interval based on this sample is (128 minutes, 147 minutes), which is based on the normal model for the mean. Determine whether the following statement is true or false, and explain your reasoning: 95% of random samples have a sample mean between 128 and 147 minutes.
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False, the confidence interval is not about a sample mean. The true interpretation of the confidence level would be that 95% of random samples produce confidence intervals that include the true population mean.
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A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a simple random sample of 64 patients and determines the time (in minutes) between when they checked in to the ER until they were first seen by a doctor. A 95% confidence interval based on this sample is (128 minutes, 147 minutes), which is based on the normal model for the mean. Determine whether the following statement is true or false, and explain your reasoning: A 99% confidence interval would be narrower than the 95% confidence interval since we need to be more sure of our estimate.
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False. To be more confident that we capture the parameter, we need a wider interval.
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A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a simple random sample of 64 patients and determines the time (in minutes) between when they checked in to the ER until they were first seen by a doctor. A 95% confidence interval based on this sample is (128 minutes, 147 minutes), which is based on the normal model for the mean. Determine whether the following statement is true or false, and explain your reasoning: The margin of error is 9.5 and the sample mean is 137.5.
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True, since the normal model was used to model the sample mean. The margin of error can be calculated as half the width of the interval, and the sample mean is the midpoint of the interval. ME = (146 – 128) / 2 = 9.5; x-bar = 128 + 9.5 = 137.5
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A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a simple random sample of 64 patients and determines the time (in minutes) between when they checked in to the ER until they were first seen by a doctor. A 95% confidence interval based on this sample is (128 minutes, 147 minutes), which is based on the normal model for the mean. Determine whether the following statement is true or false, and explain your reasoning: In order to decrease the margin of error of a 95% confidence interval to half of what it is now, we would need to double the sample size. (Hint: the margin of error for a mean scales in the same way with sample size as the margin of error for a proportion.)
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False, since in calculation of the standard error we divide the standard deviation by square root of the sample size. In order to cut the standard error in half (and hence the margin of error) we would need to sample 22 = 4 times the number of people in the initial sample.
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The General Social Survey asked the question: “For how many days during the past 30 days was your mental health, which includes stress, depression, and problems with emotions, not good?” Based on responses from 1,151 US residents, the survey reported a 95% confidence interval of 3.40 to 4.24 days in 2010. Interpret this interval in context of the data.
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We are 90% confident that US residents experience poor mental health 3.40 to 4.24 days per month.
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The General Social Survey asked the question: “For how many days during the past 30 days was your mental health, which includes stress, depression, and problems with emotions, not good?” Based on responses from 1,151 US residents, the survey reported a 95% confidence interval of 3.40 to 4.24 days in 2010. What does “95% confident” mean? Explain in the context of the application.
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90% of random samples of size 1,151 will yield a confidence interval that contains the true average number of bad mental health days that US residents experience per month.
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The General Social Survey asked the question: “For how many days during the past 30 days was your mental health, which includes stress, depression, and problems with emotions, not good?” Based on responses from 1,151 US residents, the survey reported a 95% confidence interval of 3.40 to 4.24 days in 2010. Suppose the researchers think a 99% confidence level would be more appropriate for this interval. Will this new interval be smaller or wider than the 95% confidence interval?
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To be more sure they capture the actual mean, they require a wider interval, unless they collect more data.
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The General Social Survey asked the question: “For how many days during the past 30 days was your mental health, which includes stress, depression, and problems with emotions, not good?” Based on responses from 1,151 US residents, the survey reported a 95% confidence interval of 3.40 to 4.24 days in 2010. If a new survey were to be done with 500 Americans, do you think the standard error of the estimate be larger, smaller, or about the same.
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Less data means less precision. The estimate will probably be less accurate with less data, so the interval will be larger.
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A website is trying to increase registration for first-time visitors, exposing 1% of these visitors to a new site design. Of 752 randomly sampled visitors over a month who saw the new design, 64 registered. Check any conditions required for constructing a confidence interval.
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The visitors are from a simple random sample, so independence is satisfied. The success-failure condition is also satisfied, with both 64 and 752 - 64 = 688 above 10. Therefore, we can use a normal distribution to model p^ and construct a confidence interval.
