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A 2014 study by the reputable Gallup organization estimates that 44% of U.S. adults are underemployed. Underemployed means the person wants to work full time but is employed part time or unemployed. We want to know if the proportion is smaller this year. We select a random sample of 100 U.S. adults this year and find that 40% are underemployed. After carrying out the hypothesis test for p = 0.44 compared to p < 0.44, we obtain a P-value of 0.21. Which of the following interpretations of the P-value is correct? - ANSWER-There is a 21% chance that a sample of 100 U.S. adults will have 40% or fewer underemployed if 44% of the population is underemployed this year. A group of 71 college students from a certain liberal arts college were randomly sampled and asked about the number of alcoholic drinks they have in a typical week. The purpose of this study was to compare the drinking habits of the students at the college to the drinking habits of college students in general. In particular, the dean of students, who initiated this study, would like to check whether the mean number of alcoholic drinks that students at his college in a typical week differs from the mean of U.S. college students in general, which is estimated to be 4.73. The group of 71 students in the study reported an average of 3.91 drinks per with a standard deviation of 3.88 drinks. Find the p-value for the hypothesis test - ANSWER- A medical researcher is studying the effects of a drug on blood pressure. Subjects in the study have their blood pressure taken at the beginning of the study. After being on the medication for 4 weeks, their blood pressure is taken again. The change in blood pressure is recorded and used in doing the hypothesis test. Change: Final Blood Pressure - Initial Blood Pressure The researcher wants to know if there is evidence that the drug affects blood pressure. At the end of 4 weeks, 36 subjects in the study had an average change in blood pressure of 2.4 with a standard deviation of 4.5. Find the p-value for the hypothesis test. - ANSWER- A politician claims that a larger proportion of members of the news media are Democrats when compared to the general public. Let p1 represent the proportion of the news media that is Democrat and p2 represent the proportion of the public that is Democrat. What are the appropriate null and alternative hypotheses that correspond to this claim? - ANSWER-H0: p1 - p2 = 0; Ha: p1 - p2 > 0 A quality control engineer at a potato chip company tests the bag filling machine by weighing bags of potato chips. Not every bag contains exactly the same weight. But if more than 15% of bags are over-filled then they stop production to fix the machine. They define over-filled to be more than 1 ounce above the weight on the package. The engineer weighs 100 bags and finds that 21 of them are over-filled. He plans to test the hypotheses H0:p=0.15 versus Ha:p>0.15.
What is the test statistic? - ANSWER-z = 1.68 (multiple choice) A quality control engineer at a potato chip company tests the bag filling machine by weighing bags of potato chips. Not every bag contains exactly the same weight. But if more than 15% of bags are over-filled then they stop production to fix the machine. They define over-filled to be more than 1 ounce above the weight on the package. The engineer weighs 102 bags and finds that 32 of them are over-filled. He plans to test the hypotheses H0: p = 0.15 versus Ha: p > 0.15. What is the test statistic? - ANSWER- z= A researcher conducts an experiment on human memory and recruits 15 people to participate in her study. She performs the experiment and analyzes the results. She uses a t-test for a mean and obtains a p-value of 0.17. Which of the following is a reasonable interpretation of her results? - ANSWER-If there is a treatment effect, the sample size was too small to detect it. A scientist claims that a smaller proportion of members of the National Academy of Sciences are women when compared to the proportion of women nationwide. Let p represent the proportion of women in the National Academy of Sciences and p represent the proportion of women nationwide. Which is the correct alternative hypotheses that corresponds to this claim? - ANSWER-Ha: p1 - p2 < 0 A teacher is experimenting with computer-based instruction. In which situation could the teacher use a hypothesis test for a population mean? - ANSWER-She gives each student a pretest. Then she teaches a lesson using a computer program. Afterwards, she gives each student a posttest. The teacher wants to see if the difference in scores will show an improvement. A tire manufacturer has a 60,000 mile warranty for tread life. The manufacturer considers the overall tire quality to be acceptable if less than 5% are worn out at 60, miles. The manufacturer tests 250 tires that have been used for 60,000 miles. They find that 3.6% of them are worn out. With this data, we test the following hypotheses. H0: The proportion of tires that are worn out after 60,000 miles is equal to 0.05. Ha: The proportion of tires that are worn out after 60,000 miles is less than 0.05. In order to assess the evidence, which question best describes what we need to determine? - ANSWER-NEED ANSWER Incorrect. In a hypothesis test we are trying to estimate the chance that random sampling produces sample results are as or more extreme than the result observed in the data if the null hypothesis is true. The statement you chose asks, in essence, "What is the chance that the population proportion is less than the null value?" In other words, the statement you chose asks, "What is the chance that the alternative hypothesis is true?" But a hypothesis test cannot determine this. A tire manufacturer has a 60,000 mile warranty for tread life. The manufacturer considers the overall tire quality to be acceptable if less than 5% are worn out at 60, miles. The manufacturer tests 250 tires that have been used for 60,000 miles. They find
