Test 4 Stats Help

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.)

(-.233,-.024)

The difference between teenage female and male depression rates estimated from two samples is 0.06. The estimated standard error of the sampling distribution is 0.04. What is the 95% confidence interval? Use the critical value z = 1.96.

-.02,.14

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

-.233,-.024

Of the 1303 patients who were mailed surveys, 261 patients responded. For various reasons, researchers used only 235 of the completed surveys. 31 out of 121 female cancer patients reported being past smokers, and 59 out of 114 male cancer patients reported being past smokers. Calculate the difference between the corresponding sample proportions ^p1−^p2 (female minus male). Round the answer to 4 decimal places.

-.2613

Among Cancer Survivors (January 2018, Journal of Oncology Practice). In this study, the researchers sent a 22-item survey to 1282 cancer patients. They collected demographic information (age, sex, ethnicity, zip code, level of education), clinical and smoking history, and information about quitting smoking. Of the 1282 patients who were mailed surveys, 256 patients responded. For various reasons, researchers used only 231 of the completed surveys. 35 out of 131 female cancer patients reported being past smokers, and 55 out of 100 male cancer patients reported being past smokers. Calculate the difference between the corresponding sample proportions ^p1−^p2 (female minus male). Round the answer to 4 decimal plac

-.2828

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 1400 cancer patients. They collected demographic information (age, sex, ethnicity, zip code, level of education), clinical and smoking history, and information about quitting smoking. Of the 1400 patients who were mailed surveys, 280 patients responded. For various reasons, researchers used only 252 of the completed surveys. 34 out of 132 female cancer patients reported being past smokers, and 67 out of 120 male cancer patients reported being past smokers. Calculate the difference between the corresponding sample proportions ^p1−^p2 (female minus male). Round the answer to 4 decimal places.

-.3008

Of the 1406 patients who were mailed surveys, 281 patients responded. For various reasons, researchers used only 253 of the completed surveys. 34 out of 135 female cancer patients reported being past smokers, and 73 out of 118 male cancer patients reported being past smokers. Calculate the difference between the corresponding sample proportions ^p1−^p2 (female minus male). Round the answer to 4 decimal places.

-.3668

The difference between teenage female and male depression rates estimated from two samples is 0.06. The estimated standard error of the sampling distribution is 0.03. What is the 95% confidence interval? Use the critical value z = 1.96

.00,.12

The difference between teenage female and male depression rates estimated from two samples is 0.08. The estimated standard error of the sampling distribution is 0.04. What is the 95% confidence interval? Use the critical value z = 1.96.

.00,.16

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. Medical researchers are conducting a double-blind experiment treating high blood pressure patients with a new vitamin supplement. The researchers are conducting a hypothesis test in which a Type I error is very serious, and the Type II error is not very serious. Which level of significance is the best choice? α = 0.01 α = 0.05 α = 0.10

.01

Medical researchers are conducting a double-blind experiment treating high blood pressure patients with a new vitamin supplement. The researchers are conducting a hypothesis test in which a Type I error is very serious, and the Type II error is not very serious. Which level of significance is the best choice? α = 0.01 α = 0.05 α = 0.10

.01

In March 2015, the Public Policy Institute of California (PPIC) surveyed 7,525 likely voters living in California. In the survey, respondents were asked about global warming. PPIC results show that 47% of adults age 55 and older view global warming as a serious problem, and 65% of adults age 18 to 34 view global warming as a serious problem. PPIC researchers are interested in the difference in proportions of adults age 55 and over and adults age 18 to 34. The standard error for the difference in proportions is 0.011. Which of the following is the margin of error for a 95% confidence interval?