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A website is trying to increase registration for first-time visitors, exposing 1% of these visitors to a new site design. Of 752 randomly sampled visitors over a month who saw the new design, 64 registered. Compute the standard error.
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The sample proportion is p^ = 64 / 752 = 0.085. The standard error is SE = sqrt(0.085(1 – 0.085) / 752) = 0.010
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A website is trying to increase registration for first-time visitors, exposing 1% of these visitors to a new site design. Of 752 randomly sampled visitors over a month who saw the new design, 64 registered. Construct and interpret a 90% confidence interval for the fraction of first-time visitors of the site who would register under the new design (assuming stable behaviors by new visitors over time).
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For a 90% confidence interval, use z* = 1.65. The confidence interval is therefore 0.085 ± 1:65 x 0.010 = (0.0685; 0.1015). We are 90% confident that 6.85% to 10.15% of first-time site visitors will register using the new design.
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A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they’d received in the mail. Construct a 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they’d received in the mail.
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First, we check conditions: Independence. Since the data come from a random sample, independence is satisfied. Success-failure. We also have at least 10 successes (142) and 10 failures (603 - 142 = 461), so the success-failure condition is satisfied. With the conditions satisfied, p^ = 142 / 603 = 0.235 can be modeled using a normal distribution. Next, we compute the standard error: SE = sqrt(0.235(1 – 0.235) / 603) = 0.0173. For a 95% confidence level, z* = 1.96, and the confidence interval is 0.235 ± 1.96 x 0.0173 = (0:201; 0:269). We are 95% confident that 20.1% to 26.9% of the store's shoppers during the year made their visit because they had received a coupon in the mail.
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Write the null and alternative hypotheses in words and then symbols for the following situation: A tutoring company would like to understand if most students tend to improve their grades (or not) after they use their services. They sample 200 of the students who used their service in the past year and ask them if their grades have improved or declined from the previous year.
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H0: p = 0.5 (The number of students with grades that improve after the program is equal to the number of students who do not have their grades improve) HA: p ≠ 0.5 (Either a majority or a minority of students' grades improved)
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Write the null and alternative hypotheses in words and then symbols for the following situation: Employers at a firm are worried about the effect of March Madness, a basketball championship held each spring in the US, on employee productivity. They estimate that on a regular business day employees spend on average 15 minutes of company time checking personal email, making personal phone calls, etc. They also collect data on how much company time employees spend on such non-business activities during March Madness. They want to determine if these data provide convincing evidence that employee productivity changed during March Madness.
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H0: µ = 15 (The average amount of company time spent not working is 15 minutes) HA: µ ≠ 15 (The average amount of company time spent not working is different than 15 minutes)
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Write the null and alternative hypotheses in words and using symbols for the following situations: Since 2008, chain restaurants in California have been required to display calorie counts of each menu item. Prior to menus displaying calorie counts, the average calorie intake of diners at a restaurant was 1100 calories. After calorie counts started to be displayed on menus, a nutritionist collected data on the number of calories consumed at this restaurant from a random sample of diners. Do these data provide convincing evidence of a difference in the average calorie intake of a diners at this restaurant?
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H0: µ = 1100 (The current average calorie intake is 1100 calories) HA: µ ≠ 1100 (The current average calorie intake is different than 1100 calories.)
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Write the null and alternative hypotheses in words and using symbols for the following situations: The state of Wisconsin would like to understand the fraction of its adult residents that consumed alcohol in the last year, specifically if the rate is different from the national rate of 70%. To help them answer this question, they conduct a random sample of 852 residents and ask them about their alcohol consumption.
