Week 6 Math Knowledge Check: Analyzing Statistics Concepts

1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB=… 1/19 Week 6 Knowledge Check Homework Practice Questio… Attempt 1 of 4 Written Jan 14, 2024 1:16 PM - Jan 14, 2024 3:26 PM Attempt Score 9 / 20 - 45 % Overall Grade (Highest Attempt) 9 / 20 - 45 % 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: H 0 : μ D = 0 H 1 : μ D > 0 t-Test: Paired Two Sample for Means Non-Dominant Dominant Mean 63.33 56.28 Variance 218.9643158 128.7522105 Observations 20 20 Pearson Correlation 0.9067

1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB=… 2/19 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 decision? Hide ques±on 1 feedback 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 α = 0.01. The hypotheses are: Accept H 0 Reject H 1 Accept H 1 Do not reject H 0 Reject H 0 The p-value for a one tailed test is 0.0001. This is given to you in the output. No calculations are needed. 0.0001 < .05, Reject Ho, this is significant.

1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB=… 3/19 H 0 : μ D = 0 H 1 : μ D ≠ 0 t-Test: Paired Two Sample for Means Placebo Nicotine Mean 16.75 10.3125 Variance 64.46667 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? Accept H 0 Do not reject H 0

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1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB=… 4/19 Hide ques±on 2 feedback Question 3 0 / 1 point ___ 0.9249___ (0.2958, .2958) Hide ques±on 3 feedback Accept H 1 Reject H 1 Reject H 0 The p-value for a two tailed test is 0.0011. This is given to you in the output. No calculations are needed. 0.0011 < .01, Reject Ho, this is significant. A math teacher tells her students that eating a healthy breakfast on a test day will help their brain function and perform well on their test. During finals week, she randomly samples 45 students and asks them at the door what they ate for breakfast. She categorizes 25 students into Group 1 as those who ate a healthy breakfast that morning and 20 students into Group 2 as those who did not. After grading the final, she finds that 48% of the students in Group 1 earned an 80% or higher on the test, and 40% of the students in Group 2 earned an 80% or higher. Can it be concluded that eating a healthy breakfast improves test scores? Use a 0.05 level of significance. Hypotheses: H 0 : p 1 = p 2 H 1 : p 1 > p 2 In this scenario, what is the p-value? (Round to 4 decimal places) p-value = z =

1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB=… 5/19 Question 4 0 / 1 point ___ -0.249___ (-0.356, -.356) ___ 0.166___ (0.139, .139) Hide ques±on 4 feedback z = 0.536656 Use NORM.S.DIST(0.536656,TRUE) to find the for the lower tailed test. 0.704247, then 1 - 0.704247, this is the p-value you want to use for an upper tailed test. 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 is a significant difference in the satisfaction level of patients. Enter the confidence interval - round to 3 decimal places. < p 1 - p 2 < Z-Critical Value =NORM.S.INV(.995) = 2.575 LL = (.625-.7333) - 2.575* UL = (.625-.7333) + 2.575*

1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB=… 6/19 Question 5 0 / 1 point In a 2-sample z-test for two proportions, you find the following: p̂ 1 = 0.32 n 1 =50 p̂ 2 = 0.36 n 2 =50 Find the test statistic you will use while executing this test. Hide ques±on 5 feedback Question 6 1 / 1 point If a teacher is trying to prove that a new method of teaching economics is more effective than a traditional one, he/she will conduct a: z = -0.42 z = -0.67 z = -1.96 z = ± 1.64 z = 0.34 p = (16+18)/(50+50) = .34 z = two-tailed test one-tailed test, lower tailed test

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1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB=… 7/19 Hide ques±on 6 feedback Question 7 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. Hide ques±on 7 feedback point estimate of the population parameter a one tailed, upper tailed test Upper tailed test because the keyword is more. t = 1.916 z = 1.645 z = 1.916 t = -1.916 Use T, because the sample size is small and this is referring to a sample with a mean and SD. T stat =