.021

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 190 smokers taking a placebo, 21 stopped smoking by the 8th day. Of the 264 smokers taking only the antidepressant buproprion, 76 stopped smoking by the 8th day. Calculate the estimated standard error for the sampling distribution of differences in sample proportions. The estimated standard error

.036

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 164 smokers taking a placebo, 26 stopped smoking by the 8th day. Of the 284 smokers taking only the antidepressant buproprion, 77 stopped smoking by the 8th day. Calculate the estimated standard error for the sampling distribution of differences in sample proportions. The estimated standard error

.039

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 172 smokers taking a placebo, 27 stopped smoking by the 8th day. Of the 257 smokers taking only the antidepressant buproprion, 89 stopped smoking by the 8th day. Calculate the estimated standard error for the sampling distribution of differences in sample proportions. The estimated standard error

.041

Calculate the standard error for a survey comparing proportions of cognition-enhancing drug use of fraternity members to non-fraternity members, where p1 = 0.34, n1 = 141, p2 = 0.22, n2 = 95. Round all calculations to the thousandth decimal place.

.058

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 group of males 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 = 126, p2 = 0.26, n2 = 93. Round all calculations to the thousandth decimal place.

.061

Of the 1303 patients who were mailed surveys, 261 patients responded. For various reasons, researchers used only 235 of the completed surveys. 31 out of 121 female cancer patients reported being past smokers, and 59 out of 114 male cancer patients reported being past smokers. Calculate the difference between the corresponding sample proportions ^p1−^p2 (female minus male). Round the answer to 4 decimal places.

.061

Calculate the standard error for a survey comparing proportions of cognition-enhancing drug use of fraternity members to non-fraternity members, where p1 = 0.3, n1 = 110, p2 = 0.26, n2 = 97. Round all calculations to the thousandth decimal place.

.062

Calculate the standard error for a survey comparing proportions of cognition-enhancing drug use of fraternity members to non-fraternity members, where p1 = 0.33, n1 = 128, p2 = 0.26, n2 = 91. Round all calculations to the thousandth decimal place.

.062

Gardeners on the west coast of the United States are investigating the difference in survival rates of two flowering plants in drought climates. Plant A has a survival rate of 0.68 and plant B has a survival rate of 0.46. The standard error of the difference in proportions is 0.082. What is the margin of error for a 99% confidence interval? Use critical value z = 2.576. MOE =

.211

Gardeners on the west coast of the United States are investigating the difference in survival rates of two flowering plants in drought climates. Plant A has a survival rate of 0.73 and plant B has a survival rate of 0.42. The standard error of the difference in proportions is 0.088. What is the margin of error for a 99% confidence interval? Use critical value z = 2.576. MOE=

.227

Gardeners on the west coast of the United States are investigating the difference in survival rates of two flowering plants in drought climates. Plant A has a survival rate of 0.78 and plant B has a survival rate of 0.42. The standard error of the difference in proportions is 0.098. What is the margin of error for a 99% confidence interval? Use critical value z = 2.576

.252

Which of the following is the best estimate of the population proportions?

.7,.55

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? 10 green and 6 orange 4 green and 8 orange 10 green and 10 orange

10 green and 6 orange

95% 99.7

2 stdeb 3

Find p value from t

Double

Ho: p1-p2=0 (p1=p2) Ha: p1-p2>0 p1>p2 The p-value is 0.06. If we conduct this test at a 5% level of significance, what would be an appropriate conclusion? Reject H0 , and support Ha . Support H0 , and reject Ha . Fail to Reject H0 , do not support Ha . Feedback

Fail to Reject H0 , do not support Ha .

The p-value is 0.06. If we conduct this test at a 5% level of significance, what would be an appropriate conclusion?

Fail to Reject H0 , do not support Ha .

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? H0: p1 - p2 = 0; Ha: p1 - p2 < 0 H0: p1 - p2 = 0; Ha: p1 - p2 > 0 H0: p1 - p2 = 0; Ha: p1 - p2 ≠ 0

H0: p1 - p2 = 0; Ha: p1 - p2 > 0

If p1 and p2 are the proportions of conservatives and liberals, respectively, who smoke cannabis, what are the appropriate null and alternative hypotheses?H0: p1− p2= 0 and Ha: p1− p2≠ 0.H0: p1− p2= 0 and Ha: p1− p2< 0.H0: p1− p2= 0 and Ha: p1− p2> 0.

