Week 6 Knowledge

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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 Knowledge Check Homework Practice Questio... X Attempt 1 of 4 Written Oct 7, 2023 5:48 PM - Oct 7, 2023 10:48 PM Attempt Score 0/20-0% Overall Grade (Highest Attempt) 19/20-95% Question 1 0/ 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 a = 0.01. The hypotheses are: H0:HD=0 Hi:pp#0 t-Test: Paired Two Sample for Means Placebo Nicotine Mean 16.75| 10.3125 Variance 64.4666/33.29583 Observations 16 16

Pearson Correlation 0.6105 Hypothesized Mean Difference 0 df 15 t Stat 4.0119 P(T<=t) one-tail 0.0006 t Critical one-tail 2.6025 P(T<=t) two-tail 0.0011 t Critical two-tail 2.9467 What is the correct decision? = > Reject Hy ( > Accept Hy ‘} Accept H, > Do not reject Hy \> Reject Hy Question 2 0/ 1 point An adviser 1s 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 a = 0.05. For the context of this problem, pp=p,.w—Ho1q 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. HO: Uup = 0 HII uD<O

You obtain the following paired sample of 19 students that took the placement test before and after the learning module: New Choose the correct decision and summary and state the p-value. LM Old LM 58.1 55.8 58.3 53.7 83.6 76.6 49.5 47.5 51.8 48.9 20.6 11.4 35.2 30.6 46.7 54 22.5 21 47.7 58.5 51.5 42.6 76.6 61.2 29.6 26.8 14.5 12.5 43.7 56.3

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New | o1d LM LM 57 43.1 66.1 [12.8 38.1 [42.2 024 [51.3 > Reject Hy), there 1s enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.4533. > Reject Hy, there 1s not enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.2266. = > Do not reject Hy, there 1s not enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.2266 > Do not reject Hy, there 1s not enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.4533. Question 3 0/ 1 point Two competing toothpaste brands both claim to produce the best toothpaste for whitening. A dentist randomly samples 48 patients that use Brand A (Group 1) and finds 30 of them are satisfied with the whitening results of the toothpaste. She then randomly samples 45 patients that use Brand B (Group 2) and finds 33 of them are satisfied with the whitening results of the toothpaste. Construct a 99% confidence interval for the difference in proportions and use it to decide if there 1s a significant difference in the satisfaction level of patients. Enter the confidence interval - round to 3 decimal places. ® (-0.356,-.356) <pq-py< ® (0.139,.139) w Hide question 3 feedback Z-Critical Value =NORM.S.INV(.995) = 2.575

LL = (.625-.7333) - 2.575* 625 * .375 N 1333 % .2667 48 45 UL =(.625-.7333) + 2.575* 625 * .375 N 1333 % .2667 48 45 Question 4 0/ 1 point The manager at a sports radio station will be covering several football games over the weekend. She knows that most of her listeners are at least 22 years old and wants to know what age group she should gear her advertisements to for the games. She takes a random sample of 45 listeners from age 22-39 and 55 listeners who are 40 and older and asks them if they are likely to tune in to football games. The "yes" responses are recorded below. Age 22-39 (Group 1) Age 40+ (Group 2) Responded "Yes" to Football 32 44 Sample Size n 45 55 At the 0.05 level of significance, we are attempting to investigate if there 1s a significant difference in the proportion of listeners based on age. Enter the P-Value - round to 4 decimal places. p-value = ® (0.3005, .3005) w Hide question 4 feedback This is a two tailed test because you want to find significant difference. Z=

71111 — .80 \/-T6 % .24 % (1/45 + 1/55) z=-1.03543 Use NORM.S.DIST(-1.03543,TRUE) to find the for the lower tailed test. This value is smaller of the two, thus you will multiply it by 2 for a two tailed test 0.150233*2, this is the p-value you want to use for the conclusion. Question 5 0/ 1 point In a 2-sample z-test for two proportions, you find the following: X1=24n;=200 Xy =17 ny =150 You decide to run a test for which the alternative hypothesis is Hy: p; > p,. Find the appropriate test statistic for the test. Enter the test statistic - round to 4 decimal places. Z: ¥ (0.1919,.1919) ¥ Hide question 5 feedback pl =24/200 p2=17/150 p = (24+17)/(200+150) Z=

