Week 6 Test

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American Public University *

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302

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Statistics

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Feb 20, 2024

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Week 6 Test - Results X Attempt 1 of 2 Written Oct 13, 2023 3:50 PM - Oct 13, 2023 5:51 PM Attempt Score 19/20-95% Overall Grade (Highest Attempt) 19/20-95% Question 1 1/ 1 point A researcher is testing reaction times between the dominant and non-dominant hand. They randomly start with each hand for 20 subjects and their reaction times in milliseconds are recorded. Test to see if the reaction time is faster for the dominant hand using a 5% level of significance. The hypotheses are: HO:LLD=0 Hi:pp>0 t-Test: Paired Two Sample for Means Non-Dominant | Dominant Mean 63.33 56.28 Variance 218.9643158128.7522105 Observations 20 20 Pearson Correlation 0.9067

Hypothesized Mean Difference 0 df 19 t Stat 4.7951 P(T<=t) one-tail 0.0001 t Critical one-tail 1.7291 P(T<=t) two-tail 0.0001 t Critical two-tail 2.0930 What is the correct test statistic? ) 0.9067 v 47951 () 0.0001 () 2.0930 ~ 17291 ¥ Hide question 1 feedback The Test stat is 4.7951. This is given to you in the output. No calculations are needed. The Test stat is 4.7951. This is given to you in the output. No calculations are needed. Question 2 1/ 1 point A physician wants to see if there was a difference in the average smokers' daily cigarette consumption after wearing a nicotine patch. The physician set up a study to track daily smoking consumption. In the study, the patients were given a placebo patch that did not contain nicotine for 4

weeks, then a nicotine patch for the following 4 weeks. Test to see if there was a difference in the average smoker's daily cigarette consumption using o = 0.01. The hypotheses are: Hop:up=0 Hi:pup#0 t-Test: Paired Two Sample for Means Placebo | Nicotine Mean 18.7512.3125 Variance 66.59934.666/ Observations 16 16 Pearson Correlation 0.6105 Hypothesized Mean Difference 0 df 15 t Stat 3.5481 P(T<=t) one-tail 0.0126 t Critical one-tail 2.6025 P(T<=t) two-tail 0.0252 t Critical two-tail 2.9467 What is the correct p-value?

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v( )0.0252 ¢ 0.0126 () 0.6105 () 2.9467 w Hide question 2 feedback The p-value for a two tailed test is 0.0252. This is given to you in the output. No calculations are needed. Question 3 1/ 1 point The makers of a smartphone have received complaints that the face recognition tool often doesn't work, or takes multiple attempts to finally unlock the phone. They've upgraded to a new version and are claiming the tool has improved. To test the claim, a critic takes a random sample of 80 users of the new version (Group 1) and 75 users of the old version (Group 2). He finds that the face recognition tool works on the first try 75% of the time in the new version and 60% of the time 1n the old version. Can it be concluded that the new version 1s performing better? Test at a=0.10. Hypotheses: Hy: p1 =P Hy:py>p) In this scenario, what is the test statistic? Round to four decimal places. Z: 1.9964 v w Hide question 3 feedback .75 — .60 \/-677419 % .322581 x (1/80 + 1/75)

z=1.9964 Question 4 1/ 1 point In a 2-sample z-test for two proportions, you find the following: X{ =27 n; =200 Xp =18 ny= 150 You decide to run a test for which the alternative hypothesis is H;: p; > p,. Find the appropriate p- value for the test. Round to 4 decimal places. p-value = 0.3391 v w Hide question 4 feedback 135 — .12 \/-128571 * .871429 * (1/200 + 1/150) 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. Question 5 1/ 1 point In a 2-sample z-test for two proportions, you find a p-value of 0.0081. In a two-tailed test with a=0.01, what decision should be made?

