MATH302 Week 6 Test

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American Military University *

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302

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Apr 3, 2024

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1/21 Week 6 Test - Results Attempt 1 of 2 Attempt Score 19 / 20 - 95 % Overall Grade (Highest Attempt) 19 / 20 - 95 % Question 1 1 / 1 point ___ 0.1987___ New LM Old LM 57.1 55.8 58.3 51.7 83.6 76.6 An adviser is testing out a new online learning module for a placement test. They wish to test the claim that on average the new online learning module increased placement scores at a significance level of α = 0.05. For the context of this problem, μ D =μ new –μ old where the first data set represents the new test scores and the second data set represents old test scores. Assume the population is normally distributed. H 0 : μ D = 0 H 1 : μ D < 0 You obtain the following paired sample of 19 students that took the placement test before and after the learning module: Find the p-value. Round answer to 4 decimal places. p-value =

2/21 New LM Old LM 50.5 47.5 51.5 48.6 20.6 15.5 35.2 29.9 46.7 54 23.5 21 48.8 58.5 53.1 42.6 76.6 61.2 29.6 26.3 14.5 11.4 43.7 56.3 57 46.1 66.1 72.9 38.1 43.2 44.4 51.1

3/21 Hide ques±on 1 feedback Question 2 1 / 1 point Copy and paste the data into Excel. Use the Data Analysis Toolpak in Excel. Data - > Data Analysis -> scroll to where is says t:Test: Paired Two Samples for Means -> OK Variable 1 Range: is New LM Variable 2 Range: is Old LM The Hypothesized Mean Difference is 0 and make sure you click Labels in the first row and click OK. You will get an output and this is the p-value you are looking for. P(T<=t) one-tail 0.1987 A manager wishes to see if the time (in minutes) it takes for their workers to complete a certain task is different if they are wearing earbuds. A random sample of 20 workers' times were collected before and after wearing earbuds. Test the claim that the time to complete the task will be different, i.e. meaning has production differed at all, at a significance level of α = 0.01 For the context of this problem, μ D = μ before −μ after where the first data set represents before earbuds and the second data set represents the after earbuds. Assume the population is normally distributed. The hypotheses are: H 0 : μ D = 0 H 1 : μ D 0 You obtain the following sample data: Before After 69 65.3 69.5 61.6 39.3 21.4

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4/21 ___ 0.0038___ Hide ques±on 2 feedback Before After 66.7 60.4 38.3 46.9 85.9 76.6 70.3 77.1 59.8 51.3 72.1 69 79 83 61.7 58.8 55.9 44.7 56.8 50.6 71 63.4 80.6 68.9 59.8 35.5 73.1 77 49.9 38.4 56.2 55.4 64.3 55.6 Find the p-value. Round answer to 4 decimal places. p-value: Copy and paste the data into Excel. Use the Data Analysis Toolpak in Excel. Data - > Data Analysis -> scroll to where is says t:Test: Paired Two Samples for Means -> OK Variable 1 Range: is Before

5/21 Question 3 1 / 1 point ___ 0.3391___ Hide ques±on 3 feedback Variable 2 Range: is After The Hypothesized Mean Difference is 0 and make sure you click Labels in the first row and click OK. You will get an output and this is the p-value you are looking for. P(T<=t) two-tail 0.0038 In a 2-sample z-test for two proportions, you find the following: X 1 = 27 n 1 = 200 X 2 = 18 n 2 = 150 You decide to run a test for which the alternative hypothesis is H 1 : p 1 > p 2 . Find the appropriate p - value for the test. Round to 4 decimal places. p-value = z = z = 0.4149 Use NORM.S.DIST(0.4149,TRUE) to find the for the lower tailed test. 0.660888, to get the upper tailed test you take 1 - 0.660888, this is the p- value you want to use for the conclusion.