Child Health and Development Studies (CHDS) has been collecting data about expectant mothers in Oakland, CA since 1959. One of the measurements taken by CHDS is the weight increase (in pounds) for expectant mothers in the second trimester. In a fictitious study, suppose that CHDS finds the average weight increase in the second trimester is 14 pounds. Suppose also that, in 2015, a random sample of 41 expectant mothers have mean weight increase of 15.4 pounds in the second trimester, with a standard deviation of 5.8 pounds. A hypothesis test is done to see if there is evidence that weight increase in the second trimester is greater than 14 pounds. Find the p-value for the hypothesis test - ANSWER-. Child Health and Development Studies (CHDS) has been collecting data about expectant mothers in Oakland, CA since 1959. One of the measurements taken by CHDS is the weight increase (in pounds) for expectant mothers in the second trimester. In a fictitious study, suppose that CHDS finds the average weight increase in the second trimester is 14 pounds. Suppose also that, in 2015, a random sample of 40 expectant mothers have mean weight increase of 16 pounds in the second trimester, with a standard deviation of 6 pounds. At the 5% significance level, we can conduct a one-sided T-test to see if the mean weight increase in 2015 is greater than 14 pounds. Statistical software tells us that the p-value = 0.021. Which of the following is the most appropriate conclusion? - ANSWER- College students and STDs: A recent report estimated that 25% of all college students in the United States have a sexually transmitted disease (STD). Due to the demographics of the community, the director of the campus health center believes that the proportion of students who have a STD is lower at his college. He tests H0: p = 0.25 versus Ha: p < 0.25. The campus health center staff select a random sample of 50 students and determine that 18% have been diagnosed with a STD. Is the sample size condition for conducting a hypothesis test for a population proportion satisfied? - ANSWER-Yes, because (50)(.25) and (50)(1 - 0.25) are both at least 10. This means we can use the normal distribution to model the distribution of sample proportions. Commute times in the U.S. are heavily skewed to the right. We select a random sample of 200 people from the 2000 U.S. Census who reported a non-zero commute time. In this sample the mean commute time is 27.5 minutes with a standard deviation of 18. minutes. Can we conclude from this data that the mean commute time in the U.S. is less than half an hour? Conduct a hypothesis test at the 5% level of significance. What is the p-value for this hypothesis test? - ANSWER- Dean Halverson recently read that full-time college students study 20 hours each week. She decides to do a study at her university to see if there is evidence to show that this is not true at her university. A random sample of 32 students were asked to keep a diary of their activities over a period of several weeks. It was found that the average number
of hours that the 32 students studied each week was 19.1 hours. The sample standard deviation of 4.2 hours. Find the p-value. - ANSWER- Does secondhand smoke increase the risk of a low weight birth? A baby is "low birth weight" if it weighs less than 5.5 pounds at birth. According to the National Center of Health Statistics, about 7.8% of all babies born in the U.S. are categorized as low birth weight. Researchers randomly select 1200 babies whose mothers had extensive exposure to secondhand smoke during pregnancy. 10.4% of the sample are categorized as low birth weight. Which of the following are the appropriate null and alternative hypotheses for this research question - ANSWER-H0: p = 0.078; Ha: p > 0.078 MAYBE Every simulation in this module is based on an assumption about the difference between two population proportions. The population proportions affect the mean and the standard error of the differences in sample proportions. The sample size also affects the standard error. The distribution of differences between sample proportions shown below has mean 0.35, and a standard error of about 0.10. Which of the following did we use to generate this sampling distribution? - ANSWER- Population proportions of 0.85 and 0.50 with samples of size 35. Facebook friends: According to Facebook's self-reported statistics, the average Facebook user has 130 Facebook friends. For a statistics project a student at Contra Costa College (CCC) tests the hypothesis that CCC students will average more than 130 Facebook friends. She randomly selects 3 classes from the schedule of classes and distributes a survey in these classes. Her sample contains 45 students. From her survey data she calculates that the mean number of Facebook friends for her sample is: ¯x= 138.7 with a standard deviation of: s=79.3. She chooses a 5% level of significance. What can she conclude from her data? - ANSWER-We cannot conclude that the average number of Facebook friends for CCC students is greater than 130. The sample mean of 138.7 is not significantly greater than
