AMS 310 Homework 5 (1)

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Jan 9, 2024

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AMS 310 Homework 5, Fall 2023 Prof. Rispoli Reminder: Show your reasoning! Material is based on Chapters 8 and 9 from Ahn. Upload your solutions into Blackboard. There are 15 problems, each is worth 6.5 points, 2.5 points for trying. Type up solutions to problems 1 and 2 for Chapter 8 (5-point penalty for not doing this). Chapter 8 1. The shelf life of a certain ointment is known to have a normal distribution. A sample of size 120 tubes of ointment gives X = 36.1 and s = 3.7. a) Construct a 90% confidence interval of the population mean shelf life. a = 1 − 0.9 = 0.1 t a 2 ,df = 1.66 36.1 ± 1.658 ∗ 3.7 √ 120 ¿ ( 35.54,36.66 ) b) Suppose a researcher believed that before an experiment  = 4. What would be the required sample size to estimate the population mean to be within 1.5 months with 99% confidence? a 2 = 1 − 0.99 2 = 0.005 Z a / 2 = 2.576 n = ( 2.576 ∗ 4 1.5 ) 2 = 47 2. The foot size of each of 16 men was measured, resulting in the sample mean of 11.5 inches. Assume that the distribution of foot sizes is normal with  = 1.5 inches. a) Test if the population mean of men’s foot sizes is 11 inches using  = 0.01. z = x − μ σ √ n = 11.5 − 11 1.5 √ 16 = 4 3 2 ∗ p ( z < 4 3 ) = 0.1824

b) If  = 0.01 is used, what is the probability of a type II error when the population mean is 10.75 inches? z = μ − x σ √ n = 10.75 − 11 1.5 4 = 0.67 z crit = 2..576 z B = 2..576 + 0.67 = 3.24 B = NORM .S. DIST ( 3.24 ) = 0.9994 c) Find the sample size required to ensure that the type II error probability  (10.75) = 0.1 when  = 0.01. n = α ( Z 0.005 + Z 0.1 ) 10.75 − 11 2 = 1.5 ( 2.576 + 1.28 ) 0.25 ¿ 2 = 535 3. A computer company claims that the batteries in its laptops last 4 hours on average. A consumer report firm gathered a sample of 16 batteries and conducted tests on this claim. The sample mean was 3 hours 50 minutes, and the sample standard deviation was 20 minutes. Assume that the battery time distribution as normal. a) Test if the average battery time is shorter than 4 hours at  = 0.05. t = sample mean − populationmean samplestd / √ n t = 230 − 240 5 =− 2 p ¿ 0.032 is less than the significance level so the null hypothesis is rejected. So, the average battery life is shorter than 4 hours. b) Construct a 95% confidence interval of the mean battery time. Df = 15 T.INV.2T(0.05, 15) = 2.13 Margin of error = 2.13 ¿ ( 20 4 ) = 10.66 200 ± 10.66 =( 219.34 , 240.66 ) 4. BMI is obtained as weight (in kg) divided by the square of height (in M 2 ). Adults with BMI over 25 are considered overweight. A trainer at a health club measured the BMI of

people who registered for his program this week. Assume that the population is normal. The numbers are given below. 29.4 24.2 25.6 23.6 23.0 22.4 27.4 27.8 a) Construct a 95% confidence interval for the mean BMI. x = ¿ 25.425 Std = 2.54 Df = 7 T.INV.2T(0.05, 7) = 2.36 25.425 += 2.13 = (23.295, 27.555) b) To find if newly registered people for the program are overweight on average, conduct an appropriate test using  = 0.05. 25.425 − 25 2.54 √ 8 = 0.473 P ( 0.472 ) = T .DIST . RT ( 0.473,8 ) = 0.324 Since 0.324 is greater than the significance level, don’t reject the null hypothesis. 5. A toxicologist wishes to investigate the relationship between the probability of a developmental defect and the level of a certain acid in the environment. Twenty mg/kg/day of the acid is given to mice during pregnancy, and the malformation rate of the fetuses is observed. a) How large a sample should be chosen to estimate the proportion with 95% error margin of 0.01? Use p = 0.15. Z = NORM .S .INV ( 0.975 ) = 1.96 n =( 1.96 ¿¿ 2 ∗ 0.15 ∗( 0.85 ))/ 0.01 2 = 4898.04 ¿ b) A random sample of 400 mice is taken, and 37 mice were found to have malformed fetuses. Construct a 95% confidence interval for the population proportion. E = Z ∗ √ 0.0925 ∗ 1 − 0.0925 400 = 0.0284 0.0925 ± 0.0284 =( 0.0641,0.1209 ) 6. A random sample of size 13 from a normal population is given below. 69 74 75 76 78 79 81 83 83 85 86 87 80 a) Someone claimed that  = 10. Does this data set support the claim? Test at  = 0.05. std = 5.25