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12 —.113333 \/-1177143 % .882857 * (1/200 + 1/150) Question 6 0/ 1 point A lab technician is tested for her consistency by taking multiple measurements of cholesterol levels from the same blood sample. The target accuracy in the past has been 2.5 or less with a standard deviation of 1.15. If the lab technician takes 16 measurements and the variance of the measurements in the sample is 2.2, does this provide enough evidence to reject the claim that the lab technician's accuracy is within the target accuracy? At the a = .05 level of significance, what 1s your conclusion? > Reject Hp . At the ot = .05 level of significance, there is enough evidence to support the claim that this technician's average is less than the target accuracy. () Cannot determine = () Do not reject Hp. At the o = .05 level of significance there is not sufficient evidence to suggest that this technician's true average is less than the target accuracy. Reject Hp. At the ot = .05 level of significance, there is not enough evidence to support the claim that this technician's true average is less than the target accuracy. Question 7 0/ 1 point The hypothesis that an analyst is trying to prove is called the: = () alternative hypothesis > elective hypothesis ) guality of the researcher ) level of significance

Question 8 0/ 1 point Smaller p-values indicate more evidence in support of the: () qguality of the researcher ( > null hypothesis = () alternative hypothesis () the reduction of variance Question 9 0/ 1 point The form of the alternative hypothesis can be: () one-tailed ) two-tailed () neither one nor two-tailed = ‘> one or two-tailed Question 10 0/ 1 point Suppose that the mean time for a certain car to go from O to 60 miles per hour was 7.7 seconds. Suppose that you want to test the claim that the average time to accelerate from O to 60 miles per hour is longer than 7.7 seconds. What would you use for the alternative hypothesis? () Hq: L > 7.7 seconds " JHq: L = 7.7 seconds " JHq: b < 7.7 seconds " JHy: [ 2 7.7 seconds Question 11 0/ 1 point

A new over-the-counter medicine to treat a sore throat 1s to be tested for effectiveness. The makers of the medicine take two random samples of 25 individuals showing symptoms of a sore throat. Group 1 receives the new medicine and Group 2 receives a placebo. After a few days on the medicine, each group is interviewed and asked how they would rate their comfort level 1-10 (1 being the most uncomfortable and 10 being no discomfort at all). The results are below. Find the 95% confidence interval in the difference of the mean comfort level. Is there sufficient evidence to conclude that the new medicine is effective? Assume the data 1s normally distributed with unequal variances. (Round answers to 4 decimal places) Make sure you put the 0 in front of the decimal. Average Group 1 =5.84, SD Group 1 =2.211334, nl1 =25 Average Group 2 = 3.96, SD Group 2 = 2.35372, n2 =25 _ SHp-Hps<_ Answer for blank #1: 3 (0.5813, 0.5812) Answer for blank #2: ¢ (3.1787,3.1788) w Hide question 11 feedback T-Critical Value =T.INV.2T(.05,48) = 2.010635 LL = (5.84 - 3.96) - 2.010635 * 2.21133472 N 2.35372° 25 25 UL =(5.84 - 3.96) + 2.010635 * 2.21133472 N 2.353722 25 25

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Question 12 0/ 1 point You are testing the claim that the mean GPA of night students is less than the mean GPA of day students. You sample 30 night students and 30 day students. Test the claim using a 10% level of significance. Assume the population standard deviations are unequal. t-Test: Two-Sample Assuming Unequal Variances Night Day Mean 2.8967|, 3.1872 Variance 0.2669(0.277582 Observations 30 30 Hypothesized Mean Difference 0 df 58 t Stat -2.1557 P(T<=t) one-tail 0.01763 t Critical one-tail 1.6716 P(T<=t) two-tail 0.03526 t Critical two-tail 2.0017 The Hypotheses for this problem are: Ho: py =1, Hp:pg<my,

Find the p-value. Round answer to 2 decimal places. Make sure you include the 0 in front of the decimal. p-value= Answer: 3 (0.02) w Hide question 12 feedback This is a lower tailed test and no calculations are needed. The answer is in the output. P(T<=t) one-tail 0.01763 Question 13 0/ 1 point A movie theater company wants to see if there 1s 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 221 209 221 214 206 223 213 221 226 218 243 214 182 222

229 220 214 223 233 222 231 233 217 219 219 226 234 226 239 219 235 228 211 211 239 216 221 219 234 209 The Hypotheses for this problem are: Ho: py =1, Hy:py # p, Find the p-value. Round answer to 4 decimal places. Make sure you put the O in front of the decimal. p-value=

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Answer: % (0.2853) w Hide question 13 feedback 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 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 0.2853 Question 14 0/ 1 point A survey found that the average daily cost to rent a car in Los Angeles is $102.24 and in Las Vegas is $97.35. The data were collected from two random samples of 40 in each of the two cities and the population standard deviations are $5.98 for Los Angeles and $4.21 for Las Vegas. At the 0.05 level of significance, construct a confidence interval for the difference in the means and then decide if there 1s a significant difference in the rates between the two cities. Let the sample from Los Angeles be Group 1 and the sample from Las Vegas be Group 2. Confidence Interval (round to 4 decimal places): <MW mUp<_ Answer for blank # 1: % (2.6236) Answer for blank #2: x (7.1564) w Hide question 14 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96