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H0: p = 0.7 (The fraction of Wisconsin adults who consume alcohol is 0.7) HA: p ≠ 0.7 (The fraction of Wisconsin adults who consume alcohol is different from 0.7)
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A study suggests that 60% of college student spend 10 or more hours per week communicating with others online. You believe that this is incorrect and decide to collect your own sample for a hypothesis test. You randomly sample 160 students from your dorm and find that 70% spent 10 or more hours a week communicating with others online. A friend of yours, who offers to help you with the hypothesis test, comes up with the following set of hypotheses. Indicate any errors you see. H0: p^ < 0.6; HA: p^ > 0.7
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First, the hypotheses should be about the population proportion (p), not the sample proportion. Second, the null hypothesis should have an equal sign. Third, the alternative hypothesis should have a not-equals sign and reference the null value, p0 = 0.6, not the observed sample proportion. The correct way to set up these hypotheses is: H0: p = 0.6; HA: p ≠ 0.6
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A study suggests that the 25% of 25 year olds have gotten married. You believe that this is incorrect and decide to collect your own sample for a hypothesis test. From a random sample of 25 year olds in census data with size 776, you find that 24% of them are married. A friend of yours offers to help you with setting up the hypothesis test and comes up with the following hypotheses. Indicate any errors you see. H0: p^ = 0.24; HA: p^ ≠ 0.24
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First, the hypotheses should be about the population proportion (p), not the sample pro- portion. Second, the null value should be what we are testing (0.25), not the observed value (0.24). The correct way to set up these hypotheses is: H0: p = 0.25; HA: p ≠ 0.25
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Teens were surveyed about cyberbullying, and 54% to 64% reported experiencing cyberbullying (95% confidence interval). Answer the following questions based on this interval. A newspaper claims that a majority of teens have experienced cyberbullying. Is this claim supported by the confidence interval? Explain your reasoning.
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This claim is reasonable, since the entire interval lies above 50%.
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Teens were surveyed about cyberbullying, and 54% to 64% reported experiencing cyberbullying (95% confidence interval). Answer the following questions based on this interval. A researcher conjectured that 70% of teens have experienced cyberbullying. Is this claim supported by the confidence interval? Explain your reasoning.
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The value of 70% lies outside of the interval, so we have convincing evidence that the re- searcher's conjecture is wrong.
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Teens were surveyed about cyberbullying, and 54% to 64% reported experiencing cyberbullying (95% confidence interval). Answer the following questions based on this interval. Without actually calculating the interval, determine if the claim of the researcher that 70% of teens have experienced cyberbullying would be supported based on a 90% confidence interval?
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A 90% confidence interval will be narrower than a 95% confidence interval. Even without calculating the interval, we can tell that 70% would not fall in the interval, and we would reject the researcher's conjecture based on a 90% confidence level as well.
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Exercise 5.11 provides a 95% confidence interval for the mean waiting time at an emergency room (ER) of (128 minutes, 147 minutes). Answer the following questions based on this interval. A local newspaper claims that the average waiting time at this ER exceeds 3 hours. Is this claim supported by the confidence interval? Explain your reasoning.
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This claim does is not supported since 3 hours (180 minutes) is not in the interval.
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Exercise 5.11 provides a 95% confidence interval for the mean waiting time at an emergency room (ER) of (128 minutes, 147 minutes). Answer the following questions based on this interval. The Dean of Medicine at this hospital claims the average wait time is 2.2 hours. Is this claim supported by the confidence interval? Explain your reasoning.
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2.2 hours (132 minutes) is in the 95% confidence interval, so we do not have evidence to say she is wrong.
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Exercise 5.11 provides a 95% confidence interval for the mean waiting time at an emergency room (ER) of (128 minutes, 147 minutes). Answer the following questions based on this interval. Without actually calculating the interval, determine if the claim of the Dean that the average wait time is 2.2 hours would be supported based on a 99% confidence interval?
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A 99% confidence interval will be wider than a 95% confidence interval. Even without calculating the interval, we can tell that 132 minutes would be in it, and we would not reject her claim based on a 99% confidence level as well.
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Do a majority of US adults believe raising the minimum wage will help the economy, or is there a majority who do not believe this? A Rasmussen Reports survey of a random sample of 1,000 US adults found that 42% believe it will help the economy. Conduct an appropriate hypothesis test to help answer the research question.