6/21 Question 4 1 / 1 point ___ -0.028___ (50 %) ___ 0.096___ (50 %) Hide ques±on 4 feedback Question 5 1 / 1 point A car insurance company suspects that the younger the driver is, the more reckless a driver he/she is. They take a survey and group their respondents based on age. Of 217 respondents between the ages of 16-25 (Group 1), 183 claimed to wear a seat belt at all times. Of 398 respondents who were 26+ years old (Group 2), 322 claimed to wear a seat belt at all times. Find a 95% confidence interval for the difference in proportions. Can it be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group? Enter the confidence interval - round to 3 decimal places. < p 1 - p 2 < Z-Critical Value =NORM.S.INV(.975) = 1.96 LL = (0.843318 - 0.809045) - 1.96* UL = (0.843318 - 0.809045) +1.96* TDaP is a booster shot that prevents Diphtheria, Tetanus, and Pertussis in adults and adolescents. It should be administered every 8 years for it to remain effective. A random sample of 550 people living in a town that experienced a pertussis outbreak this year were divided into two groups. Group 1 was made up of 145 individuals who had not had the TDaP booster in the past 8 years, and Group

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7/21 ___ 0.00012___ Hide ques±on 5 feedback Question 6 1 / 1 point A two-tailed test is one where: 2 consisted of 355 individuals who had. In Group 1, 18 individuals caught pertussis during the outbreak, and in Group 2, 13 individuals caught pertussis. Is there evidence to suggest that the proportion of individuals who caught pertussis and were not up to date on their booster shot is higher than those that were? Test at the 0.05 level of significance. Enter the p-value - round to 5 decimal places. p-value = This is an upper tailed test because keyword is higher. z = z = 3.6823 Use NORM.S.DIST(3.6823,TRUE) to find the for the lower tailed test. 0.999884, to get the upper tailed test you take 1 - 0.999884, this is the p- value you want to use for the conclusion. results in only one direction can lead to rejection of the null hypothesis results in either of two directions can lead to rejection of the null hypothesis negative sample means lead to rejection of the null hypothesis no results lead to the rejection of the null hypothesis

8/21 Question 7 0 / 1 point You conduct a hypothesis test and you observe values for the sample mean and sample standard deviation when n = 25 that do not lead to the rejection of H 0 . You calculate a p-value of 0.0667. What is the highest possible significance level you can have that will lead to you Rejecting Ho if the p- value is 0.0667? Hide ques±on 7 feedback Question 8 1 / 1 point In an article appearing in Today's Health a writer states that the average number of calories in a serving of popcorn is 76. To determine if the average number of calories in a serving of popcorn is different from 76, a nutritionist selected a random sample of 20 servings of popcorn and computed the sample mean number of calories per serving to be 80 with a sample standard deviation of 6.3. State the null and alternative hypotheses. 90% 94% 93% 95% 93% = 1 - .93 = .07 Alpha = .07 .0667 < .07, Reject Ho μ μ

9/21 Hide ques±on 8 feedback Question 9 1 / 1 point Which of the following statements are true of the null and alternative hypotheses? Question 10 1 / 1 point The plant-breeding department at a major university developed a new hybrid boysenberry plant called Stumptown Berry. Based on research data, the claim is made that from the time shoots are planted 90 days on average are required to obtain the first berry. A corporation that is interested in marketing the product tests 60 shoots by planting them and recording the number of days before each plant produces its first berry. The sample mean is 92.3 days and the SD is 6.66 days. The corporation wants to know if the mean number of days is different from the 90 days claimed? Use alpha = .01. μ μ Hypothesized value is 76 and this is a two tailed test because the keyword is different. Both hypotheses must be true It is possible for both hypotheses to be true Exactly one hypothesis must be true It is possible for neither hypothesis to be true Yes, because the p-value = .0048 No, because the p-value =.0048

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10/21 Hide ques±on 10 feedback Question 11 1 / 1 point A pet store owner believes that dog owners spend more on average than cat owners on their pets. The owner randomly records the sales of 40 customers that said they only owned dogs and 40 customers that said they only owned cats. What is the correct decision and summary based on the Excel output below? Assume the populations are approximately normally distributed with unequal variances. t-Test: Two-Sample Assuming Unequal Variances No, because the p-value = .0097 Yes, because the p-value = .0097 No, because the p-value = .9873 The hypothesized value is 90 and this is a two tailed test because the keyword is different. T Stat = T stat = 2.675 Use =T.DIST.2T(2.675,59) to find the p-value.

