Assignment3-Solution

1 Assignment 3 Due: Friday, 11:59 PM, July 14, 2023 in Crowdmark Q1 [7 pts] Answer the following question and explain or justify your answer. (a) [3 pts] Average income. A researcher wants to estimate the average income of all households in a city. He randomly selected 200 households and records their incomes. Identify whether each of. The following statements refers to a parameter or statistic. (1) The average income of the 200 households is $50,000. (2) The standard deviation of incomes in this city is $10,000. (3) The proportion of households in the sample with an income above %60,000 is 35%. (b) [2 pts] Choose the right choice and explain your answer. An October 20, 2021, poll of Canadian adults who were registered voters found that 32.7% said they would vote conservative in an upcoming election. Election records show that 37.8% voted conservative. The boldface number is a (A) sampling distribution. (B) statistic. (C) parameter. (c) [2 pts] Choose the right choice and explain your answer. All scores of SAT exam in 2020 were roughly N(1560, 300), i.e. mean 1560 and standard deviation 300. You choose an SRS of 256 students and average their SAT scores. If you do this sampling and finding average many times, the standard deviation of the average scores you get will be close to (A) 300 (B) 18.75 (C) 1.172 How to prepare your solution: • For (a), just indicating each statement refers to a parameter or statistic. No justification is needed. • For (b), choose the correct choice and explain your answer. Prepare your solution in whatever way you preferred, but make sure you upload them as PDF or PNG files in the Crowdmark. • Part (c), the justification should be the detailed calculation that you have to find the answer. Marking: • For (a): 1 point for correct answer for each statement ( statistic/ parameter/statistic ). • For (b) or (c): 1 point for indicating that 37.9% is a population parameter . 1 point for the explanation . Solution: (a) [3 pts]: (a)(1) statement refers to a statistic . (a)(2) statement refers to a population parameter . (a)(3) is a statement about a statistic . (b) [2 pts]: (c). 37.9% is obtained from the actual election results, so this number is a proportion of all registered voters, and it describes the population and therefore it is a parameter . (c) [2 pts ] Answer is (B). The SD of the sampling distribution of ࠵?̅ is ! √# = $%% √&’( = $%% )( = 18.75

2 Q2 [5 pts] Glucose testing Sarah’s doctor is concerned that she may suffer from gestational diabetes (high blood glucose levels during pregnancy). There is variation both in the actual glucose level and in the blood test that measures the level. In a test to screen for gestational diabetes, a patient is classified as needing further testing for gestational diabetes if the glucose level is above 131 milligrams per deciliter (mg/dL) one hour after having a sugary drink. Sarah’s measured glucose level one hour after the sugary drink varies according to the Normal distribution with ࠵? = 122 ࠵?࠵?/࠵?࠵? and ࠵? = 9 ࠵?࠵?/࠵?࠵? . Answer the following questions and round your answer up to 3 decimal places. (a) [1 pt] If a single glucose measurement is made, what is the probability of Sarah being diagnosed as no needing further testing for gestational diabetes? (b) [3 pts] If measurements are made on 5 separate days and the mean result is compared with the criterion 131 mg/dL, what is the probability that Sarah is diagnosed as needing further testing for gestational diabetes? (c) [1 pt] Sarah’s measured glucose level one hour after having a sugary drink varies according to the Normal distribution with 122 mg/dL and 9 mg /dL. What is the level L such that there is probability only 0.05 that the mean glucose level of four test results falls above L ? How to prepare your solution: • Prepare the solution in your preferred way, but make sure to convert your final solution file to either PDF or PNG format for further submission with Crowdmark. • For (a), note that the population distribution, N(122, 9), is used for all single glucose measurement. So, (a) is looking for P(X<131)=? To find this probability, you could either use R or Shiny app as you did in your assignment 1 . • For (b), the question is based on the average of 5 separate single measurements, and the question is asking for P( ࠵? 4 > 131) =? You need to find the distribution of ࠵? 4 first, then to find this probability, you could either use R or Shiny app as you did in your assignment 1 . • For (c), again, this question is based on the average of 4 separate single measurements, and the questions is asking for a boundary point, L, such that P( ࠵? 4 > ࠵?) = 0.05. You need to find out the distribution of ࠵? 4 , then use R or Shiny app to pin down the L value. Marking: • (a) [1 pt]: 1 point for correct answer = 0.841. • (b) [3 pts]: o 2 points for giving the correct distribution form for ࠵? 4 (1 pt for the distribution name is Normal, 1 pt for the mean 122 and sd = 9/root(5) ). o 1 point for the correct P( ࠵? 4 > 131) = 0.013 or 0.01267 • (c) [1 pt] : 1 point for the correct L value = 128.612. Solution: (a) [1 pt] P(X<131)= 0.841 where X~N(122,9) (using Find probability Shiny App or running R code: pnorm(131,122,9,lower.tail=TRUE) ) (b) [3 pts] ࠵? 4 ~࠵? =122, * √’ > = ࠵?(122,4.025), ࠵?(࠵? 4 >131) = 0.013 . (using find probability Shiny App or running R code: pnorm(131,122,4.025,lower.tail=FALSE) ) (c) [1 pt] ࠵? 4 ~࠵?(122, 4.025) , P( ࠵? 4 > ࠵?) = 0.05 , with finding percentile app, we have L = 128.621 . (using find percentile Shiny App or running R code: qnorm(0.05, 122,4.025, lower.tail=FALSE) )

4 Q4. [7 pts] Determining ࠵? ! ࠵?࠵?࠵? ࠵? " for the following scenario, define the population before you form the ࠵? ! ࠵?࠵?࠵? ࠵? " . a) [1.5 pts] A 2015 study reported that 92% students owned a cell phone. You plan to take an SRS of students to see if the percent has increased. State the hypotheses Ho and H a . b) [1.5 pts] The average income of American women who work full-time and have only a high school diploma is $48,000. You wonder whether the mean income of female graduates from your local high school who work full-time but have only a high school diploma is different from the national average. You obtain income information from an SRS of 100 female graduates of your high school who work full-time and have only a high school diploma and find that ࠵?̅ = $45,453 . What are your null and alternative hypotheses? c) [2.5 pts] In planning a study on the number of days in the past 30 days that adult driver aged 20- 29 texted while driving sometime during the day, a researcher states the hypotheses as ࠵? % : ࠵?̅ = 16 ࠵?࠵?࠵?࠵?; ࠵? - : ࠵?̅ > 16 ࠵?࠵?࠵?࠵? What’s wrong with the hypotheses? Could you also give the correct H0 and Ha? (d) [1.5] In a matched-pairs study of 18 workers, we assign each worker to both assembly lines at machine-paced or to a self-paced settings in random order. After weeks in each work setting, we collect the job stratification from all workers. Using this data, we would like to see if the machined-paced setting leads to lower level of justification. What are the null and alternative hypotheses? How to prepare your solution: • Prepare the solution in your preferred way, but make sure to convert your final solution file to either PDF or PNG format for further submission with Crowdmark. • In part (a), (b) and (d), you need to clearly define the population parameter to be tested based on the context, then state the null hypothesis (Ho) and the alternative hypothesis (Ha). • In part (c), point out the mistake and give the correct H0 and Ha. Marking: • For (a), (b) and (d): 0.5 pts for defining the target parameter; 0.5 pts for correct Ho; 0.5 pts for correct Ha. • For (c): 1 pt for pointing out the mistake; 0.5 pts for defining the target parameter; 0.5 pts for correct Ho; 0.5 pts for correct Ha. Solution: • (a) [1.5 pts] Let ࠵? be the population proportion of students owned a cell phone ࠵? % : ࠵? = 0.92, ࠵?࠵? ࠵? - : ࠵? > 0.92 • (b) [1.5 pts] Let ࠵? be the mean full-time income for women high school graduates in your school. ࠵? % : ࠵? = $48,000 ࠵?࠵? ࠵? - : ࠵? ≠ $48,000 (Note that this is a two-sided test, because you wonder whether the full-time income for women high school graduates in your school differs from the national average). • (c) [2.5 pts] Hypotheses are statements about parameters ( ࠵?) , not statistics ( ࠵?̅) . Let ࠵? be the number of days in the past 30 days that adult driver aged 20-29 texted while driving. So the research questions should not about the sample mean ( ࠵?̅) but the population mean, ࠵? . Correct one: ࠵? % : ࠵? = 16 ࠵?࠵?࠵?࠵?, ࠵?࠵? ࠵? - : ࠵? > 16 ࠵?࠵?࠵?࠵? • (d) [1.5 pts] Let ࠵? be the true mean job satisfaction scores between machine-paced and self-paced of all workers, i.e. ࠵? = Score(machine-paced)-score(self-paced). The testing is: ࠵? % : ࠵? = 0, ࠵?࠵? ࠵? - : ࠵? < 0

5 Q5. [8 pts] Confidence interval mistakes and misunderstandings. Supposed that 100 randomly selected members of the Karaoke Channel were asked how many times they typically spend on the site during the week. The sample mean ࠵?̅ was found to be 3.7 hours and the sample SD, s = 2.8 hours. Assume that the population standard deviation is known to be ࠵? = 2.9 ℎ࠵?࠵?࠵?࠵? 。 (a) [3 pts] Cary Oakey computes the 95% confidence interval for the average time on the site as 3.7 ± 1.96(2.9/100 ), what is this mistake? What is the correct 95% confidence interval for ࠵?. (b) [2 pts] He corrects this mistake and then states that “95% of the members spend between 3.13 and 4.27 hours a week on the site”. What is wrong with his interpretation of this interval? (c) [3 pts] The margin of error is slightly larger than half an hour. To reduce this to roughly 15 minutes, Cary say that the sample size needs to be doubled to 200. What is wrong this statement? What sample size do we need to reduce the margin of error to 15 minutes? How to prepare your solution: • Prepare the solution in your preferred way, but make sure to convert your final solution file to either PDF or PNG format for further submission with Crowdmark. • In part (a), show calculation details of the correct way to find the 95% CI. • In part (c), show calculation details of finding the required sample size to reduce the margin of error to 15 minutes. Marking: • For (a) [3 pts] : 1 pt for pointing out the mistake; 1 pt for the correct CI formula and 1 pt for the correct lower and upper bounds. • For (b) [2 pts] : 1 pt for pointing out the mistake about the given interpretation; 1 pt for giving the correct interpretation of the 95% CI. • For (c) [3 pts]: 1 pt for pointing out the mistake that Cary claimed the sample size should be doubled to 200; 1 pt for showing the correct way to find the required sample size; 1 pt for the correct value of the sample size needed to reduce the margin of error to 15 minutes. Solution: • (a) [3 pts] The margin error should be . ∗ ! √# ࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵? . ∗ ! # . The correct margin of error = 1.96 = &.* )% > = 0.5684 , so the 95% CI should be : ࠵?