The Marist Poll published a report stating that 66% of adults nationally think licensed drivers should be required to retake their road test once they reach 65 years of age. It was also reported that interviews were conducted on 1,018 American adults, and that the margin of error was 3% using a 95% condence level. (a) Verify the margin of error reported by The Marist Poll. First, we must verify the independence and normality of the distribution. Since the random sample represents less than a / 10% of the population, independence is a / satised. Since the sample size is 1,018, the success-failure condition is a / satised. The margin of error is therefore (in the final step round your answer to three decimal places) l ' 1 ' ME=Z. 9.1K n p) This is approximately equal to 3%, so the margin of error reported by The Marist Poll is verified. (b) Based on a 95% confidence interval, does the poll provide convincing evidence that more than 70% of the population think that licensed drivers should be required to retake their road test once they turn 65? (Round your answers to two decimal places.) A 95% confidence interval would be from to . Since this does not a / contain 70%, the poll does not a / provide convincing evidence that more than 70% of the population think that licensed drivers should be required to retake their road test once they turn 65. Greece has faced a severe economic crisis since the end of 2009. A Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that 25% of them said they would rate their lives poorly enough to be considered "suffering." (a) Describe the population parameter of interest. the proportion of all Greeks who say they were negatively affected by the economic crisis all Greeks 0 the proportion of all Greeks who would rate their lives as poorly enough to be considered "suffering" / What is the value of the point estimate of this parameter? (Enter your answer to two decimal places.) i; = 0.0165 x (b) Check if the conditions required for constructing a condence interval based on these data are met. The data must be independent / . Since the sample is a / random, and 1,000 represents less than a / 10% of all Greeks, this condition is met. The successifailure / condition must also be met. Since 1,000 x 0.25 = 250, which is greaterthan a / 10, and 1,000 x 0.75 = 750, which is greaterthan / 10, this condition is also met. (c) Construct a 95% confidence interval for the proportion of Greeks who are "suffering." (Round your answers to four decimal places.) (0.2522 X , 0.3078 x ) (d) Without doing any calculations, describe what would happen to the confidence interval if we decided to use a higher confidence level. 0 Increasing the condence level would increase the margin of error, and hence widen the interval. Increasing the condence level would increase the margin of error, and hence make the interval narrower. Increasing the condence level would decrease the margin of error, and hence make the interval narrower. Increasing the condence level would decrease the margin of error, and hence widen the interval. A random survey asked 1,600 US residents: "Do you think the use of marijuana should be made legal, or not?" 62% of the respondents said it should be made legal. (a) Is 62% a sample statistic or a population parameter? Explain. It is a population parameter, since it is a measure based on the population of US residents. 0 It is a sample statistic, since it is the observed sample proportion. (b) Construct a 95% confidence interval for the proportion of US residents who think marijuana should be made legal, and interpret the interval (in percent) in the context of the data. (Round your (C) answers to two decimal places.) We are 95% confident that approximately 0.4422 x % to 0.5223 x % of Americans think marijuana should be legalized. A critic points out that this 95% condence interval is only accurate if the statistic follows a normal distribution, or if the normal model is a good approximation. 15 this true for these data? Explain. The sample is / random and comprises lessthan a / 10% of the American population, therefore we can / assume that the individuals in this sample are independent of each other. The number of successes is and the number of failures is , so the successfailure condition is a / met. Therefore the distribution of the sample proportion is 8 / expected to be approximately normal. (d) A news piece on this survey's findings states, "Majority of Americans think marijuana should be legalized." Based on your condence interval, is this news piece's statement justied? Yes / , because the interval does not a / contain 50%, suggesting that the true proportion could a / be 50% or higher. A popular news agency covering a local election will declare a winner when their sample data provides enough evidence that the proportion of all voters who will vote for that candidate is a majority more than 50%. IA usas (a) State the appropriate null and alternative hypotheses. H0: p ? a 0.50 Ha: p 7 6 0.50 (b) If data from a random sample of 200 voters indicates that 108 of them are voting for a certain candidate, what value ofz is the test statistic? (Round your answer to two decimal places.) 2: (c) At a signicance level of 0.05, what decision does the resulting P-value lead you to make about this election? Because the P-value is less than our alpha level of 0.05, we reject the null hypothesis and conclude that there's enough evidence to confidently state that our candidate will have the support of a majority of voters. Declare this candidate the winner! 