11/21 Dog Cat Mean 57.169553.82225 Variance 588.088573.7205 Observations 40 40 Hypothesized Mean Difference 0 df 78 t Stat 0.6211 P(T<=t) one-tail 0.2682 t Critical one-tail 1.6646 P(T<=t) two-tail 0.5363 t Critical two-tail 1.9908 What is the correct test statistic? Hide ques±on 11 feedback 0.5860 1.6646 0.2682 0.6211 1.9908

12/21 Question 12 1 / 1 point Two random samples are taken from private and public universities (out-of-state tuition) around the nation. The yearly tuition is recorded from each sample and the results can be found below. Test to see if the mean out-of-state tuition for private institutions is statistically significantly higher than public institutions. Assume unequal variances. Use a 1% level of significance and round answers to 4 decimal places. Private Institutions (Group 1 ) Public Institutions (Group 2) 43,120 26,469 28,190 21,450 34,490 18,347 20,893 28,560 42,984 32,592 34,750 23,871 44,897 24,120 32,198 27,450 18,432 29,100 33,981 23,870 29,498 22,650 31,980 29,143 22,764 25,379 54,190 23,450 37,756 23,871 30,129 28,745 33,980 30,120 47,909 21,190 32,200 21,540 No Calculations are needed. Looking at the output the t Stat is 0.6211

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13/21 38,120 27,346 Hypotheses: H 0 : μ 1 = μ 2 H 1 : μ 1 > μ 2 Identify the p-value. Round to 4 decimal places. Make sure you put the 0 in front of the decimal. p-value=___ ___ Answer: 0.0001 Hide ques±on 12 feedback Question 13 1 / 1 point A movie theater company wants to see if there is a difference in the average movie ticket sales in San Diego and Portland per week. They sample 20 sales from San Diego and 20 sales from Portland over a week. Test the claim using a 5% level of significance. Assume the variances are unequal and that movie sales are normally distributed. Copy and Pasta data into Excel. Use the Data Analysis Toolpak in Excel. Data - > Data Analysis -> scroll to where is says t:Test 2 Samples Assuming Unequal Variances -> OK Variable Input 1: is Private Institution Variable Input 2: is Public Institution The Hypothesized Mean Difference is 0 and make sure you click Labels in the first row and click OK. You will get an output and this is the p-value you are looking for. P(T<=t) one-tail 0.0001

14/21 San Diego Portland 234 211 221 214 202 228 214 222 228 218 244 216 182 222 245 220 215 228 233 224 227 234 217 219 219 226 234 226 255 219 235 228

15/21 211 212 248 216 232 217 233 214 Choose the correct decision and summary based on the p-value. Hide ques±on 13 feedback Do not reject H 0. There is evidence that the average movie ticket sales in San Diego and Portland per week differ. Reject H 0 . There is no evidence that the average movie ticket sales in San Diego and Portland per week differ. Reject H 0 . There is evidence that the average movie ticket sales in San Diego and Portland per week differ. Do not reject H 0 . There is no evidence that the average movie ticket sales in San Diego and Portland per week differ. The p-value for the test is 0.169070589. This is greater than .05. Data -> Data Analysis -> scroll to where is says t:Test 2 Samples Assuming Unequal Variances -> OK Variable Input 1: is San Diego Variable Input 2: is Portland The Hypothesized Mean Difference is 0 and make sure you click Labels in the first row and click OK. You will get an output and this is the p-value you are looking for. P(T<=t) two-tail 0.1691

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16/21 Question 14 1 / 1 point A liberal arts college in New Hampshire implemented an online homework system for their introductory math courses and wanted to know whether or not the system improved test scores. In the Fall semester, homework was completed the old fashioned way – with pencil and paper, checking answers in the back of the book. In the Spring semester, homework was completed online – giving students instant feedback on their work. The results are summarized below. Population standard deviations were used from past studies. Online (Group 1) Pen and Paper (Group 2) Number of Students 144 127 Mean Test Score 78.4 75 Population Standard Dev. Test Score 11.98 10.56 Is there evidence to suggest that the online system improves test scores? Use α=0.05. Select the correct alternative hypothesis and state the p-value. Hide ques±on 14 feedback H1: µ1 > µ2; p-value = .0065 H1: µ1 ≠ µ2; p-value = .0130 H1: µ1 < µ2; p-value = .0130 H1: µ1 > µ2; p-value = .0130 H1: µ1 < µ2; p-value = .0065 H1: µ1 ≠ µ2; p-value = .0065 z-stat = z-stat = 2.4832

