Sp23 Homework 4

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Feb 20, 2024

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Homework 4 (non-Excel portion) Spring 2023 Must be uploaded into Canvas by 11:55 pm on Wednesday, March 15, 2023. Upon submitting this homework, I affirm that I have not given or received unauthorized aid on this quiz, and I completed this work honestly and according to the instructor’s guidelines. Answers may be typed or handwritten. Homeworks are graded on completion, not accuracy. Work must be shown on the homework to receive full credit. Credit for completion will not be awarded if adequate effort has not been shown. Each homework is worth 2.5 points. So, for example, if a student gets a score of 2.5, it means that they did all the work, it does not mean the answers are all correct. In order to promote professional development of students in learning the importance of appearance and presentation of submitted work, 0.1 point will be deducted from homework scores for each the following conditions that are not met (students could lose a total of 0.2 points for not meeting these conditions): • Questions numbered and done in order (all work and answers for each question in one place) • Increased space between each question if the answers are included on the posted homework assignment If you take any definitions/concepts directly from the packet, please cite the page number in the packet. YOU MUST SHOW WORK TO GET CREDIT. 1. Define sampling distribution. - is a probability distribution that shows all possible sample results for a given sample situation 2. Define standard error. - the standard error of a statistic is the standard deviation of the sample distribution for a statistic (Questions 1-2 can be completed after Lecture 14.) 3. A random sample of size 64 is taken from a population with mean equal to 12 and the standard deviation equal to 2. Calculate the standard error of the mean. 64-12/2=58 4. An exam is given to 36 randomly selected individuals. The population mean score is 72 and the population standard deviation is equal to 9. Calculate the following probabilities: a. Probability that the sample mean is greater than 76. z = 76 − 72/1.5 =2.67 P( Z> 2.67) = 0.0038 b. Probability that the sample mean is less than 69. z = (69 - 72) / 1.5 = -2 P(Z<-2.00) = 0.0228 c. Probability that the sample mean is between 70 and 75.

z1 = (70 - 72) / 1.5 = -1.33 z2 = (75 - 72) / 1.5 = 2 P( -1.33<Z< 2) = 0.8855 (Questions 3-4 can be completed after Lecture 15.) 5. Historically, 15% of patients at a hospital have been unhappy with their care. A poll of 60 patients is to be conducted regarding their hospital stay. What is the probability that at most 6 patients will be unhappy with their care? p=0.15 A poll of 60 patients is to be conducted regarding their hospital stay n=60 P(z>-1.083) 1-P(z <1.083) 1-.8599=.1401 probability that at most 6 patients will be unhappy with their care is 0.1401 (Question 5 can be completed after Lecture 16.) 6. In using the standard normal distribution to establish a confidence interval for the average time to complete a bank transaction, what is the appropriate z-value to use for a 94.52% level of confidence. .9452/2 = .47526 Z = 2.83 7. A hamburger fast-food chain wants to make sure that the mean waiting time for orders to be filled is not longer than the average time for orders to be filled at a competing chain. The company randomly samples 81 customers and the mean waiting time is 4.8 minutes. The population standard deviation is assumed to be .6 minutes. Find a 96.92% confidence interval of the mean waiting time. You must conclude your work by stating in words what this confidence interval means in the context of waiting times. No credit will be given if this statement is not included. (Questions 6-7 can be completed after Lecture 17.) 8. A confidence interval for a true population mean is to be constructed from sample data with degrees of freedom equal to 24. Find the t-value to use for setting a 99% level of confidence. 24-1=23 1-.99= .01/2=.005 T value= -2.065 9. A random sample of four glass rods is tested and reveals the following breaking strength in pounds: 7, 8, 8, 9. Construct a 95% confidence interval for the true mean breaking strength. You must conclude your work by stating in words what this confidence interval means in terms of breaking strength. No credit will be given if this statement is not included. Calculate the degrees of freedom:df = n - 1 = 4 - 1 = 3 Calculate the significance level: α = 1 - CL = 1 - 0.95 = 0.05 Calculate the probability (p):p= 1 - α /2 = 1 - 0.05/2 = 0.975