Find the p-value for the hypothesis test. A random sample of size 50 is taken. The sample has a mean of 420 and a standard deviation of 81. H0: μ = 400 Ha: μ ≠ 400 The p-value for the hypothesis test is - ANSWER- Find the p-value for the hypothesis test. A random sample of size 51 is taken. The sample has a mean of 392 and a standard deviation of 83. H0: μ = 400 Ha: μ< 400 The p-value for the hypothesis test is - ANSWER-
In 2011, the Institute of Medicine (IOM), a non-profit group affiliated with the US National Academy of Sciences, reviewed a study measuring bone quality and levels of vitamin-D in a random sample from bodies of 675 people who died in good health. 8.5% of the 82 bodies with low vitamin-D levels (below 50 nmol/L) had weak bones. Comparatively, 1% of the 593 bodies with regular vitamin-D levels had weak bones. Is a normal model a good fit for the sampling distribution? - ANSWER- In 2014, students in an advanced Statistics course at UC Berkeley conducted an anonymous survey about use of cognition-enhancing drugs among college males. One survey group of males included members from a fraternity, and the other survey of males group included no fraternity members. The graph below represents the sampling distribution of differences between the proportion of fraternity members who have used cognition-enhancing drugs and the proportion of non-fraternity members who have used cognition-enhancing drugs. The difference in proportions is p1−p2=0.05 (fraternity members minus non-fraternity members). Consider simulating this data with 5000 samples from each of the fraternity member group and the non-fraternity member group. Based on the sampling distribution of difference in proportions, which of the following results for p1−p2 would be most unusual? - ANSWER-0. In 2014, students in an advanced Statistics course at UC Berkeley conducted an anonymous survey about use of cognition-enhancing drugs among college males. One survey group of males included members from a fraternity, and the other survey of males group included no fraternity members. The standard error formula for the difference between sample proportions is Calculate the standard error for a survey comparing proportions of cognition-enhancing drug use of fraternity members to non-fraternity members, where p1 = 0.31, n1 = 148, p2 = 0.21, n2 = 94. Round all calculations to the thousandth decimal place. - ANSWER- In 2014, students in an advanced Statistics course at UC Berkeley conducted an anonymous survey about use of cognition-enhancing drugs among college males. One survey group of males included members from a fraternity, and the other survey of males group included no fraternity members. We can use a simulation model to estimate the true difference between the proportion of fraternity members who have used cognition-enhancing drugs and the proportion of non-fraternity members who have used cognition-enhancing drugs. The graph below represents the sampling distribution of differences. The difference in proportions is p1 - p2 = 0.05 (fraternity members minus non-fraternity members), and the standard error is 0.08. The z-score for this difference in sample proportions is 0.885. Which of the following is the most appropriate conclusion? - ANSWER-There is not a statistically significant difference between the proportion of fraternity members who
have used cognition-enhancing drugs and the proportion of non-fraternity members who have used cognition-enhancing drugs. In 2014, students in an advanced Statistics course at UC Berkeley conducted an anonymous survey about use of cognition-enhancing drugs among college males. One survey group of males included members from a fraternity, and the other survey of males group included no fraternity members. The difference between the proportion of fraternity members who have used cognition-enhancing drugs and the proportion of non-fraternity members who have used cognition-enhancing drugs is equal to 0.054. Which one of the following conclusions is true? - ANSWER-One possibility is that the proportion of fraternity members who have used cognition-enhancing drugs is 0.295 and the proportion of non-fraternity members who have used cognition-enhancing drugs is 0.241. In a fictional study, suppose that a psychologist is studying the effect of daily meditation on resting heart rate. The psychologist believes the patients who not meditate have a higher resting heart rate. For a random sample of 45 pairs of identical twins, the psychologist randomly assigns one twin to one of two treatments. One twin in each pair meditates daily for one week, while the other twin does not meditate. At the end of the week, the psychologist measures the resting heart rate of each twin. Assume the mean resting heart rate is 80 heart beats per minute. The psychologist conducts a T-test for the mean of the differences in