17/21 Question 15 1 / 1 point In a random sample of 50 Americans five years ago (Group 1), the average credit card debt was $5,598. In a random sample of 50 Americans in the present day (Group 2), the average credit card debt is $6,499. Let the population standard deviation be $1,155 five years ago, and let the current population standard deviation be $1,699. Using a 0.01 level of significance, test if there is a difference in credit card debt today versus five years ago. What is the test statistic? (Round to 4 decimal places) z =___ ___ Answer: -3.1011 Hide ques±on 15 feedback Question 16 1 / 1 point The CEO of a large manufacturing company is curious if there is a difference in productivity level of her warehouse employees based on the region of the use NORM.S.DIST(2.4832,true) , this is the p-value for a lower tailed test. To find the p-value for an upper tailed test take 1 - this value. z -stat =

18/21 country the warehouse is located. She randomly selects 35 employees who work in warehouses on the East Coast (Group 1) and 35 employees who work in warehouses in the Midwest (Group 2) and records the number of parts shipped out from each for a week. She finds that East Coast group ships an average of 1287 parts and knows the population standard deviation to be 348. The Midwest group ships an average of 1449 parts and knows the population standard deviation to be 298. Using a 0.01 level of significance, test if there is a difference in productivity level. What are the correct hypotheses for this problem? Hide ques±on 16 feedback Question 17 1 / 1 point A physical therapist believes that at 30 years old adults begin to decline in flexibility and agility. To test this, he randomly samples 25 of his patients who are less than 30 years old and 27 of his patients are 30 or older and measures each patient's flexibility in the Sit-and-Reach test. Assume the populations are normally distributed. The results are below. Less Than 30 (Group 1) 30 or Older (Group 2) n 25 27 Mean Sit-and-Reach Score 20.21 18.99 H0: μ1= μ2 ; H1: μ1 ≠ μ2 H0: μ1 ≤ μ2 ; H1: μ1 ≥ μ2 H0: μ1 ≠ μ2 ; H1: μ1 = μ2 H0: μ1 ≥ μ2 ; H1: μ1 ≤ μ2 H0: μ1 = μ2 ; H1: μ1 = μ2 H0: μ1 > μ2 ; H1: μ1 = μ2 This is a two tailed test because of the keyword difference.

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19/21 Population Standard Dev. Sit-and-Reach Score 2.222 2.121 Is there evidence to suggest that adults under the age of 30 are more flexible? Round to 4 decimal places. Make sure you put the 0 in front of the decimal. Use α=0.05 . p-value =___ ___ Answer: 0.0216 Hide ques±on 17 feedback Question 18 1 / 1 point A type I error occurs when the: Question 19 1 / 1 point Which of the following symbols represent a confidence level? z-stat = z-stat = 2.0218 Use =NORM.S.DIST(2.0218,true) = 0.9784. This is the p-value for a lower tailed test. To find the p-value for an upper tailed test take 1 - this value. test is biased sample mean differs from the population mean null hypothesis is incorrectly rejected when it is true null hypothesis is incorrectly accepted when it is false

20/21 Hide ques±on 19 feedback Question 20 1 / 1 point ___ 2.1999___ α (1- α )*100% µ 1- β β If you have a 95% confidence level then alpha = 1 .95 = .05. Alpha and a confidence level are two different values. The workweek for adults in the US that work full time is normally distributed with a mean of 47 hours. A newly hired engineer at a start-up company believes that employees at start-up companies work more on average then working adults in the US. She asks 15 engineering friends at start-ups for the lengths in hours of their workweek. Their responses are shown in the table below. Test the claim using a 5% level of significance. See Excel for Data. Work Hour Data The hypotheses for this problem are: H 0 : μ = 47 H 1: μ > 47 Find the test statistic. Round answer to 4 decimal places. Test Statistic t =

21/21 Hide ques±on 20 feedback Done T-stat = T-stat = 2.1999

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MATH302 Week 6 Test

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