1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB=… 8/19 Question 8 1 / 1 point Match the Symbol with the correct phrase. __ 5__ __ 4__ __ 1__ __ 2__ __ 3__ 1 . 2 . 3 . 4 . 5 . Hide ques±on 8 feedback Question 9 0 / 1 point Results from previous studies showed 79% 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. Compute the z or t value of the sample test statistic. Different from More than Reduced At most At least < ≤ ≥ > ≠ Reduced is less than At most means you can not have any more than that amount. That is the most you are allowed to have. At least, this is the least amount you can have but you can have more than the least amount. More than is greater Different from is not equal t = 1.645

1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB=… 9/19 Hide ques±on 9 feedback Question 10 1 / 1 point Results from previous studies showed 79% 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. What is the p-value associated with your test of hypothesis? Hide ques±on 10 feedback z = 0.69 z = 0.62 z = 1.96 Hypothesized value = .79 Z stat = 0.2437 0.6874 0.4874 0.7563

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1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB… 10/19 Question 11 0 / 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 2.35 with a standard deviation of 0.46. You sample 25 day students, and the sample mean GPA is 2.58 with a standard deviation of 0.47. Test the claim using a 5% level of significance. Assume the sample standard deviations are unequal and that GPAs are normally distributed. Hypotheses: H 0 : μ 1 = μ 2 H 1 : μ 1 ≠ μ 2 What is the p-value for this scenario? Round to four decimal places. Make sure you put the 0 in front of the decimal. p-value =___ ___ Answer: 0.0208 (0.0737, 0.0739, 0.0738) Hide ques±on 11 feedback Hypothesized value is .79 and this is an upper tailed test because of the keyword increased z = z = 0.694419 NORM.S.DIST(0.694419,TRUE) = 0.75629. This is the p-value for a lower tailed test. To find the p-value for an upper tailed take 1 - 0.75629. Use this value for the conclusion

1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB… 11/19 Question 12 0 / 1 point A new over-the-counter medicine to treat a sore throat is 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 90% 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 is 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, n1 = 25 Average Group 2 = 3.96, SD Group 2 = 2.35372, n2 = 25 ___< μ 1 - μ 2 <___ ___ Answer for blank # 1: -1.1262 (0.7967, 0.7966) Answer for blank # 2: 3.3062 (2.9633, 2.9634) Hide ques±on 12 feedback T-Stat = T-stat = -1.82463 To find the p-value use =T.DIST.2T(abs(-1.82463),53) T-Critical Value =T.INV.2T(.10,48) = 1.677224 LL = (5.84 - 3.96) - 1.677224 *

1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB… 12/19 Question 13 0 / 1 point A new over-the-counter medicine to treat a sore throat is 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 is 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, n1 = 25 Average Group 2 = 3.96, SD Group 2 = 2.35372, n2 = 25 ___< μ 1 - μ 2 <___ ___ Answer for blank # 1: -1.5515 (0.5813, 0.5812) Answer for blank # 2: 3.7315 (3.1787, 3.1788) Hide ques±on 13 feedback UL = (5.84 - 3.96) + 1.677224 * T-Critical Value =T.INV.2T(.05,48) = 2.010635

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1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB… 13/19 Question 14 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? Question 15 0 / 1 point A professor wants to know whether or not there is a difference in comprehension of a lab assignment among students depending on if the instructions are given all in text, or if they are given primarily with visual illustrations. She randomly divides her class into two groups of 15, gives one group instructions in text and the second group instructions with visual illustrations. The following data summarizes the scores the students received on a test given after the lab. Let the populations be normally distributed with a population standard deviation of 5.32 points for both the text and visual illustrations. LL = (5.84 - 3.96) - 2.010635 * UL = (5.84 - 3.96) + 2.010635 * 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.

