Concept explainers
Dr. Vegapunk thinks that watching anime (Japanese animated shows) decreases social skills in college students. To test this, Dr. Vegapunk randomly selected 30 brooklyn college students and assigned them to watch an episode of anime everyday for a week. After the week, each of the students answered a questionnaire about their social skills. The results showed that the sample had a mean social skills score of 7.3 and a standard deviation of 2.4. A previous study showed that the overall population of brooklyn college students had a mean social skills score of 6.2, but the standard deviation was not reported. Dr. Vegapunk decides to use an alpha level of 0.05.
Should it be one or two tailed?
Trending nowThis is a popular solution!
Step by stepSolved in 2 steps
- You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7. For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. What is the mean square within samples?arrow_forwardResearch was being done on who makes more money right after college, a person with a sociology degree or a psychology degree. To test this, the researcher randomly selected 45 recent soclology graduates and found that the mean income they are recelving Is $53,900 with a standard devlation of $2,050. The researcher also randomly selected 40 psychology majors and found that the mean income they are receiving is $55,200 with a standard deviation of $3,850.Test the claim that the mean Income for psychology majors is more than the mean Income for soclology majors at the a-0.05 level of significance. Calculator Function [ Select] P-Value [ Select] Conclusionarrow_forwardFor a study conducted by the research department of a pharmaceutical company, 235 randomly selected individuals were asked to report the amount of money they spend annually on prescription allergy relief medication. The sample mean was found to be $17.80 with a standard deviation of $4.30 . A random sample of 250 individuals was selected independently of the first sample. These individuals reported their annual spending on non-prescription allergy relief medication. The mean of the second sample was found to be $18.20 with a standard deviation of $4.00 . As the sample sizes were quite large, it was assumed that the respective population standard deviations of the spending for prescription and non-prescription allergy relief medication could be estimated as the respective sample standard deviation values given above. Construct a 90% confidence interval for the difference −μ1μ2 between the mean spending on prescription allergy relief medication ( μ1 ) and the mean…arrow_forward
- A pharmaceutical company needs to know if its new cholesterol drug, Praxor, is effective at lowering cholesterol levels. It believes that people who take Praxor will average a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 38 participants in the treatment group lowered their cholesterol levels by a mean of 23.2 points with a standard deviation of 4.4 points. The 44 participants in the control group lowered their cholesterol levels by a mean of 21.5 points with a standard deviation of 2.9 points. Assume that the population variances are not equal and test the company's claim at the 0.01 level. Let the treatment group be Population 1 and let the control group be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.arrow_forwardAccording to the College Board, scores on the math section of the SAT Reasoning college entrance test for the class of 2010 had a mean of 516 and a standard deviation of 116. Assume that they are roughly normal.One of the quartiles of the scores from the math section of the SAT Reasoning test is 438. The other quartile is _______.arrow_forwardFive students took a self-confidence test. Their deviation scores were –4, +3, +1, –2, and +2. Assuming that these scores com form a normaldistribution, which deviation represents the highest raw score?arrow_forward
- A nationwide job recruiting firm wants to compare the annual incomes for childcare workers in Utah and Oregon. Due to recent trends in the childcare industry, the firm suspects that the mean annual income of childcare workers in Utah is less than the mean annual income of childcare workers in Oregon. To see if this is true, the firm selected a random sample of 20 childcare workers from Utah and an independent random sample of 20 childcare workers from Oregon and asked them to report their mean annual income. The data obtained were as follows. Annual income in dollars 27060 , 42320 , 36445 , 37895, 44564 , 35325 , 30313 , 28698, 33706 , 48365 , 26946 , 25063 , 28339 , 29265 , 40051 , 37026 , 40150 , 42753 , 27820, 39471 Utah 29769 , 48859 , 32480 , 36288,45102 , 46219 ,51098 , 45516 , 38834 , 44659 , 40755 , 38006 , 39947 , 29801 , 32200, 38731 , 41370 , 28059 , 33753 , 35271 Oregon Send data to calc... v Send data to Excel The population standard deviation for the annual incomes of…arrow_forwardIn preparation for the upcoming school year, a teacher looks at raw test scores on the statewide standardized test for the students in her class. Instead of looking at the scores relative to the norms in the state, the teacher wants to understand the scores relative to the students who will be in the class. To do so, she decides to convert the test scores into z-scores relative to the mean and standard deviation of the students in the class. The mean test score in her upcoming class is 49, and the standard deviation is 20.7. The teacher now wants to identify those students who may need extra help. She decides to look at students who have z-scores below z = -2.00 Identify the test score corresponding to a z-score of below z = -2.00. Round to the nearest whole number.arrow_forwardAccording to the Vivino website, suppose the mean price for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is $32.48. A New England-based lifestyle magazine wants to determine if red wines of the same quality are less expensive in Providence, and it has collected prices for 64 randomly selected red wines of similar quality from wine stores throughout Providence. The mean and standard deviation for this sample are $30.15 and $12, respectively. (a) Develop appropriate hypotheses for a test to determine whether the sample data support the conclusion that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48. (Enter != for # as needed.) Ho: H: (b) Using the sample from the 64 bottles, what is the test statistic? (Round your answer to three decimal places.) Using the sample from the 64 bottles, what is the p-value? (Round your answer to four decimal places.) p-value =…arrow_forward
- A pharmaceutical company needs to know if its new cholesterol drug, Praxor, is effective at lowering cholesterol levels. It believes that people who take Praxor will average a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 44 participants in the treatment group lowered their cholesterol levels by a mean of 18.7 points with a standard deviation of 3.3 points. The 39 participants in the control group lowered their cholesterol levels by a mean of 18.1 points with a standard deviation of 2.1 points. Assume that the population variances are not equal and test the company’s claim at the 0.02 level. Let the treatment group be Population 1 and let the control group be Population 2. Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ1=μ2 Ha : μ1⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯μ2 Step 2 of 3: what is the test statistic Step 3 of 3: draw a conclusion, fail or reject. Is…arrow_forwardA report about how American college students manage their finances includes data from a survey of college students. Each person in a representative sample of 793 college students was asked if they had one or more credit cards and if so, whether they paid their balance in full each month. There were 500 who paid in full each month. For this sample of 500 students, the sample mean credit card balance was reported to be $825. The sample standard deviation of the credit card balances for these 500 students was not reported, but for purposes of this exercise, suppose that it was $205. Is there convincing evidence that college students who pay their credit card balance in full each month have a mean balance that is lower than $907, the value reported for all college students with credit cards? Carry out a hypothesis test using a significance level of 0.01. State the appropriate null and alternative hypotheses. H0: ? = 907 Ha: ? < 907 H0: ? = 907 Ha: ? > 907 H0: ? < 907…arrow_forwardLogan and Brian began arguing about who did better on their tests, but they couldn't decide who did better given that they took different tests. Logan took a test in Art History and earned a 76.8, and Brian took a test in English and earned a 68.9. Use the fact that all the students' test grades in the Art History class had a mean of 71.9 and a standard deviation of 11.6, and all the students' test grades in English had a mean of 61 and a standard deviation of 9.4 to answer the following questions.a) Calculate the z-score for Logan's test grade.z=z= [Round your answer to two decimal places.] b) Calculate the z-score for Brian's test grade.z=z= [Round your answer to two decimal places.]c) Which person did relatively better? Logan Brian They did equally well.arrow_forward
- MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
- Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman