AMS 310 Homework 5 (1)

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