H0: p1− p2= 0 and Ha: p1− p2< 0.

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 p1 represent the proportion of women in the National Academy of Sciences and p2 represent the proportion of women nationwide. Which is the correct alternative hypotheses that corresponds to this claim? Ha: p1 - p2 < 0 Ha: p1 - p2 > 0 Ha: p1 - p2 ≠ 0

Ha: p1 - p2 < 0

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. Researchers conduct a hypothesis test to determine if the proportion of U.S. residents consuming recommended levels of calcium is different among women and men. The p-value is 0.035, and researchers conduct this test at a 5% level of significance. What does a p-value of 0.035 mean? There is a 3.5% chance that there is no difference in calcium consumption for women and men. If the calcium consumption rates are different for the women and men, there is a 3.5% chance that future experiments will show the same difference in calcium consumption as observed in this experiment. If calcium consumption is the same for women and men, there is a 3.5% chance that future studies will show differences in calcium consumption greater than observed in this study. If calcium consumption is the same for women and men, there is a 3.5% chance that calcium consumption will be different in future experiments.

If calcium consumption is the same for women and men, there is a 3.5% chance that future studies will show differences in calcium consumption greater than observed in this study.

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? 1. n this experiment the nicotine treatment had a higher success rate than the placebo group, but the improvement was less than 1%. 2. In this experiment, the placebo group had a higher success rate than the nicotine group by 0.8%. 3. Nicotine patches will produce a slightly higher success rate when compared to a placebo, but the difference is not statistically significant. 4. Nicotine patches and the antidepressant bupropion work equally well on "quit smoking" rates.

In this experiment the nicotine treatment had a higher success rate than the placebo group, but the improvement was less than 1%.

he Public Policy Institute of California (PPIC) surveyed 7,525 likely voters living in California. In the survey, respondents were asked about global warming. PPIC researchers are interested in the difference in viewpoints across racial/ethnic groups. PPIC results show that 75% of Latinos view global warming as a serious problem, and 46% of whites view global warming as a serious problem. Using the data from the survey, we calculate the sample difference in global warming viewpoints for Latino respondents and white respondents to be 29% = 0.29. Researchers can use the 29% sample difference to draw a conclusion about which populations? 1. Latinos and whites living in the United States who are likely voters 2. Latinos and whites who live in California 3. Latinos and whites living in California who are likely to vote 4. Latinos and whites who live in California and are concerned about global warming

Latinos and whites living in California who are likely to vote

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 rates are the true parameters for all cancer patients. Can we use a normal model for the sampling distribution of differences in proportions?

No a normal model is not a good fit for this sampling distribution. For male cancer patients, np = 0.104(67) = 7 < 10.

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? Yes, there are close to equal numbers in each group. Yes, there are at least 10 people with weak bones and 10 people with strong bones in each group. No, the groups are not the same size. No, there are not at least 10 people with weak bones and 10 people with strong bones in each group.

No, there are not at least 10 people with weak bones and 10 people with strong bones in each group.

For the following question, a coffee drinker is a woman who drinks four or more cups each day. Suppose that the difference between the proportions of coffee drinkers and non-coffee drinkers who are depressed is equal to −0.02. (So the statistic is p-hat for coffee drinkers minus p-hat for non-coffee drinkers.) Which group has a higher depression rate? Coffee Drinkers Non-Coffee Drinkers

Non coffee

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?0.0 -0.06 0.29 -0.1

Not -.1, not 0 Not .65 and .6 .7,.55

Ho: p1-p2=0 (p1=p2) Ha: p1-p2>0 p1>p2 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? The data do not support the claim that a lower proportion of conservatives smoke cannabis when compared to liberals. The data support the claim that the proportion of conservatives who smoke cannabis is no different that the proportion for liberals. The data support the claim that a lower proportion of conservatives smoke cannabis when compared to liberals.

Not 1st choice because small a means rejet null and support alt

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 group of males 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? Students who are not fraternity members use cognition-enhancing drugs at higher rates than fraternity members. 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. Students could have discovered that the proportion of fraternity members who have used cognition-enhancing drugs is 0.315 and the proportion of non-fraternity members who have used cognition-enhancing drugs is 0.369.