resting heart rate of patients who do not meditate minus resting heart rate of patients who do meditate. Which of the following is the correct null and alternative hypothesis for the psychologist's study? - ANSWER-H0: μ = 0; Ha: μ > 0 In a study at West Virginia University Hospital, researchers investigated smoking behavior of cancer patients to create a program to help patients stop smoking. They published the results in Smoking Behaviors Among Cancer Survivors (January 2009 issue of the Journal of Oncology Practice.) In this study, the researchers sent a 22-item survey to 1,000 cancer patients. They collected demographic information (age, sex, ethnicity, zip code, level of education), clinical and smoking history, and information about quitting smoking. The questionnaire filled out by cancer patients at West Virginia University Hospital also asked patients if they were current smokers. The current smoker rate for female cancer patients was 11.6%. 95 female respondents were included in the analysis. For male cancer patients, the current smoker rate was 10.4%, and 67 male respondents were included in the analysis. Suppose that these current smoker - ANSWER-No, a normal model is not a good fit for this sampling distribution. In a study at West Virginia University Hospital, researchers investigated smoking behavior of cancer patients to create a program to help patients stop smoking. They published the results in Smoking Behaviors Among Cancer Survivors (January 2018, Journal of Oncology Practice). In this study, the researchers sent a 22-item survey to 1299 cancer patients. They collected demographic information (age, sex, ethnicity, zip
Correct. The counts of successes (answered "Yes") and failures (answered "No") must be at least 10 for a normal model to be a good fit. Both Democrats and Republicans have large sample sizes, with n1=3266 and n2=2137 people. The calculations n1p1, n1(1-p1), n2p2, and n2(1-p2) are all greater than 10. In the article "Attitudes About Marijuana and Political Views" (Psychological Reports, 1973), researchers reported on the use of cannabis by liberals and conservatives during the 1970s. To test the claim (at 1% significance) that the proportion of voters who smoked cannabis frequently was lower among conservatives, the hypotheses were In this hypothesis test which of the following errors is a Type II error? - ANSWER-We conclude that there is no difference between the proportions of conservatives and liberals that smoke cannabis, when the proportion is actually lower for conservatives. In the article "Attitudes About Marijuana and Political Views" (Psychological Reports, 1973), researchers reported on the use of cannabis by liberals and conservatives during the 1970s. To test the claim (at 1% significance) that the proportion of voters who smoked cannabis frequently was lower among conservatives, the hypotheses were Suppose that we conduct a hypothesis test in which a Type II error is very serious. But the Type I error is not very serious. Which level of significance is the best choice? - ANSWER-α = 0. In the article "Coffee, Caffeine, and Risk of Depression Among Women" in the September 2011 edition of the Archives of Internal Medicine, researchers investigated the relationship between caffeine consumption and depression among women. The participants in this study were older, with substantially lower rates of depression when compared to female teens. Researchers compared two groups of women (among others) in this study: those who do not drink coffee and those who routinely drink 4 or more cups of coffee each day. For the following question, a coffee drinker is a woman who drinks four or more cups each day. One of the graphs below represents the sampling distribution of differences between the sample proportions for depressed coffee and non-coffee drinking women. Under the assumption that both women who drink coffee and do not drink coffee have a 6% depression rate, which distribution of differences in sample propo - ANSWER-Graph A- last number is. In the article Foods, Fortificants, and Supplements: Where Do Americans Get Their Nutrients? researchers analyze the nutrient and vitamin intake from a random sample of 16,110 U.S. residents. Researchers compare the level of daily vitamin intake for vitamin A, vitamin B-6, vitamin B-12, vitamin C, vitamin D, vitamin E and calcium. Unless otherwise stated, all hypothesis tests in the study are conducted at the 5% significance level. For the claim that the proportion of U.S. residents who consumed recommended levels of vitamin A is higher among women than men, the null and alternative hypotheses are:
The p-value is 0.08, and researchers conduct this test at a 5% level of significance. Which of the following is the correct conclusion? - ANSWER-Fail to Reject H0 , do not support Ha. In the article Foods, Fortificants, and Supplements: Where Do Americans Get Their Nutrients? researchers analyze the nutrient and vitamin intake from a random sample of 16,110 U.S. residents. Researchers compare the level of daily vitamin intake for vitamin A, vitamin B-6, vitamin B-12, vitamin C, vitamin D, vitamin E and calcium. Unless otherwise stated, all hypothesis tests in the study are conducted at the 5% significance level. To test the claim (at 5% significance) that the proportion of U.S. residents who consume recommended levels of vitamin A is higher among women than men, researchers set up the following hypotheses: In this hypothesis test which of the following errors is a Type I error? - ANSWER- Researchers conclude that a larger proportion of women consume the recommended daily intake of vitamin A when there is actually no difference between vitamin A consumption for women and men. In the article, Attitudes About Marijuana and Political Views (Psychological Reports, 1973), researchers reported on the use of cannabis by liberals and conservatives during the 1970's. To test the claim (at 1% significance) that the proportion of voters who smoked cannabis frequently was lower among conservatives, the hypotheses were In the hypothesis test about cannabis use by conservatives and liberals, the test statistic was z = -4.27, with a corresponding p-value of about 0.00001. Which conclusion is most appropriate in the context of this situation? - ANSWER-The data support the claim that a lower proportion of conservatives smoke cannabis when compared to liberals. maybe?? Living with parents: A Pew Research analysis stated that in 2012, 36.6% of the nation's young adults ages 18-31—the so-called Millennial generation—were living in their parents' home. After reading the analysis, a statistics student wanted to design a study to determine if the percentage was higher for the Millennial students who attend his college. Which of the following is an appropriate statement of the null hypothesis? - ANSWER- The percentage of Millennial students at his college who live in their parents' home is the same as the percentage of Millennials nationwide, i.e., H0: p = 36.6% Correct. We assume that the percentage of Millennial students at his college who live in their parents' home is the same as the percentage of Millennials nationwide. Thus H states that the percentage is 36.6%.
Based on the hypotheses above, which of the following statements is considered aType I error? - ANSWER-The student is not pregnant, but the test result shows she is pregnant. Pregnancy testing: A college student hasn't been feeling well and visits her campus health center. Based on her symptoms, the doctor suspects that she is pregnant and orders a pregnancy test. The results of this test could be considered a hypothesis test with the following hypotheses: H0: The student is not pregnant Ha: The student is pregnant. Based on the hypotheses above, which of the following statements is considered a Type II error? - ANSWER-The student is pregnant, but the test result shows she is not pregnant. Quit Smoking: Previous studies suggest that use of nicotine-replacement therapies and antidepressants can help people stop smoking. The New England Journal of Medicine published the results of a double-blind, placebo-controlled experiment to study the effect of nicotine patches and the antidepressant bupropion on quitting smoking. The target for quitting smoking was the 8th day of the experiment. In this experiment researchers randomly assigned smokers to treatments. Of the 162 smokers taking a placebo, 28 stopped smoking by the 8th day. Of the 272 smokers taking only the antidepressant buproprion, 82 stopped smoking by the 8th day. Calculate the 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment). (The standard error is about 0.0407. Use critical value z = 2.576.) (, ) - ANSWER- Quit Smoking: Previous studies suggest that use of nicotine-replacement therapies and antidepressants can help people stop smoking. The New England Journal of Medicine published the results of a double-blind, placebo-controlled experiment to study the effect of nicotine patches and the antidepressant bupropion on quitting smoking. The target for quitting smoking was the 8th day of the experiment. In this experiment researchers randomly assigned smokers to treatments. Of the 162 smokers taking a placebo, 31 stopped smoking by the 8th day. Of the 256 smokers taking only the antidepressant buproprion, 81 stopped smoking by the 8th day. Calculate the estimated standard error for the sampling distribution of differences in sample proportions. The estimated standard error = - ANSWER- Quit Smoking: The New England Journal of Medicine published the results of a double- blind, placebo-controlled experiment to study the effect of nicotine patches and the antidepressant bupropion on quitting smoking. With the data from the experiment we calculate the sample difference in the "quit smoking" rates for the nicotine treatment group and the placebo group ("treatment" minus "placebo"). We get 0.8% = 0.008. Which of the following is an appropriate conclusion based on this finding? - ANSWER-In this experiment the nicotine treatment