̅ ± . ∗ ! √# = 3.7 ± 0.5684 , further calculation we have: (3.7 − 0.5684, 3.7 + 0.5684) = (3.13, 4.27) • (b) [2 pts] Let ࠵? be true average time that all members of the Karaoke Channel spent on the site during the week. The interpretation of the 95% CI , (3.13,4.27), in incorrect, it is not about 95% of the members spend between 3.13 and 4.27 hours a week on the site, it is in the repeated sampling sense that 95% of CIs we constructed in this way will capture the true population mean, or it is equivalently to say that, this CI, (3.13,4.27) will capture the population mean with a success rate of 0.95. • (c) [3 pts] To halve the margin of error, the sample size needs to be 4 times the original sample size which is 400 instead of 200. To find the sample size that the margin of error is 15 minutes, we have: 1.96 = &.* √# > = )’ (% = 0.25 ⇒ ࠵? = (1 .96*2.9/0.25)^2 = 516.9257. So, it needs 517 members to make the margin of error to be 15 minutes. Q6. [6 pts] Who is the author? Statistics can help decide the authorship of literacy works. Sonnets by a certain Elizabethan poet are known to contain an average of 8.9 new words (words not used in the poet’s other works). The standard deviation of the number of new works is ࠵? = 2.5. Now a manuscript with 6 new sonnets has come to

7 (c) T he coach of a Canadian university’s women’s soccer team records the resting heart rates of the 25 team members. You should not trust a confidence interval for the mean resting heart rate of all female students at this Canadian university based on these data because A. the members of the soccer team can’t be considered a random sample of all female students at this university. B. heart rates may not have a Normal distribution. C. with only 25 observations, the margin of error will be large. (d): Many sample surveys use well-designed random samples, but half or more of the original sample can’t be contacted or refuse to take part. Any errors due to this nonresponse A. have no effect on the accuracy of confidence intervals. B. are included in the announced margin of error. C. are in addition to the random variation accounted for by the announced margin of error. (e): A medical experiment compared zinc supplements with a placebo for reducing the duration of colds. Let ࠵? denote the mean decrease, in days, in the duration of a cold. A decrease to ࠵? = 2 is a practically important decrease. The significance level of a test of H0: ࠵? = 0 versus Ha: ࠵? > 0 is defined as: A. the probability that the test fails to reject H0 when ࠵? = 2 is true. B. The probability that the test rejects H0 when ࠵? = 2 is true. C. The probability that the test rejects H0 when ࠵? = 0 ࠵?࠵? ࠵?࠵?࠵?࠵?. (f) The type II error of the test in (e) against the specific alternative ࠵? = 2 is defined as A. the probability that the test fails to reject H0 when ࠵? = 2 is true. B. The probability that the test rejects H0 when ࠵? = 2 is true. C. The probability that the test rejects H0 when ࠵? = 0 ࠵?࠵? ࠵?࠵?࠵?࠵?. (g): The power of the test in (e) against the specific alternative ࠵? = 2 is defined as A. the probability that the test fails to reject H0 when ࠵? = 2 is true. B. The probability that the test rejects H0 when ࠵? = 2 is true. C. The probability that the test rejects H0 when ࠵? = 0 ࠵?࠵? ࠵?࠵?࠵?࠵?. (h): The power of a test is important in practice because power A. Describe how well the test performs when the null hypothesis is actually true. B. Describes how sensitive the test is to violations of conditions such as normal population distribution. C. Describes how well the test performs when the null hypothesis is actually not true. How to prepare your solution: • Prepare the solution in your preferred way, but make sure to convert your final solution file to either PDF or PNG format for further submission with Crowdmark. • In you solution, indicate the problem and choice clearly. Marking: For each part, from (a) to (h): 1 pt for correct choice. 0 pt for wrong choice.