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x = 79.69 X 2 = 12 ∗ 5.25 2 100 = 3.3075 b) Test if the population mean is 85 using  = 0.05. E = 5.25 √ 79.69 = 0.5881 t = 5.25 − 85 79.69 √ 13 =− 3.61 P ( 3.608 ) = T .DIST .2 T ( 3.6,12 ) = 0.00359 c) Find the p-value of the test. Use R > 2 * (1-pt(3.608265958,12)) 0.003591028 7. An airline claims that only 6% of all lost luggage is never found. If, in a sample, 17 of 200 pieces of lost luggage are not found, test the null hypothesis p = 0.06 against the alternative p > 0.06 at the 0.05 level of significance. Use the 5 step method. Null hypothesis : p = 0.06 alt = p > 0.06 z = 17 200 − 0.06 √ 0.06 ( 1 − 0.06 ) 200 = 1.4887 p = 1 − NORM .S .DIST ( 1.4887 ) = 0.06828 Chapter 9 8. Imagine a researcher wants to determine whether or not a given drug has any effect on the scores of human subjects performing a task of ESP sensitivity. He randomly assigns his subjects to one of two groups. Nine hundred subjects in group 1 (the experimental group) receive an oral administration of the drug prior to testing. In contrast, 1000 subjects in group 2 (control group) receive a placebo. For the drug group, the mean score on the ESP test was 9.78, the standard deviation was 4.05, and n is 900. For the no-drug group, the mean was 15.10, the standard deviation was 4.28, and n is 1000.

Determine if there is a significant difference between the mean scores. Use the 5 step procedure and label each step. Make sure to provide a practical interpretation of the result and give a p-value. √ ( ( m − 1 ) S 2 + 4.28 √ 1000 )= 0.004388 t = 9.78 − 15.1 0.004388 =− 1212.397 df = 1898 p = T .DIST .2 T ( 1212.397 , 1898 ) = 0 We reject the null hypothesis since the p-value is less than the significance value. 9. In a study on the healing process of bone fractures, researchers observed the time required to completely heal a bone fracture. Ten rats were randomly divided into two groups of five. No treatment was done to one group and a medical treatment was given to the other group. The time spent (in days) to completely heal the fracture for each rat is given below: Control: 30, 22, 28, 35, 45 Treatment: 33, 40, 24, 25, 24 a) Test to see if the mean duration of healing can be reduced by the medical treatment at  = 0.01. Assume equal variances. Null: meal in greater than or equal than control Alt: the mean is less than the control SE = 62.6 32 − 29.2 √ s 2 p ( 1 n 1 + 1 n 2 ) = 2.8 √ 62.6 ( 2 5 ) = 0.5596 df = 10 − 2 = 8 p = T .DIST .2 T ( 0.5596,8 ) = 0.59107 Since p is less than the significance interval, we do not reject the null hypothesis. b) Construct a 99% confidence interval for the difference of the means of the two groups.

2.8 ± ( 3.355 ∗ 5.004 ) = ( − 13.9884 , 19.5884 ) 10. The following table shows summary data on mercury concentration in salmons (in ppm) from two different areas of the Pacific Ocean. Area 1: m = 15, x = 0.0860, s 1 = 0.0032 Area 2: n = 15, y = 0.0884, s 1 = 0.0028 a. Do the data suggest that the mercury concentration is higher in salmons from Area 2? Assume equal variance and use  = 0.05. Give the 5 steps used in the 5- step method. t = 0.086 − 0.0884 0.003 √ 2 15 =− 2.191 crit : TINV ( 0.1,28 ) =− 1.701 P = TDIST ( 2.191 , 28,1 ) We reject the null hypothesis. b. Suppose actual data from this experiment is given below (note, different means and standard deviations) Use R to run the test performed in part (a). The coded needed is in the class notes. Are the results the same as in part a? Area 1 Area 2 # Data for Area 1 and Area 2