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Here's the answer in the Prepare, Check, Calculate, and Conclude framework. Prepare. Set up hypotheses. H0: p = 0.5; HA: p ≠ 0.5 We will use a significance level of α = 0:05. Check. Simple random sample gets us independence, and the success-failure conditions is satisfied since 0.42 x 1000 = 420 and (1 – 0.42) x 1000 = 580 are both at least 10. Calculate. SE = sqrt(0.5(1 – 0.5) / 1000) = 0.016. Z = (0.42-0.5) / 0.016 = -5, which has a one-tail area of about 0.0000003, so the p-value is twice this one-tail area at 0.0000006. Conclude. Because the p-value is less than α = 0:05, we reject the null hypothesis and conclude that the fraction of US adults who believe raising the minimum wage will help the economy is not 50%. Because the observed value is less than 50% and we have rejected the null hypothesis, we can conclude that this belief is held by fewer than 50% of US adults. For reference, the survey also explores support for changing the minimum wage, which is itself a different question.
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Getting enough sleep. 400 students were randomly sampled from a large university, and 289 said they did not get enough sleep. Conduct a hypothesis test to check whether this represents a statistically significant difference from 50%, and use a significance level of 0.01.
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Here's the answer in the Prepare, Check, Calculate, and Conclude framework. Prepare. Set up hypotheses. H0: p = 0.5, HA: p ≠ 0.5. We are told to use a significance level of α = 0:01. Check. Simple random sample gets us independence, and the success-failure conditions is satisfied since 289 and 400 - 289 = 111 are both at least 10. Calculate. p^ = 289/400 = 0.7225. SE = sqrt(0.5(1 – 0.5) / 400) = 0.025. Z = (0.7225 – 0.5) / 0.025 = 8.9, which has a one-tail area of about 3e-19 (0.0000000000000000003). So the p-value is twice this one-tail area at 6e-19 (0.0000000000000000006). Conclude. Because the p-value is less than _ = 0:01, we reject the null hypothesis and conclude that the fraction of students who don't get enough sleep is different than 50%. Because the observed value is greater than 50% and we have rejected the null hypothesis, we can conclude that a majority of students at the surveyed university don't get enough sleep.
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You are given the following hypotheses: H0: p = 0.3; HA: p ≠ 0.3. We know the sample size is 90. For what sample proportion would the p-value be equal to 0.05? Assume that all conditions necessary for inference are satisfied.
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If the p-value is 0.05, this means the test statistic would be either Z = -1.96 or Z = 1.96. We'll show the calculations for Z = 1.96. The standard error would be SE = sqrt(0.3(1 – 0.3) / 90) = 0.048. Finally, we set up the test statistic formula and solve for p^: Z = (p^ - 0.3) / SE; 1.96 = (p^ - 0.3) / 0.048; p^ = 0.394. Alternatively, if Z = -1.96 was used: p^ = 0.206.
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You are given the following hypotheses: H0: p = 0.9; HA: p ≠ 0.9. We know that the sample size is 1,429. For what sample proportion would the p-value be equal to 0.01? Assume that all conditions necessary for inference are satisfied.
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If the p-value is 0.01, this means the test statistic would be either Z = -2.58 or Z = 2.58. If either of these Zs is chosen, it is okay. We'll use Z = 2.58 for our calculations. The standard error would be SE = sqrt(0.9(1 – 0.9) / 90) = 00079. Finally, we set up the test statistic formula and solve for p^: Z = (p^ - 0.9) / SE; 2.58 = (p^ - 0.9) / 0.0079; p^ = 0.92 Alternatively, if Z = -2.58 was used: p^ = 0.88.
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A patient named Diana was diagnosed with Fibromyalgia, a long-term syndrome of body pain, and was prescribed anti-depressants. Being the skeptic that she is, Diana didn’t initially believe that anti-depressants would help her symptoms. However after a couple months of being on the medication she decides that the anti-depressants are working, because she feels like her symptoms are in fact getting better. Write the hypotheses in words for Diana’s skeptical position when she started taking the anti-depressants.
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H0: Anti-depressants do not affect the symptoms of Fibromyalgia. HA: Anti-depressants do affect the symptoms of Fibromyalgia (either helping or harming).
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A patient named Diana was diagnosed with Fibromyalgia, a long-term syndrome of body pain, and was prescribed anti-depressants. Being the skeptic that she is, Diana didn’t initially believe that anti-depressants would help her symptoms. However after a couple months of being on the medication she decides that the anti-depressants are working, because she feels like her symptoms are in fact getting better. What is a Type 1 Error in this context?