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 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: Ho= pq-p2=0 p1=p2 Ha: p1-p2<0 (p1<p2) In this hypothesis test which of the following errors is a Type I error?

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.

H0: μ = 50, Ha: μ > 50 Suppose that a sample of size 30 has a t-statistic of 2.43. Find the corresponding p-value. 0.0108 0.9892 0.0216

Right tailed test T.DIST ( 2.43,1) = 0.9892

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? The data do not support the claim that a lower proportion of conservatives smoke cannabis when compared to liberals. The data support the claim that the proportion of conservatives who smoke cannabis is no different that the proportion for liberals. The data support the claim that a lower proportion of conservatives smoke cannabis when compared to liberals.

The data support the claim that a lower proportion of conservatives smoke cannabis when compared to liberals.

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. Which of the following is a valid conclusion? 1. We are 99% confident that the proportion of smokers who will quit smoking with a nicotine patch treatment is about 8.8 to 10.4% higher than the proportion who quit smoking with a placebo treatment. 2. The nicotine patch treatment is not effective in helping smokers quit. We are 99% confident that smokers using a nicotine patch treatment have a "quit smoking" rate that is anywhere from 10.4% lower to 8.8% higher than the success rate of smokers with a placebo treatment.

The nicotine patch treatment is not effective in helping smokers quit. We are 99% confident that smokers using a nicotine patch treatment have a "quit smoking" rate that is anywhere from 10.4% lower to 8.8% higher than the success rate of smokers with a placebo treatment.

Which of the following is the best estimate of the population proportions? (fq of differences)

The population proportions are 0.65 and 0.60.

(fq of diffeence from 0-25000) Which of the following is the best estimate of the standard error? The standard error is -0.15. The standard error is 0.15. The standard error is -0.05. The standard error is 0.05.

The standard error is 0.05.

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?

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.

, the Public Policy Institute of California (PPIC) surveyed 7525 likely voters living in California. This is the 148th PPIC research poll, and is part of a survey series that started in 1998. PPIC researchers find that 3266 survey participants are registered Democrats and 2137 survey participants are registered Republicans. PPIC is interested in the difference in approval rate for the governor between registered Democrats and registered Republicans in California. Participants were asked: "Overall, do you approve or disapprove of the way the governor of California is handling his job?" 2450 of the registered Democrats answered "Yes" and 684 of the registered Republicans answered "Yes." True or false? A normal model is an appropriate fit for the sampling distribution of sample differences. True False

True

Does Involving a Statistician Improve the Chance That a Medical Research Paper Will Be Published? The following excerpt from "How Statistical Expertise Is Used in Medical Research" (Altman, D. G., S. N. Goodman, and S. Schroter, Journal of the American Medical Association 287(21):2817-20, 2002) describes the data collection method for this study. Authors of original research articles who submitted to BMJ [British Medical Journal]and Annals of Internal Medicine from May through August 2001 were sent a short questionnaire....Authors were asked if they received assistance from a person with statistical expertise. Of the 190 who did not work with a statistician, 134 had papers rejected without peer review. Of the 514 who did work with a statistician, 293 had papers rejected without peer review. Give the 90% confidence interval to estimate the difference in the proportion of medical papers rejected without peer review when a statistician was and was not involved. Interpret your interval in context

We are 90% confident that the authors who do not work with a statistician are 7 to 20% more likely to have their paper rejected without peer review. I got .1325, .1378

In March 2015, the Public Policy Institute of California (PPIC) surveyed 7525 likely voters living in California. In the survey, respondents were asked about global warming. PPIC researchers are interested in the difference in viewpoints across racial/ethnic groups. PPIC results show that 75% of Latinos view global warming as a serious problem, and 46% of whites view global warming as a serious problem. Using the data from the survey, we calculate the sample difference in global warming viewpoints for Latino respondents and white respondents to be 29% = 0.29. - The 95% confidence interval based on this sample difference is (0.275, 0.305). Which of the following is a valid conclusion? Check all that apply. We are 95% confident that the proportion of Latinos who view global warming as a serious problem is about 27.5% to 30.5% lower than the proportion of whites who view global warming as a serious problem. We are 95% confident that the true difference in proportion of Latinos who view global warming as a serious problem and whites who view global warming as a serious problem is 27.5% to 30.5%. We are 95% confident that the difference between Latino and white opinions about the severity of global warming is 29% with a margin of error of 1.5%.