0 Because the Pvalue is greater than our alpha level of 0.05, we fail to reject the null hypothesis and conclude that there's not enough evidence to condently state that our candidate will have the support of a majority of voters. Do not declare this candidate the winner yet. Because the Pvalue is greater than our alpha level of 0.05, we reject the null hypothesis and conclude that there's enough evidence to condently state that our candidate will have the support of a majority of voters. Declare this candidate the winner! Because the P-value is less than our alpha level of 0.05, we fail to reject the null hypothesis and conclude that there's not enough evidence to confidently state that our candidate will have the support of a majority of voters. Do not declare this candidate the winner yet. Algebra students in a local high school use a computer-based program to brush up on algebra skills to be assessed by a test. The program claims that 80% of students using the program will pass the test. After an early pilot of the new program, only 33 of 50 students passed the test. A 1-proportion 2 test is going to be conducted to see if there's evidence that the true passing rate is less than what was claimed. The null hypothesis is p = 0.80. Ia USE SALT (a) What is the alternative hypothesis? Ha: p = 0.80 Ha: p > 0.80 Ha: p i 0.80 0 Ha: p 0.5 H0: p = 0.5 HA: p 0.5 H0: p at 0.5 HA: p = 0.5 0 H0: p = 0.5 HA: p 0.5 / Check the relevant conditions. The sample is a / random and the sample represents less than 6 / 10% of all Americans. Therefore whether or not one person in the sample decided not to go to college because they can't afford it is 8 / independent of another. The success-failure condition is / met since the number of successes and failures is greater than B / 10. Calculate the test statistic and determine the p-value. (Round your test statistic to two decimal places and your p-value to four decimal places.) 2 = p-value = Suppose a survey asked 821 randomly sampled registered voters in California \"Do you support? Or do you oppose? Drilling for oil and natural gas off the Coast of California? Or do you not know enough to say?" Below is the distribution of responses, separated based on whether or not the respondent graduated from college. (a) (b) College Grad Yes No Support 153 131 Oppose 179 125 Do not know 103 130 Total 435 386 What percent of college graduates in this sample do not know enough to have an opinion on drilling for oil and natural gas off the Coast of California? (Round your answer to two decimal places.) 23.68 J % What percent of the non-college graduates in this sample do not know enough to have an opinion on drilling for oil and natural gas off the Coast of California? (Round your answer to two decimal places.) 33.68 / % Conduct a hypothesis test to determine if the data provide strong evidence that the proportion of college graduates who do not have an opinion on this issue is different than that of non-college graduates. (Use a significance level of 0.05. Use college graduates non-college graduates for your test.) State the null and alternative hypothesis. (Use the subscripts 1 for college graduates and 2 for non-college graduates. Enter l: for s as needed.) Ho: P1 =P2 x HA: P1 =P2 x Check the relevant conditions. Both samples are a / random and the samples represent lessthan a / 10% of their populations, so independence is a / satised. The success-failure condition is / met since the number of successes and failures in each group is greater than / 10. State the test statistic. (Round your answer to two decimal places.) -3.18 X State the p-value. (Round your answer to four decimal places.) Consider the following data on sleep deprivation rates of Californians and Oregonians. The proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 9.0%, while this proportion is 9.6% for Oregon residents. These data are based on simple random samples of 11,556 California and 4,691 Oregon residents. (Use a significance level of 0.05. Use I304 _ 70,2-) (a) Conduct a hypothesis test to determine if these data provide strong evidence that the rate of sleep deprivation is different for the two states. Check the relevant conditions. The sample is / random and the sample represents less than a / 10% of all California and Oregon residents. Therefore whether or not one person in the sample reported insufcient rest or sleep is a / independent of another. The success-failure condition is 8 / met since the number of successes and failures is greater than 8 / 10. State the appropriate null and alternative hypotheses. 0 Ho: \"CA = \"0R HA: \"CA * \"0R Ho: \"CA * \"0R HA: \"CA = \"0R \"0' pCA 5 \"0R HA: \"CA > \"0R Ho: \"CA 5 \"0R HA: \"CA * \"0R Ho: \"CA > \"on HA: \"CA S \"0R / Calculate the test statistic and determine the p-value. (Round your test statistic to two decimal places and your p-value to four decimal places.) 2 = -1.20 / p-value = 0.2303 x Interpret the p-value in context of the hypothesis test and the data. Reject Ho. The data provide convincing evidence that the rate of sleep deprivation is different for the two states. 0 Fail to reject H0. The data do not provide convincing evidence that the rate of sleep deprivation is different for the two states. Reject Ho. The data do not provide convincing evidence that the rate of sleep deprivation is different for the two states