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Concluding that anti-depressants either help or worsen Fibromyalgia symptoms when they actually do neither.
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A patient named Diana was diagnosed with Fibromyalgia, a long-term syndrome of body pain, and was prescribed anti-depressants. Being the skeptic that she is, Diana didn’t initially believe that anti-depressants would help her symptoms. However after a couple months of being on the medication she decides that the anti-depressants are working, because she feels like her symptoms are in fact getting better. What is a Type 2 Error in this context?
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Concluding that anti-depressants do not affect Fibromyalgia symptoms when they actually do.
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In the scenario, there is a value of interest and two scenarios (I and II). For each part, report if the value of interest is larger under scenario I, scenario II, or whether the value is equal under the scenarios: The standard error of p^ when (I) n = 125 or (II) n = 500.
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Scenario I is higher. Recall that a sample mean based on less data tends to be less accurate and have larger standard errors.
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In the scenario, there is a value of interest and two scenarios (I and II). For each part, report if the value of interest is larger under scenario I, scenario II, or whether the value is equal under the scenarios: The margin of error of a confidence interval when the confidence level is (I) 90% or (II) 80%.
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Scenario I is higher. The higher the confidence level, the higher the corresponding margin of error.
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In the scenario, there is a value of interest and two scenarios (I and II). For each part, report if the value of interest is larger under scenario I, scenario II, or whether the value is equal under the scenarios: The p-value for a Z-statistic of 2.5 calculated based on a (I) sample with n = 500 or based on a (II) sample with n = 1000.
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They are equal. The sample size does not affect the calculation of the p-value for a given Z-score.
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In the scenario, there is a value of interest and two scenarios (I and II). For each part, report if the value of interest is larger under scenario I, scenario II, or whether the value is equal under the scenarios: The probability of making a Type 2 Error when the alternative hypothesis is true and the significance level is (I) 0.05 or (II) 0.10.
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Scenario I is higher. If the null hypothesis is harder to reject (lower α), then we are more likely to make a Type 2 Error when the alternative hypothesis is true.
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The General Social Survey asked the question: “After an average work day, about how many hours do you have to relax or pursue activities that you enjoy?” to a random sample of 1,155 Americans. A 95% confidence interval for the mean number of hours spent relaxing or pursuing activities they enjoy was (1.38, 1.92). Interpret this interval in context of the data.
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We are 95% confident that Americans spend an average of 1.38 to 1.92 hours per day relaxing or pursuing activities they enjoy.
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The General Social Survey asked the question: “After an average work day, about how many hours do you have to relax or pursue activities that you enjoy?” to a random sample of 1,155 Americans. A 95% confidence interval for the mean number of hours spent relaxing or pursuing activities they enjoy was (1.38, 1.92). Suppose another set of researchers reported a confidence interval with a larger margin of error based on the same sample of 1,155 Americans. How does their confidence level compare to the confidence level of the interval stated above?
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Their confidence level must be higher as the width of the confidence interval increases as the confidence level increases.
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The General Social Survey asked the question: “After an average work day, about how many hours do you have to relax or pursue activities that you enjoy?” to a random sample of 1,155 Americans. A 95% confidence interval for the mean number of hours spent relaxing or pursuing activities they enjoy was (1.38, 1.92). Suppose next year a new survey asking the same question is conducted, and this time the sample size is 2,500. Assuming that the population characteristics, with respect to how much time people spend relaxing after work, have not changed much within a year. How will the margin of error of the 95% confidence interval constructed based on data from the new survey compare to the margin of error of the interval stated above?
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The new margin of error will be smaller, since as the sample size increases, the standard error decreases, which will decrease the margin of error.
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We learned that a Rasmussen Reports survey of 1,000 US adults found that 42% believe raising the minimum wage will help the economy. Construct a 99% confidence interval for the true proportion of US adults who believe this.