¢<> Reject the null. () Do not reject the null. ) Not enough information to tell Question 6 1/ 1 point A two-tailed test is one where: v() results in either of two directions can lead to rejection of the null ~ hypothesis > negative sample means lead to rejection of the null hypothesis () results in only one direction can lead to rejection of the null hypothesis [ )no results lead to the rejection of the null hypothesis Question 7 1/ 1 point Results from previous studies showed 76% of all high school seniors from a certain city plan to attend college after graduation. A random sample of 200 high school seniors from this city reveals that 162 plan to attend college. Does this indicate that the percentage has increased from that of previous studies? Test at the 5% level of significance. State the null and alternative hypotheses. > Hy: p=.76, Hi: p > .76 > Ho:p > .76, Hq:p > .76 > Hy:p=.76, H;: p #.76 \/<> Hy:p<.76,H;: p>.76 w Hide question 7 feedback Hypothesized value is .76 and this is an upper tailed test because of the keyword increased

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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 75. To determine if the average number of calories in a serving of popcorn is different from 75, a nutritionist selected a random sample of 20 servings of popcorn and computed the sample mean number of calories per serving to be 78 with a sample standard deviation of 7. Compute the z or t value of the sample test statistic. (Ht= -1.916 J() t=1.916 ) z=1.916 ¥ Hide question 8 feedback Use T, because the sample size is small and this is referring to a sample with a mean and SD. T stat = Question 9 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. CHy:p > 76, Hy:u < 76 “Hy:u — 76, T u # 76 80, Hy : pn > 80 ) HO : l,l, w Hide question 9 feedback Hypothesized value is 76 and this is a two tailed test because the keyword is different. Question 10 1/ 1 point The null and alternative hypotheses divide all possibilities into: ¢<> two non-overlapping sets () two sets that may or may not overlap () as many sets as necessary to cover all possibilities () two sets that overlap Question 11 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 1f 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 38,120 27,346 Hypotheses: Ho: 1 = 1y Hy:pg >y

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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 w Hide question 11 feedback 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 O 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 Question 12 1/ 1 point You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 30 night students, and the sample mean GPA is 2.34 with a standard deviation 0f 0.456. You sample 25 day students, and the sample mean GPA is 2.59 with a standard deviation of 0.482. Test the claim using a 10% level of significance. Assume the sample standard deviations are unequal and that GPAs are normally distributed. Hypotheses: Ho: p1 =1, Hy:ipy # What is the test statistic for this scenario? Round to 4 decimal places.

Answer: -1.9627 w Hide question 12 feedback T-stat = 2.34 — 2.99 4562 4822 \// 30 + 25 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. San Diego | Portland 234 211 221 214 202 228 214 222 228 218 244 216

182 222 245 220 215 228 233 224 227 234 217 219 219 226 234 226 255 219 235 228 211 212 248 216 232 217 233 214 Choose the correct decision and summary based on the p-value. > Do not reject Hy There 1s evidence that the average movie ticket sales in San Diego and Portland per week differ. > Reject Hy. There 1s no evidence that the average movie ticket sales in San Diego and Portland per week differ.

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> Reject Hy. There is evidence that the average movie ticket sales in San Diego and Portland per week differ. ¢< > Do not reject Hy. There 1s no evidence that the average movie ticket sales in San Diego and Portland per week differ. w Hide question 13 feedback 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 O 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 Question 14 1/ 1 point A researcher 1s curious if age makes a difference in whether or not students make use of the gym at a university. He takes a random sample of 30 days and counts the number of upperclassmen (Group 1) and underclassmen (Group 2) that use the gym each day. The data are below. The population standard deviation for underclassmen 1s known to be 22.57 and the population standard deviation for upperclassmen 1s known to be 13.57. Upper Classmen average = 202.4, population SD = 13.57, n = 30 Under Classmen average = 191.3, population SD = 22.57, n = 30 Is there evidence to suggest that a difference exists in gym usage based on age? Construct a confidence interval for the data above to decide. Use a=0.10. Confidence Interval (round to 4 decimal places):