Calculate the t-score:T0.975(3) = 3.18 confidence interval = (6.7 < U <9.3) (Questions 8-9 can be completed after Lecture 18.) 10. Historically it has taken workers an average 12 minutes to assemble a product. New training procedures have been implemented. Write the null and alternative hypotheses for the following three questions that management might be interested in. Your null will be the same for each (H 0 : µ =12). The alternative is changing. a.) Have the new procedures changed the efficiency of the workers? Null hypothesis: new procedures have not changed the efficiency of workers Alternative hypothesis: new procedures have changed the efficiency of workers b.) Have the new procedures made the workers more efficient? Null hypothesis: new procedures had no effect on efficiency of workers Alternative hypothesis: new procedures improved the efficiency of worker c.) Have the new procedures made the workers less efficient? Null hypothesis: new procedures had no effect on efficiency of workers Alternative hypothesis: new procedures reduced the efficiency of workers 11. True or False: If we accept the null, we are proving that the null is true. Why or why not? FLASE, we are not proving it is true. We would have to prove it and proving is not the same as accepting in business/ (Questions 10-11 can be completed after Lecture 19.) 12. For a z-value of -2.47, compute a p-value for an upper-tail test, a lower-tail test, AND a two-tail test. p(z<2.47)=.9222 p(z>2.47)=.0778 p(z ≠ 2.47)=.1546 13. To test that the mean lifetime of light bulbs is no more than 900 hours (population is normally distributed and population standard deviation is 30), a random sample of 49 bulbs is tested, yielding a sample mean of 910 hours. What conclusion would you reach if alpha = .05? Show all the steps of hypothesis testing. You must state your conclusion in terms of the context of the mean lifetime of light bulbs. X-bar-910 hours N=49 bulbs Pop st.dev-35 Sigma-35 Hypothesis: Ho: u ≤ 900 hours Ha: u> 900 hours alpha = 0.05 Z stat = (Xbar- µ )/(o/SQRT(n)) = (910-900)/(35/ √ 100)) = 2.857

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P-value = .9978 (From Z table) Conclusion: P-value> 0.05, There is insufficient evidence to warrant rejection of claim that the mean lifetime of light bulbs is at most 900 hours (Questions 12-13 can be completed after Lecture 20.) Excel Homework 4 Spring 2023 All Excel tutorials have a red box titled “Are you in the right tutorial?” Read the box before you start– it will help you make sure that you are using the correct tutorial for your Excel version. Use the appropriate Excel Homework 4 Tutorial in Canvas or any other sources to answer all of the following questions. There is no Excel data set for this assignment. Open a blank worksheet in Excel to complete these questions. A retail store has a goal that the mean wait time of customers in line is not more than 3 minutes. A random sample of wait times collected for 16 customers resulted in a sample mean of 3.15 minutes and a sample standard deviation of 0.5 minutes. The level of significance for this problem is 0.10. 1. Write the null and alternative for this problem. (.1) Ho: µ ≤ 3 (Claim) Ha: µ > 3 Pg. 126 2. What assumption must be true about your population since your sample size is small? (.1) The one-sample t -test has four main assumptions: The dependent variable must be continuous (interval/ratio). The observations are independent of one another. The dependent variable should be approximately normally distributed. The dependent variable should not contain any outliers. (Questions 1-2 can be completed after Lecture 20.) 3. Decide what distribution is appropriate and solve for your test statistic in Excel. Following the pattern in the Excel tutorial, write your own equation for your test statistic in an empty cell in Excel and then solve in Excel. You must handwrite or type the entire equation (including the equal sign) following the

pattern in the tutorial. No credit will be given without the equation following the pattern in the tutorial and the test statistic value. (.1) Test statistic. t = (x - µ )/s/ √ n t = (3.15 - 3)/0.5/ √ 16 t = 1.2 4. Using the appropriate T.DIST function, find the p-value. (You will have to decide if you need area to the right, left, or both tails.) Do not round, make sure you write down all of the decimal places. If you use tables to get the p-value, you will not get credit. You must handwrite or type the entire function equation (including the equal sign, the function name, and the argument) and the p-value. No credit will be given without the entire equation and the p-value. (.1) The degrees of freedom = n- 1 = 16 - 1 = 15 The p-value is 0.1244. [=T.DIST(3.15,16,TRUE) 5. Compare your p-value to alpha and state your decision about rejecting/accepting the null. (.1) Since the p-value (0.1244) is greater than the significance level (0.10), we fail to reject the null hypothesis. 6. Clearly state your conclusion now in terms of the context of the problem. What does this mean about the retail store’s goal? (.1) We can conclude that the mean wait time would be no more than 3 minutes for customers in line waiting. (Questions 3-6 can be completed after Lecture 21.) 14. Four runners were randomly sampled and it was found they ran 8, 12, 11, and 17 miles per week. Assuming the population is normally distributed, if we wish to test the claim that the mean running distance is at least 13 miles per week, what conclusion would you reach at the 10% level of significance? Show all the steps of hypothesis testing. You must state your conclusion in terms of the mean running distance of runners. Mean of observations, xbar = (8+12+11+17)/ 4 = 12 Standard deviation, s = [Sum of(x - xbar)^2/(4-1)]^0.5 = 3.8297 Sample size, n = 4 Standard error, se = s/n^0.5 = 3.8297/4^0.5 = 1.91485

Hypothesis, H0: Xbar > = 20, against H1: Xbar < 20 (Lower tailed test) statistic = (xbar - Xbar)/se = (15-20)/1.91485 = 2.6111706 p value for 3 (=n-1) degrees of freedom = 0.039805 or 3.9805% As pvalue < 10% (level of significance), so, we reject the null hypothesis. Hence, the mean running time of runners is not significantly atleast 20 miles. Question 14 can be completed after Lecture 21.)

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