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We could note the data and context provided: p^ = 0.42, n = 1000, and we're using a 99% confidence level. Checking conditions. While it doesn't explicitly say here whether the sample is random, we will assume the survey by this company was in fact random, which would satisfy the independence condition. The success-failure condition is also satisfied since 0.42 x 1000 and 0.58 x 1000 are both at least 10. Next, we compute the standard error for the confidence interval using p^ = 0.42: SE = sqrt(0.42(1 – 0.42) / 1000 = 0.016 Next, we can compute the confidence interval itself, where we use z* = 2.58 for a 99% confidence level: 2.42 ± 2.58 x 0:016 = (0.379; 0.461) We are 99% confident that 37.9% to 46.1% of US adults believe that increasing the minimum wage would help the economy.
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A food safety inspector is called upon to investigate a restaurant with a few customer reports of poor sanitation practices. The food safety inspector uses a hypothesis testing framework to evaluate whether regulations are not being met. If he decides the restaurant is in gross violation, its license to serve food will be revoked. Write the hypotheses in words.
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H0: The restaurant meets food safety and sanitation regulations. HA: The restaurant does not meet food safety and sanitation regulations.
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A food safety inspector is called upon to investigate a restaurant with a few customer reports of poor sanitation practices. The food safety inspector uses a hypothesis testing framework to evaluate whether regulations are not being met. If he decides the restaurant is in gross violation, its license to serve food will be revoked. What is a Type 1 Error in this context?
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The food safety inspector concludes that the restaurant does not meet food safety and sanitation regulations and shuts down the restaurant when the restaurant is actually safe.
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A food safety inspector is called upon to investigate a restaurant with a few customer reports of poor sanitation practices. The food safety inspector uses a hypothesis testing framework to evaluate whether regulations are not being met. If he decides the restaurant is in gross violation, its license to serve food will be revoked. What is a Type 2 Error in this context?
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The food safety inspector concludes that the restaurant meets food safety and sanitation regulations and the restaurant stays open when the restaurant is actually not safe.
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A food safety inspector is called upon to investigate a restaurant with a few customer reports of poor sanitation practices. The food safety inspector uses a hypothesis testing framework to evaluate whether regulations are not being met. If he decides the restaurant is in gross violation, its license to serve food will be revoked. Between a Tupe 1 and Type 2 error, which error is more problematic for the restaurant owner? Why?
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A Type 1 Error may be more problematic for the restaurant owner since his restaurant gets shut down even though it meets the food safety and sanitation regulations.
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A food safety inspector is called upon to investigate a restaurant with a few customer reports of poor sanitation practices. The food safety inspector uses a hypothesis testing framework to evaluate whether regulations are not being met. If he decides the restaurant is in gross violation, its license to serve food will be revoked. Between a Tupe 1 and Type 2 error, which error is more problematic for the diners? Why?
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A Type 2 Error may be more problematic for diners since the restaurant deemed safe by the inspector is actually not.
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A food safety inspector is called upon to investigate a restaurant with a few customer reports of poor sanitation practices. The food safety inspector uses a hypothesis testing framework to evaluate whether regulations are not being met. If he decides the restaurant is in gross violation, its license to serve food will be revoked. As a diner, would you prefer that the food safety inspector requires strong evidence or very strong evidence of health concerns before revoking a restaurant’s license? Explain your reasoning.
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A diner would prefer strong evidence as any indication of evidence might mean there may be an issue with the restaurant meeting food safety and sanitation regulations, and diners would rather a restaurant that meet the regulations get shut down than a restaurant that doesn't meet the regulations not get shut down.
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Determine if the following statement is true or false, and explain your reasoning. If false, state how it could be corrected: If a given value (for example, the null hypothesized value of a parameter) is within a 95% confidence interval, it will also be within a 99% confidence interval.
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True.
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Determine if the following statement is true or false, and explain your reasoning. If false, state how it could be corrected: Decreasing the significance level (α) will increase the probability of making a Type 1 Error.
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False. The significance level is the probability of the Type 1 Error.
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Determine if the following statement is true or false, and explain your reasoning. If false, state how it could be corrected: Suppose the null hypothesis is p = 0.5 and we fail to reject H0. Under this scenario, the true population proportion is 0.5.
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False. Failure to reject H0 only means there wasn't sufficient evidence to reject it, not that it has been confirmed.
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Determine if the following statement is true or false, and explain your reasoning. If false, state how it could be corrected: With large sample sizes, even small differences between the null value and the observed point estimate, a difference often called the effect size, will be identified as statistically significant.