had a higher success rate than the placebo group, but the improvement was less than 1%. Quit Smoking:The New England Journal of Medicine published the results of a double- blind, placebo-controlled experiment to study the effect of nicotine patches study the effect of nicotine patches versus a placebo on quitting smoking. With the data from the experiment we calculate the sample difference in the "quit smoking" rates for the nicotine treatment group and the placebo group ("treatment" minus "placebo"). We get 0.8%=0.008. The 99% confidence interval based on this sample difference is -0.1036 to 0.0882. We also calculate a 90% confidence interval. How does the 90% confidence interval compare to the 99% confidence interval? For a 90% confidence interval, the margin of error (MOE) will: - ANSWER-Decrease Researchers conducted a randomized, double-blind clinical trial to compare the herb St. John's Wort, the antidepressant drug sertraline, and a placebo for the treatment of depression. Of the 113 patients in the herb treatment group, 23.9% showed improvement, compared to 24.8% of the 113 in the sertraline group and 31.9% of the 113 patients in the placebo group. Medscape Medical News published this research in
Is a normal model a good fit for the sampling distribution? - ANSWER-Yes, there are at least 10 people who improved and 10 who did not improve in each randomized treatment group. Short-term classes: Does taking a class in a short-term format (8 weeks instead of 16 weeks) increase a student's likelihood of passing the course? For a particular course, the pass rate for the 16-week format is 59%. A team of faculty examine student data from 40 randomly selected accelerated classes and determine that the pass rate is 78%. Which of the following are the appropriate null and alternative hypotheses for this research question? - ANSWER-H0: p = 0.59; Ha: p > 0. Skittles: The popular Skittles candy comes in 5 colors. According to the Skittles website, the colors are evenly distributed in the population of Skittle candies. So each color makes up 20% of the population. Suppose that we buy 2 small bags of Skittles. We determine the percentage of green Skittles in one bag and the percentage of orange Skittles in the other bag. Suppose that each bag contains 40 candies. We define the difference in sample proportions as "green" minus "orange." Which of the following will give a sample difference of 0.10? - ANSWER-10 green and 6 orange Students in a discussion of gun control in a sociology class at Foothill Community College argue that Republicans are more likely to oppose gun control than Independents. They use data from an article titled "Gun Control Splits America," published March 23, 2010 in pewresarch.org by the Pew Research Center for the
The Food and Drug Administration (FDA) is a U.S. government agency that regulates (you guessed it) food and drugs for consumer safety. One thing the FDA regulates is the allowable insect parts in various foods. You may be surprised to know that much of the processed food we eat contains insect parts. An example is flour. When wheat is ground into flour, insects that were in the wheat are ground up as well. The mean number of insect parts allowed in 100 grams (about 3 ounces) of wheat flour is 75. If the FDA finds more than this number, they conduct further tests to determine if the flour is too contaminated by insect parts to be fit for human consumption. The null hypothesis is that the mean number of insect parts per 100 grams is 75. The alternative hypothesis is that the mean number of insect parts per 100 grams is greater than 75. Is the following a Type I error or a Type II error or neither? The test fails to sho - ANSWER-Type II error The histogram below is from a simulation of the difference between two population proportions. The mean and the standard error are calculated using the population proportion values and the sample size values. Which of the following is the best estimate of the population proportions? - ANSWER- The population proportions are 0.70 and 0.55. The histogram below is from a simulation of the difference between two population proportions. The mean and the standard error are calculated using the population proportion values and the sample size values. Which of the following is the best estimate of the standard error? - ANSWER-The standard error is 0.05. Watches and bacteria: A group of researchers investigated the contamination of medical personnel watches at a New York hospital, since there is a potential for patient exposure to potentially dangerous bacteria. They sampled watches worn by physicians, physician assistants, and medical students at a teaching hospital in New York. Nearly half (46.6%) of the watches tested harbored microorganisms that can cause illness. By comparison, only one of the 10 watches worn by security guards tested positive for a disease-carrying microorganism. The researchers want to determine if the difference is statistically significant. Which of the following is an appropriate statement of the null hypothesis, H0? - ANSWER-The proportion of contaminated wrist-watches from medical personnel is the same as the proportion of contaminated wrist-watches from security guards, i.e., H0: p = 46.6%. We conduct a study to determine whether the majority of community college students plan to vote in the next presidential election. We choose a significance level of 0.05. We survey 650 randomly selected community college students and find that 54% of them plan to vote. The p-value is 0.02. H0: 50% of community college students plan to vote in the next presidential election. Ha: More than 50% of community college students plan to vote in the next presidential election.
What can we conclude? - ANSWER-The evidence suggests that the majority of community colleges students plan to vote in the next presidential election because the p-value is less than the significance level.