8 Solution: (In the following solution, I also explain the answer which is for your review only, you don’t need to justify your answer in your solution). • (a): answer is (A) the smaller p-value, the larger the test statistics, this stands for a rare event when the H0 is true, so smaller p-value provides stronger evidence to against H0 . • (b) answer is (C) ࠵? ࠵?࠵? ࠵?࠵?࠵?࠵?࠵? . If the population SD is known, we use the Z test statistics, otherwise we use the t test statistics. • (c) answer is (A) the members of the soccer team can’t be considered a random sample of all female students at the university . A sample obtained in this way will produce systematic bias. • (d): answer is (C), (C) are in addition to the random variation accounted for by the announced margin of error. Well-designed surveys incur error due to random chance; this random variation is the only source of error accounted for in the margin of error. All forms of bias are not accounted for and are errors in addition to those due to chance. • (e). answer is (C). The significance level ( ࠵?) is the probability of type I error, which is the probability rejecting Ho when Ho is true. • (f): answer is (A). The probability of type II error is the probability of not reject H0 when Ha ( = 2) ࠵?࠵? ࠵?࠵?࠵?࠵?. • (g): answer is (B). The power of the test is the probability of rejecting Ho when Ha ( ࠵? = 2 > 0) is true • (h): answer is (C). The power of a test describes the test’s ability to reject H0 whenever it is false. Q8. [6 pts] R coding or Shiny app Writing R code or using Shiny app to answer the following questions. (a) [1 pt] P( ࠵? ’ > 2.3) (b) [1 pt] P( −1.4 < ࠵? )% ࠵?࠵? ࠵? )% > 1.4) = ࠵?( |࠵? )% | > 1.4) (c) [1 pt] P( ࠵? $ < 1.02) (d) [1 pt] Find t* such that P( | ࠵? )’ | < t*) = P(- ࠵? ∗ < ࠵? )’ < ࠵? ∗ ) = 0.90 (e) [2 pts] Z is N(0,1), find P(Z<2.3) and find z* such that P( -z* <Z <z*) = 0.90 How to prepare your solution: • Prepare the solution in your preferred way, but make sure to convert your final solution file to either PDF or PNG format for further submission with Crowdmark. • In you solution, you need to show the R code you used to get the answer. Of course, you could use Shiny app to confirm your answer, but if you hand in solution by Shiny app, you receive zero points as this question asks you to use only R. Marking: • For part (a)-(d): 1 pt for the correct R code and ouptu. • For part(e): 1 pt for fining P(Z<2.3) and 1 pt for find z* such that P(-z*<Z<z*) =0.90 Solution (this question (a) P( ࠵? ’ > 2.3) = 0.0349 R code: pt(2.3, df=5,lower.tail=FALSE) (b) P( |࠵? )% | > 1.4) = 0.1918 R code: pt(-1.4,df=10,lower.tail=TRUE)+pt(1.4,df=10,lower.tail=FALSE) Or by symmetry: 2*pt(1.4,df=10,lower.tail=FALSE)

10 (b) [2 pts] +1 pt for calculation details; +1 pt for correct test stat value. (c) [3 pts] +1 pt for correct numerator, +1 pt for correct denominator, +1 pt for correct df. (d) [2 pts] +1 pt for the p-value; +1 pt for the conclusion. (e) [3 pts] +1 pt for choosing correct t*, +1 pt for detail CI calculation; +1 pt for correct lower and upper bounds. Solution: (a) [2 pts] Let ࠵? ) be the mean job reference score for lowest-floor group, and ࠵? & be the mean job reference score for highest-floor group. Then we have ࠵? % : ࠵? & − ࠵? ) = 0, ࠵?࠵?. ࠵? - : ࠵? & − ࠵? ) > 0 (b) [2 pts] This is a two-sample t-test, so we have ࠵? = ((.1’3+.(’)3% 5 ".$% " "$ 6 ".&’ " "$ = 2.9222 (c) [3 pts] The df=37.025 based on option 1 ! 7.89 7 78 " 7.:; 7 78 # 7 < 7.89 7 78 = 7 78>? " < 7.:; 7 78 = 7 78>? = 37.025 (d) Two-sided p-value = 2 *(one-sided p-value ) =0.0059, so the one-sided p-value = 0.0059/2 = 0. 00295. The p-value is less than 5%, so we have evidence to reject Ho and conclude Ha, we therefore conclude that the highest floor group gives higher job preference score. (e) The critical value t*= 2.0261, and 95% CI for ࠵? & − ࠵? ) (6.75-4.65) ±2.0261b &.%0 " &% + &.+’ " &% = 2.10 ±1.4560 = (0.644, 3.566)