Virtual+10+Test

Test 10 AP Statistics Name: Date: Part 1: Multiple Choice. Circle the letter to the best answer. 1. A simple random sample of 40 ankle biters at a large middle school are asked, "How many hours do you study in a week? If the standard deviation for the students 1.4853, what is the standard error of twice the sample size? (a) 1.4853 (b) 0.1174 (c) 0.2348 (d) 0.1661 (e) The standard error cannot be calculated from the information provided. 2. The information in question 1 was found by using Instagram. The agency using the information is worried that there may be undercoverage as not all students have Instagram. Which of the follow methods could alleviate that concern of bias? (a) Use a lower confidence level, such as 80%. (b) Use a higher confidence level, such as 99%. (c) Throw this sample out and start over again with a better sampling method. (d) Take a larger sample to get a better representation. (e) Use a z -interval instead of a t -interval to accommodate some leeway. 3. The heights (in inches) of males in United States are believed to be Normally distributed with mean ! . The mean height of a random sample of 25 American adult males is found to be 69 .72 inches with a standard deviation of 4 .15 . Which value would not be plausible for the population height of males in the United States based on a 95% confidence interval? 68.007 to 71.433 inches. H (a) 68.9372 (b) 69.0082 (c) 69.72 (d) 67.0252 (e) 70.3852 4. What is the approximate t* that satisfies the condition that the t- distribution with a sample size of 23 has a probability of 0.05 to the right of t* ? (a) 2.074 (b) 1.717 (c) 1.714 (d) 1.645 (e) 1.960 5. In checking conditions for constructing confidence intervals for a population mean, it's important to plot the distribution of sample data. Below are dot plots describing samples from three different populations. For which of the three samples would it be safe to construct a t -interval? (a) X only (b) Y only (c) Z only (d) Y and Z (e) None of the samples Shauna George 419 21 O t E Ff O O O O

6. A 90 percent confidence interval for the mean time, in minutes, for rodents to finish this exam is estimated to be (25.7, 31.5). Which of the following is the best interpretation of the interval? (a) Ten percent of the time, the time to finish the test is less than 25.7 minutes or greater than 31.5 minutes. (b) The probability is .90 that a randomly selected rodent will have finished the test between 25.7 minutes and 31.5 minutes. (c) Ninety percent of the time the mean time for a rodent to finish the test is between 25.7 and 31.5 minutes. (d) We are 90 percent confident that the mean time for a rodent to finish the test is between 25.7 minutes and 31.5 minutes. (e) We are 90 percent confident that a randomly selected time to finish the test will be between 25.7 and 31.5 minutes. 7. A sample of 5 students is used to estimate the mean time in minutes that homework takes for AP Statistics on a given day. Their responses are 32, 57, 61, 47, and 41. Which of the following conditions must be met in order to carry out a confidence interval? I. The observations are from a Normally distributed population. II. The data can be viewed as coming from a simple random sample III. The standard deviation of the population is known. (a) I only (b) II only (c) III only (d) Both I and II (e) I, II, and III 8. A random sample of 12 chocolate chip cookies from Tiff’s Treats is inspected, and your teacher counted the number of chocolate chips per cookie. A 95% confidence interval was created to estimate the mean number of chocolate chips per chocolate chip cookie from Tiff’s Treats. If the confidence level would change to 90%, which of the following escribes how the interval would change? (a) The 90% interval would be wider. (b) The 90% interval would be the same width. (c) The 90% interval would be more narrow. (d) The standard error would change. (e) Multiple of the above are correct. O

11. We have continued to diligently track Emma’s sneezes over a very long time. Her health care provider believes her average number of daily sneezes is higher than that of the population mean for a teenage girl of 3.6, and thus suspects Emma has Chronic Respiratory Involuntary Expulsion Syndrome (CRIES). The distribution of Emma’s daily sneezes for 108 randomly selected days is summarized in the Minitab output below: MEAN STDEV SEMEAN MIN Q1 MEDIAN Q3 MAX Sneezes: 4.0185 2.9576 0.2846 0 2 3.5 6 12 (a) Construct and interpret a 95% confidence interval for the population mean number of daily sneezes by Emma. Use the six step process. (20 points) (b) Based on the confidence interval in part (a), comment on whether you agree with the health provider’s suspicion. Explain your reasoning clearly. (3 points) I sample T interval for µ M true population man www of Emma's daily sneezes Random met in question independent Emma's individual sneezes dont involve other people Normal Tint I 4.0185 Sx 2.9576 n 10,8 U 95 we are 901 confident the true mean number of Emma is daily sneezes is captured by the inland to

12. Jordan’s cat “Fern” is a finicky eater. Jordan is trying to determine which of two brands of canned cat food Fern prefers, Tab-a-Cat or Chow Lion. For two months, she flips a coin each day to decide which of the two foods to feed Fern, and weighs how much Fern eats in grams. Here is the data: (a) Construct and interpret a 90% confidence interval for the difference in mean amount of food Fern eats when she is offered Tab-a-Cat and when she is offered Chow Lion. Use the six step process! (27 points) n 1̅ s Tab-a-Cat 35 87.2 3.45 Chow Lion 33 85.1 4.26 2 sample interval for difference in mean M the amount of food ten catsfrom Tab a cat Mz the amount of food Fern eatsfrom Chow Li on Random met in question independent 3 S z 30 33 Z 30 Normal large counts met 2 sample Tnt I 87.2 SX 3.45 n 35 I 2 85 I s x 2 4.26 n z 33 CL 9 we we 90 confident the true difference in meanamount of food Ferneats between Tais o cat and Chow Lion is captured by the interval