1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB… 14/19 Text (Group 1) Visual Illustrations (Group 2) 57.3 59 45.3 57.6 87.1 72.9 61.2 83.2 43.1 64 87.3 76.7 75.2 78.2 88.2 64.4 67.5 89 86.2 72.9 67.2 88.2 54.4 43.8 93 97.1 89.2 95.1 52 84.1 Is there evidence to suggest that a difference exists in the comprehension of the lab based on the test scores? Use α=0.10 . Enter the test statistic - round to 4 decimal places. Test Statistic z =___ ___ Answer: -2.0985 (-2.4709) Hide ques±on 15 feedback Copy and paste the data into Excel. Use the Data Analysis Toolpak in Excel.

1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB… 15/19 Question 16 1 / 1 point Researchers conducted a study to measure the effectiveness of the drug Adderall on patients diagnosed with ADHD. A total of 112 patients with ADHD were randomly split into two groups. Group 1 included 57 patients and they were each given a dose of 15 mg. of Adderall daily. Group 2 was given a daily placebo. The effectiveness of the drug was measured by testing the patients before and after and observing the mean improvement in score among the patients. Group 1 was found to have a mean improvement of 11.93 points and Group 2 had a mean improvement of 9.27 points. From past studies, the population standard deviation of both groups is known to be 6.55 points. Is there evidence to suggest the patients taking Adderall had a higher mean improvement? Test at the 0.01 level of significance. Select the correct alternative hypothesis and state the p-value. Data - > Data Analysis -> scroll to where is says z:Test 2 Samples for Means - > OK Variable 1 Range: highlight Text Variable 2 Range: highlight Visual Illustrations The Hypothesized Mean Difference is 0. Variable 1 Variance: You are given the population SD you need to take 5.32 2 and use this for the population Variance for Variable 1 and 2. Make sure you click Labels in the first row and click OK. You will get an output and this is the z stat you are looking for. -2.4709 H1: µ1 < µ2; p-value = 0.0169 H1: µ1 ≠ µ2; p-value = 0.0169 Ho: µ1 > µ2; p-value = 0.0158 H1: µ1 ≠ µ2; p-value = 0.0158 H1: µ1 > µ2; p-value = 0.0158

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1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB… 16/19 Hide ques±on 16 feedback Question 17 0 / 1 point In a random sample of 50 Americans five years ago (Group 1), the average credit card debt was $5,779. 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 $1,152 five years ago, and let the current population standard deviation be $1,634. 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 p-value? Make sure you put the 0 in front of the decimal. (Round to 4 decimal places) p-value =___ ___ Answer: 0.0018 (0.0109) Hide ques±on 17 feedback H1: µ1 > µ2; p-value = 0.0169 Z-Stat = Z-Stat = 2.1486 Use =NORM.S.DIST(2.1486,TRUE) this is the p-value for the lower tailed test. Take 1 - this value to find the p-value for an upper tailed test. Z-stat =

1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB… 17/19 Question 18 1 / 1 point Match the symbol with the correct phrase. µ Hide ques±on 18 feedback Question 19 1 / 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. z-stat = -2.5465 Use NORM.S.DIST(-2.5465,TRUE). This is the p-value for a lower tailed test. To find the p-value for a two tailed test, take this value and multiply it by 2. significance level confidence level parameter power P(Type II Error) This is a parameter of the population average.

1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB… 18/19 What type of error is possible in this situation? Question 20 0 / 1 point ___ 0.0016___ (0.0097, .0097) Hide ques±on 20 feedback Type I Error Standard Error Margin of Error Type II Error No error was made 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 than 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 1% level of significance. See Excel for Data. workweek hour data.xlsx The hypotheses for this problem are: H 0 : μ = 47 H 1: μ > 47 Find the p-value. Round answer to 4 decimal places. p-value = T-stat =

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1/14/24, 3:34 PM Steven Felmlee - MATH302 D006 Fall 2023 - APEI https://myclassroom.apus.edu/d2l/lms/quizzing/user/quiz_submissions_attempt.d2l?isprv=&qi=267645&ai=3059719&isInPopup=0&cfql=0&fromQB… 19/19 Done T-stat = 2.641042 Use =T.DIST.RT(2.641042,14) to find the p-value for an upper tailed test.

knowledge check 6.1

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