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area1 <- c(0.086, 0.119, 0.112, 0.068, 0.115, 0.069, 0.064, 0.044, 0.098, 0.103, 0.083, 0.070, 0.093, 0.072, 0.113) area2 <- c(0.136, 0.106, 0.110, 0.044, 0.106, 0.075, 0.112, 0.076, 0.117, 0.099, 0.105, 0.094, 0.062, 0.106, 0.103) # Perform a t-test t_test_result <- t.test(area1, area2, var.equal = TRUE) # assuming equal variances t_test_result data: area1 and area2 t = -1.1298, df = 28, p-value = 0.2681 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -0.026630423 0.007697089 sample estimates: mean of x mean of y 0.08726667 0.09673333 11. Independent random samples are selected from two populations. The summary statistics are given below. Assume unequal variances for the questions below. m = 5 x = 12.7 s 1 = 3.2 n = 7 y = 9.9 s 2 = 2.1 a) Construct a 95% confidence interval for the difference of the means. df = 6 Crit: t 6 , 0.025 = 2.447 12.7 − 9.9 ± 2.447 ∗ √ 3.2 2 5 + 2.1 2 7 =( 1.4063 , 4.1937 ) b) Using  = 0.05, test if the means of the two populations are different based on the result from part (a). Explain your answer. Null: u1 = u2 Alt: u1 ≠ u2 We reject the null hypothesis since the difference between the 2 means is 0. 12. Within a school district, students were randomly assigned to one of two Math teachers - Mrs. Smith and Mrs. Jones. After the assignment, Mrs. Smith had 30 students, and Mrs. Jones had 25 students.

At the end of the year, each class took the same standardized test. Mrs. Smith's students had an average test score of 78, with a standard deviation of 10; and Mrs. Jones' students had an average test score of 85, with a standard deviation of 15. Test the hypothesis that Mrs. Smith and Mrs. Jones are equally effective teachers. Use a 0.10 level of significance. (Assume that student performance is approximately normal.) Assume equal variances. Null: u1 – u2 = 0 Alt: u1-u2 ≠ 0 Df = 30+25 – 2 = 53 Sp = 12.5141 t = 78 − 85 12.5141 √ 1 30 + 1 25 =− 2.07 TINV(0.1, 53) = 1.674 TDIST(2.07, 53, 2) = 0.0433 Since p value is less than the sig level, we reject null hypothesis. 13. A manufacturer of furniture claims that the company's new model of bookshelves is easier to assemble than the old model. To test the claim, 7 people were assigned to assemble bookshelves from each of the two models. Here are the times (in minutes) they needed to assemble the bookshelves. Person 1 2 3 4 5 6 7 Old model 17 22 20 14 15 21 24 New model 15 19 18 13 15 19 23 a) Test if the mean assembly time for the new model is shorter than that of the old model using  = 0.05. null = ud = 0 alt = ud > 0

t = 1.5714 0.9759 √ 7 = 4.2603 t crit onetailed at 0.05 is 0.1943 Since 4.2603 > 1.943 we reject null hypothesis which means new model is shorter than the old model/ b) Now consider the data as two independent samples and conduct the test as a two sample test. Assume equal variances. Is the decision reversed? s x 1 2 = 1 n − 1 ∑ i = 1 n ( x 1 i − x 1 ) 2 = 14 s x 2 2 = 1 n − 1 ∑ i = 1 n ( x 2 i − x 2 ) 2 = 11.29 s x 1 x 2 ❑ = √ 0.5 ¿¿ t = 19 − 17.43 3.56 √ 2 7 = 0.83 df = 2 n − 2 = 12 t crit = 1.782 As 0.8268 < 1.782, we accept null hypothesis. The means rent different. 14. In a sociology course, one group of students had access to a set of lectures online in addition to the classroom lectures, and the other group of students did not have access to the online lectures. In each case an exam was given after the lectures. Here are the scores. Classroom and online: 61, 71, 61, 100 and 94 Classroom only: 64, 61, 56, 53 and 54. Test to see if the variances of the scores are different between the two groups using α = 0.05 . Use an F test. 𝐻 0: σ1^2 = σ2^2 𝐻 1: σ1^2 ≠ σ2^2 341.30 2 22.30 2 = 15.3 Df = 4 critical values: 𝐹 α/2 , 𝑑𝑓 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 , 𝑑𝑓 𝑑𝑒𝑛𝑜𝑚inator = 9. 6045 𝐹 1−α/2 , 𝑑𝑓 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 , 𝑑𝑓 𝑑𝑒𝑛𝑜𝑚inator = 0. 1041 P = 0.0217

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We rejct the null hypothesis. The variances are not equal. 15. Using the data from question 14, run a t test to see if the means from each group are different. Use α = 0.05 and assume unequal variances. 𝑥 1 = 77. 4 𝑥 2 = 57. 6 𝑠 1 = 18. 4743 𝑠 2 = 4. 7223 n1 = 5 n2 = 5 77.4 − 57.6 √ 18.4743 2 5 + 4.7223 2 5 = 19.8 8.5 = 2.329 Df = 4 P(t>2.329) = 0.0402 WE don’t reject the null hypothesis since the p val is less than sig level. It improved test scores.

AMS 310 Homework 5 (1)

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