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True.
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A USA Today/Gallup poll asked a group of unemployed and underemployed Americans if they have had major problems in their relationships with their spouse or another close family member as a result of not having a job (if unemployed) or not having a full-time job (if underemployed). 27% of the 1,145 unemployed respondents and 25% of the 675 underemployed respondents said they had major problems in relationships as a result of their employment status. What are the hypotheses for evaluating if the proportions of unemployed and underemployed people who had relationship problems were different?
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H0: punemp = punderemp: The proportions of unemployed and underemployed people who are having relationship problems are equal. HA: punemp ≠ punderemp: The proportions of unemployed and underemployed people who are having relationship problems are different.
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A USA Today/Gallup poll asked a group of unemployed and underemployed Americans if they have had major problems in their relationships with their spouse or another close family member as a result of not having a job (if unemployed) or not having a full-time job (if underemployed). 27% of the 1,145 unemployed respondents and 25% of the 675 underemployed respondents said they had major problems in relationships as a result of their employment status. The p-value for this hypothesis test is approximately 0.35. Explain what this means in context of the hypothesis test and the data.
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If in fact the two population proportions are equal, the probability of observing at least a 2% difference between the sample proportions is approximately 0.35. Since this is a high probability we fail to reject the null hypothesis. The data do not provide convincing evidence that the proportion of unemployed and underemployed people who are having relationship problems are different.
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It is believed that nearsightedness affects about 8% of all children. In a random sample of 194 children, 21 are nearsighted. Conduct a hypothesis test for the following question: do these data provide evidence that the 8% value is inaccurate?
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Here's the answer in the Prepare, Check, Calculate, and Conclude framework. Prepare. Set up hypotheses. H0: p = 0.08, HA: p ≠ 0.08. We will use a significance level of α = 0:05. Check. Simple random sample gets us independence, and the success-failure condition is satisfied since 21 and 194 - 21 = 173 are both at least 10. Calculate. Several calculations: p^ = 21=194 = 0.108. SE = sqrt(0.08(1 – 0.08) / 194) = 0.0195. Z = (0.108 – 0.08) / 0.0195 = 1.44 which has a one-tail area of about 0.075, so the p-value is twice this one-tail area at 0.15. Conclude. Because the p-value is bigger than α = 0:05, we do not reject the null hypothesis. The sample does not provide convincing evidence that the fraction of children who are nearsighted is different from 0.08.
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The nutrition label on a bag of potato chips says that a one ounce (28 gram) serving of potato chips has 130 calories and contains ten grams of fat, with three grams of saturated fat. A random sample of 35 bags yielded a confidence interval for the number of calories per bag of 128.2 to 139.8 calories. Is there evidence that the nutrition label does not provide an accurate measure of calories in the bags of potato chips?
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Because 130 is inside the confidence interval, we do not have convincing evidence that the true average is any different than what the nutrition label suggests.
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Define the term “sampling distribution” of the sample proportion, and describe how the shape, center, and spread of the sampling distribution change as the sample size increases when p = 0.1.
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The sampling distribution is the distribution of sample proportions from samples of the same size randomly sampled from the same population. As the same size increases, the shape of the sampling distribution (when p = 0.1) will go from being right-skewed to being more symmetric and resembling the normal distribution. With larger sample sizes, the spread of the sampling distribution gets smaller. Regardless of the sample size, the center of the sampling distribution is equal to the true mean of that population, provided the sampling isn't biased.
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Determine whether the following statement is true or false, and explain your reasoning: “With large sample sizes, even small differences between the null value and the observed point estimate can be statistically significant.”
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True. If the sample size gets ever larger, then the standard error will become ever smaller. Eventually, when the sample size is large enough and the standard error is tiny, we can find statistically significant yet very small differences between the null value and point estimate (assuming they are not exactly equal).
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Same observation, different sample size. Suppose you conduct a hypothesis test based on a sample where the sample size is n = 50, and arrive at a p-value of 0.08. You then refer back to your notes and discover that you made a careless mistake, the sample size should have been n = 500. Will your p-value increase, decrease, or stay the same? Explain.
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As the sample size increases the standard error will decrease, the sample statistic will increase, and the p-value will decrease.
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