LL =(102.24 - 97.35) - 1.96 * 5.982 40 UL = (102.24 - 97.35) + 1.96 * Question 15 4.21° 40 5.982 40 4.212 40 0/ 1 point In a random sample of 29 residents living in major cities on the West Coast (Group 1) and 29 residents living in major cities on the East Coast (Group 2), each individual was asked their age. The results can be seen in the table below. The population standard deviation of the age in West Coast cities is known to be 11.2 years and in East Coast cities is known to be 9.3 years. Assume the populations are normally distributed. Run a test at a 0.05 level of significance to test if west coast cities are, on average, older. West Coast (Group | East Coast (Group 1) 2) 25 34 47 45 18 37 38 20 30 19 50 26 52 79 61 46

40 29 22 55 34 23 35 35 55 36 60 51 68 41 20 50 33 32 36 38 37 26 42 44 60 19 71 28 54 27 20 18 45 30 52 20 34 22 57 43 64 62 Enter the P-Value - round to 4 decimal places. Make sure you put the 0 in front of the decimal. p-value= Answer: 3 (0.0021) w Hide question 15 feedback

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Copy and paste the data into Excel. Use the Data Analysis Toolpak in Excel. Data - > Data Analysis -> scroll to where is says z:Test 2 Samples for Means - > OK Variable 1 Range: highlight West Coast Variable 2 Range: highlight East Coast The Hypothesized Mean Difference is O. Variable 1 Variance: You are given the population SD you need to take 11.22 and use this for the population Variance for Variable 1 and 9.32 to find the population Variance for Variable 2. Make sure you click Labels in the first row and click OK. You will get an output and this is the one tailed p-value you are looking for. 0.0021 Question 16 0/ 1 point Which of the following is a requirement that must first be satisfied before running a z-test for the difference between two means? = () The standard deviations for both populations must be known. ) The standard deviations should come from each sample. () The samples must be dependent. () The populations must be approximately normal. Question 17 0/ 1 point In a survey of 100 U.S. residents with a high school diploma as their highest educational degree (Group 1) had an average yearly income was $35,621. Another 120 U.S. residents with a GED (Group 2) had an average yearly income of $34,498. The population standard deviation for both populations is

known to be $4,150. At a 0.01 level of significance, can it be concluded that U.S. residents with a high school diploma make significantly more than those with a GED? Make sure you put the O in front of the decimal. Enter the P-Value - round to 4 decimal places. p-value =___ Answer: % (0.0228) ¥ Hide question 17 feedback Z-stat = 30621 — 34498 41502 41502 \/ 100 T 120 Z-stat = 1.9985 Use NORM.S.DIST(1.9985,TRUE). This is the p-value for a lower tailed test, to find the p-value for the upper tailed test, take 1 - this value. Question 18 0/ 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. The corporation will not market the product if the mean number of days is more than the 920 days claimed. The hypotheses are:

Holu: 90 H11u> 90 What is a type | error in the context of this problem? = ) The corporation will not market the Stumptown Berry even though the ~ berry does produce fruit within the 90 days. > The corporation will market the Stumptown Berry even though the ~ berry does produce fruit in more than 90 days. > The corporation will not market the Stumptown Berry even though the ~ berry does produce fruit in more than 90 days. > The corporation will market the Stumptown Berry even though the ~ berry does produce fruit within the 90 days. Question 19 0/ 1 point A 2011 survey, by the Bureau of Labor Statistics, reported that 91% of Americans have paid leave. In January 2012, a random survey of 1000 workers showed that 89% had paid leave. The resulting p-value is .0271; thus, the null hypothesis is rejected. It is concluded that there has been a decrease in the proportion of people, who have paid leave from 2011 to January 2012. What type of error is possible in this situation? = ) Type | Error () Standard Error ) Margin of Error ‘) Type |l Error () No error was made Question 20 0/ 1 point

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The average age of an adult's first vacation without a parent or guardian was reported to be 23 years old. A travel agent believes that the average age is different from reported. They sample 28 adults and they asked their age in years when they first vacationed as an adult without a parent or guardian, data shown below. See Excel for Data. adult vacation data.xlsx Test the claim using a 10% level of significance. The hypotheses for this problem are: HO: w= 23 HIZ o ?5 23 Find the test statistic. Round answer to 4 decimal places. test statistic t = % (1.9989) w Hide question 20 feedback 24.03571419—23 T-stat = 274175 V28 Done

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