Statistics Quiz A Chapter 18 Key

Statistics Quiz A-Chapter 18-Key 1. a. We want to find the probability that no more than 12 students in the class will say that they frequently experience stress. This is the same as asking the probability of finding less than 26.7% of "stressed" students in a class of 45 students. Check the conditions: 1. 10% condition: 45 students is less than 10% of all students who could take the class 2. Success/failure cond.: np = 45(0.33) = 14.85 nq = 45(0.67) = 30.15 which both exceed 10 We can use the N(0.33, sqrt(((0.33)(0.67))/45) = 0.07) model the sampling distribution. We need to standardize the 26.7% and the find the probability of getting a z-score less than or equal to the one we find: z = (0.267 - 0.33)/0.07 = - 0.9 P( hat p <0.267)= P(z < - 0.9) = 0.1841 , so the probability is about 18.4% that no more than 12 students will say that they frequently experience stress in their daily lives. b. From part a, we can use N( 0.33, 0.070) to model the sampling distribution. Twenty students is about 44.4% of the class. This is about 1.63 standard deviations above what we would expect, which is not a surprising result. 2. The average composite ACT score for Ohio students who took the test in 2003 was 21.4. Assume that the standard deviation is 1.05. In a random sample of 25 students who took the exam in 2003, what is the probability that the average composite ACT score is 22 or more? (Make sure to identify the sampling distribution you use and check all necessary conditions.) Check the conditions: 1. Random sampling condition: We have been told that this is a random sample. 2. Independence assumption: It's reasonable to think that the scores of the 25 students are mutually independent. 3. 10% condition: 25 students is certainly less than 10% of all students who took the exam. We're assuming that the model for composite ACT scores has mean mu = 21.4 and standard deviation sigma = 1.05 Since the sample size is large enough and the distribution of ACT scores is most likely unimodal and symmetric, CLT allows us to describe the sampling distribution of y with a Normal model with mean 21.4 and SD( overline y )= 1.05/(sqrt(25)) = 0.21 . An average score of 22 is z = (22 - 21.4)/0.21 = 2.86 SDs above the mean. P( overline x >22)= P(Z > 2.86) = 0.0021 so the probability that the average composite ACT score for a sample of 25 randomly selected students is 22 or more is 0.0021. 18-14 Statistics Quiz C-Chapter 18-Key 1. It is generally believed that nearsightedness affects about 12% of children. A school district gives vision tests to 133 incoming kindergarten children. a. Describe the sampling distribution model for the sample proportion by naming the model and telling its mean and standard deviation. Justify your answer. We can assume these kids are a random sample of all children, and certainly less than 10% of them. We expect пр 133(0.12) 15.96 successes and 117.04 failures so the sample size is large enough to use the sampling model N(0.12, 0.028). b. Sketch and clearly label the model. Proportion of 133 children who ure nearsighted 68% 95% 0.036 0.064 0.092 0.12 0.148 0.176 0.204 99,7% c. What is the probability that in this group over 15% of the children will be found to be nearsighted? 0.15-0.12 0.028 =1.07 P(p>0.15)P(2>1.07)=0.142 2. Wildlife scientists studying a certain species of frogs know that past records indicate the adults should weigh an average of 118 grams with a standard deviation of 14 grams. The researchers collect a random sample of 50 adult frogs and weigh them. In their sample the mean weight was only 110 grams. One of the scientists is alarmed, fearing that environmental changes may be adversely affecting the frogs. Do you think this sample result is unusually low? Explain. We have a random sample of frogs drawn from a much larger population. With a sample of size 50 the CLT says that the approximate sampling model for sample means will be (118, 1.98). A sample mean of only 110 grams is about 4 standard deviations below what we expect, a very unusual result.