Homework 6 Attempt 1

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 α = 0.01. The hypotheses are: 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 64.4666 7 33.2958 3 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 1 Accept H 0 Do not reject H 0 Reject H 1 Reject H 0 Hide question 1 feedback 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. Question 2 1 / 1 point 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.

A survey was given to college students nationwide. The question asked was: "Do you completely abstain from alcohol or not?" Of a random sample of 212 students at community colleges (Group 1), 87 claimed to completely abstain. Of a random sample of 390 students at 4-year colleges (Group 2), 103 claimed to completely abstain. At α=0.10 , is there evidence to claim a difference in the two proportions exists? Select the correct alternative hypothesis and decision. H 1 : p 1 ≠ p 2 ; Do not reject the null hypothesis. H 1 : p 1 ≠ p 2 ; Reject the null hypothesis. H 1 : p 1 > p 2 ; Do not reject the null hypothesis. H 1 : p 1 > p 2 ; Reject the null hypothesis. H 1 : p 1 < p 2 ; Do not reject the null hypothesis. H 1 : p 1 < p 2 ; Reject the null hypothesis. Hide question 3 feedback This is a two tailed test because you want to find significant evidence. z = .410377−.264103.315615 ∗ .684385 ∗ (1/212+1/390) z = 3.688439 Use NORM.S.DIST(3.688439,TRUE) to find the for the lower tailed test. 0.999887, then 1 - 0.999887 = 0.000113 Multiply by 2 for a two tailed test 0.000113*2, this is the p-value you want to use for the conclusion. Question 4 1 / 1 point A researcher took a random sample of 200 new mothers in the United States (Group 1) and found that 16% of them experienced some form of postpartum depression. Another random sample of 200 new mothers in France (Group 2), where mothers receive 16

weeks of paid maternity leave, found that 11% experienced some form of postpartum depression. Can it be concluded that the percent of women in the United States experience more postpartum depression than the percent of women in France? Use α=0.05 . Select the correct alternative hypothesis and decision . H 1 : p 1 ≠ p 2 ; Do not reject the null hypothesis. H 1 : p 1 ≠ p 2 ; Reject the null hypothesis. H 1 : p 1 > p 2 ; Do not reject the null hypothesis. H 1 : p 1 > p 2 ; Reject the null hypothesis. H 1 : p 1 < p 2 ; Do not reject the null hypothesis. H 1 : p 1 < p 2 ; Reject the null hypothesis. Hide question 4 feedback This is an upper tailed test because because of the keyword more. z = .16−.11.135 ∗ .865 ∗ (1/200+1/200) z = 1.463171 Use NORM.S.DIST(1.463171,TRUE) to find the for the lower tailed test. 0.92829, then 1 - 0.92829, this is the p-value you want to use for an upper tailed test. Question 5 1 / 1 point A high school is running a campaign against the over-use of technology in teens. The committee running the campaign decides to look at the difference in social media usage between teens and adults. They take a random sample of 200 teens in their city (Group 1) and find that 85% of them use social media, and then take another random sample of 180 adults in their city (Group 2) and find that 55% of them use social media. Find a 90% confidence interval for the difference in proportions. Enter the confidence interval - round to 4 decimal places. ___ 0.2262 ___ (50 %)

the reduction of variance alternative hypothesis Question 8 1 / 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 α = .05 level of significance, what is your conclusion? Do not reject H 0 . At the α = .05 level of significance there is not sufficient evidence to suggest that this technician's true average is less than the target accuracy. Reject H 0 . At the α = .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 Reject H 0 . At the α = .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. Hide question 8 feedback T-stat = 2.2−2.51.1516 T-stat = -1.04348 Lower tailed test use =T.DIST(-1.04348,15,TRUE). Use this p-value to state a conclusion.

Question 9 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: a one tailed, upper tailed test one-tailed test, lower tailed test two-tailed test point estimate of the population parameter Hide question 9 feedback Upper tailed test because the keyword is more. Question 10 1 / 1 point A type II error occurs when: the null hypothesis is incorrectly rejected when it is true the sample mean differs from the population mean the test is biased the null hypothesis is incorrectly accepted when it is false Question 11 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.

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: H 0 : μ 1 = μ 2 H 1 : μ 1 < μ 2 Find the p-value. Round answer to 2 decimal places. Make sure you include the 0 in front of the decimal. p - value =___ ___ Answer: 0.02

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 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 2.35 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: 0.0737 Hide question 13 feedback T-Stat = 2.35−2.58.46230+.47225 T-stat = -1.82463 To find the p-value use =T.DIST.2T(abs(-1.82463),53) Question 14 1 / 1 point Which of the following is a requirement that must be true in order to run a t-test over a z-test?

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: 0.1458 (-2.4709) Hide question 15 feedback 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 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 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 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 1276 parts and knows the population standard deviation to be 347. The Midwest group ships an average of 1439 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 is the test statistic? (Round to 4 decimal places) z =___ ___ Answer: -2.1083 Hide question 16 feedback Z-stat = 1276−1439347235+298235 Question 17 1 / 1 point

Hide question 18 feedback Beta is the probability of committing a Type II Error. And the power is 1 - Beta. Question 19 1 / 1 point The CDC national estimates that 1 in 68 = 0.0147 children are diagnosed with have been diagnosed with Autism Spectrum Disorder (ASD). A researcher believes that the proportion of children in their county is different from the CDC estimate. The hypotheses are: H 0 : p = 0.0147 H 1 : p ≠ 0.0147 What is a type II error in the context of this problem? The proportion of children diagnosed with ASD in the researcher's county is believed to be different from the national estimate and the proportion is different. The proportion of children diagnosed with ASD in the researcher's county is believed to be different from the national estimate, even though the proportion is the same. The proportion of children diagnosed with ASD in the researcher's county is believed to be the same as the national estimate and the proportion is the same. The proportion of children diagnosed with ASD in the researcher's county is believed to be the same as the national estimate, even though the proportion is the different. Question 20 1 / 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 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 = ___ 0.0097 ___ Hide question 20 feedback T-stat =49.7333−474.00832515 T-stat = 2.641042 Use =T.DIST.RT(2.641042,14) to find the p-value for an upper tailed test.