_ SHi-Mp<___ Answer for blank # 1: 3.1913 (50 %) Answer for blank # 2: 19.0087 (50 %) w Hide question 14 feedback Z-Critical Value = NORM.S.INV(.95) = 1.645 LL =(202.4 - 191.3) - 1.645* 13.572 N 22.573 30 30 UL = (202.4 - 191.3) + 1.645* 13.572 N 292.573 30 30 Question 15 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) Test Score Number of Students 144 127 Mean Test Score 78.4 75 Population Standard Dev. 11.98 10.56 Is there evidence to suggest that the online system improves test scores? Use a=0.05. Select the correct alternative hypothesis and state the p-value. OHE ) HI: oL ul # u2; p-value = .0065 ul <u2; p-value =.0130 ul > p2; p-value =.0130 ul # u2; p-value = .0130 ul < p2; p-value = .0065 ul > u2; p-value = .0065 w Hide question 15 feedback z-stat = 2.4832 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. Question 16 1/ 1 point In a study that followed a group of students who graduated from high school in 1997, each was monitored in progress made toward earning a bachelor's degree. The group was divided into two — those who started at community college and later transferred to a four-year college (group 1), and

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those that started out in a four-year college as freshmen (group 2). The data below summarize the findings: Community College Non-Transfers Transfers n 317 1,297 Mean Time to Graduate 5.12 4.75 Population Standard Dev. oopualon an af‘ ev 1996 1075 Time to Graduate (in years) Is there evidence to suggest that community college transfer students take longer to earn a bachelor's degree? Use 0=0.035. Select the correct alternative hypothesis and the p-value. \) H1: pl > p2; p-value = 0.001426 > HI1: pl <p2; p-value = 0.001426 > HI1: pl # p2; p-value = 0.001426 > H1: pl # u2; p-value = 0.0007 > HI1: pl < p2; p-value = 0.0007 ,/C\f) H1: pl > p2; p-value = 0.0007 ¥ Hide question 16 feedback Z-stat = 0.12 — 4.75 1.9962 4 1.075% 317 1297 z-stat = 3.1893 Use =NORM.S.DIST(3.1893,TRUE) = .9993. This is the p-value for a lower tailed test. Take 1 - take value to get the p-value for an upper tailed test.

Question 17 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 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? \/<>HO ul=p2; H1: pl # p2 > HO: ul < pn2; H1: pl1 2 p2 (RO ul # u2; HI pl = 2 > HO: ul 2 p2; H1: pl1 < u2 > HO: u1 =p2; H1: p1 = u2 f) HO: ul > 2 ; H1: p1 = p2 w Hide question 17 feedback This is a two tailed test because of the keyword difference. Question 18 1/ 1 point Which of the following symbols represent a confidence level?

v ) (1-%)*100% (M 1B B w Hide question 18 feedback If you have a 95% confidence level then alpha =1 .95 = .05. Alpha and a confidence level are two different values. Question 19 1/ 1 point According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft. In that year, Alaska had 321 complaints of identity theft out of 1,432 consumer complaints. Does this data provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%? The hypotheses are: Hy:p = 23% Hi:p <23% What is a type | error in the context of this problem? Reference: Federal Trade Commission, (2008). Consumer fraud and identity theft complaint data: January-December 2007. Retrieved from website: http:/www.ftc.gov/opa/2008/02/fraud.pdf. ) It is believed that more than 23% of Alaskans had identity theft and ~ there really were 23% or more that experience identity theft. O It is believed that less than 23% of Alaskans had identity theft and ~ there really were 23% or less that experienced identity theft.

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¢<> It is believed that less than 23% of Alaskans had identity theft even ~ though there really were 23% or more that experienced identity theft. O It is believed that more than 23% of Alaskans had identity theft even ~ though there really were less than 23% that experienced identity theft. Question 20 0/ 1 point 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: Hol n= 47 Hy. u>47 Find the test statistic. Round answer to 4 decimal places. Test Statistic t = 2.2005 ® (2.1999) w Hide question 20 feedback T-stat = 49.26667 — 47 3.990465 V15

T-stat = 2.1999 Done

Week 6 Test

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