We are 95% confident that the true difference in proportion of Latinos who view global warming as a serious problem and whites who view global warming as a serious problem is 27.5% to 30.5%. We are 95% confident that the difference between Latino and white opinions about the severity of global warming is 29% with a margin of error of 1.5%.

The 99% confidence interval for the difference in proportions of women and men who support stricter gun control is (0.02, 0.08). (Difference here is "women" minus "men.") Which of the following conclusions about the confidence interval is most appropriate? - 99% of the time the gender difference seen in a poll on this issue is between 0.02 and 0.08. - We are 99% confident that the difference between the proportions of women and men who support stricter gun control is between 0.02 and 0.08. - There is a 99% chance that the difference between the proportions of women and men who support stricter gun control is between 0.02 and 0.08.

We are 99% confident that the difference between the proportions of women and men who support stricter gun control is between 0.02 and 0.08.

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. The depression rate for women who do not drink coffee is 6%. For the moment, we assume that this is the rate for coffee drinkers as well. We plan to survey 920 women who drink coffee and 1100 women who do not. We will determine the proportion of women who are depressed in each sample. Will the sampling distribution of differences between sample proportions be approximately normal? Yes, a normal model is a good fit for this sampling distribution. No, a normal model is not a good fit for this sampling distribution.

Yes, a normal model is a good fit for this sampling distribution.s np = 920(0.06) ≈ 55 and n(1 p) = 920(0.94) ≈ 865

According to a report on academic cheating by ETS, the Educational Testing Service, college students majoring in Business and Engineering are more likely to cheat than students in other majors. For a statistics project, a community college student at Diablo Valley College (DVC) decides to investigate cheating in two popular majors at DVC: business and nursing. She convinces professors who teach business and nursing courses to distribute a short anonymous survey in their classes. The question about cheating is one of many other questions about college life. From this data, the student plans to calculate a confidence interval for the difference in two population proportions. Which is the most reasonable description of the two populations for her conclusion? 1. business and nursing students in the classes that were surveyed 2. business and nursing students at DVC 3. business and nursing students at community colleges 4. business and nursing students at community colleges the year of the stud

business and nursing students at DVC

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: Increase Decrease Stay the same

decrease

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? We conclude that a smaller proportion of conservatives smoke cannabis, in comparison to liberals, when there is actually no difference. 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.

e conclude that there is no difference between the proportions of conservatives and liberals that smoke cannabis, when the proportion is actually lower for conservatives.

So the statistic is the sample proportion for coffee drinkers minus the sample proportion for non-coffee drinkers.Graph A Graph B Graph C

graph a Each population proportion is equal to 0.06 so their difference is 0

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? α = 0.005 α = 0.01 α = 0.05 Feedback Incorrect. α is the probability of a Type I error. A choice of small α means we want to reduce the possibility of a type I error. But this will increase the probability of a Type II error.

i put .005 said incorrect A choice of small α means we want to reduce the possibility of a type I error. But this will increase the probability of a Type II erro

Probability of type 1 error

is the level of significance

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? α = 0.005 α = 0.01 α = 0.05 Ho= pq-p2=0 p1=p2 Ha: p1-p2<0 (p1<p2)

not a= .005 , wA choice of small α means we want to reduce the possibility of a type I error. But this will increase the probability of a Type II erro

In the hypothesis test about cannabis use by conservatives and liberals, the P-value is extremely small. Which of the following errors is possible in this situation? Type I only Type II only Type I and Type II Neither Type I nor Type II

type I