STAT 361 Final Exam Practice: Key Concepts and Problems

STAT 361 Final Exam Practice Problems Materials for the Final • You may use a calculator on the final. (Bring an extra set of batteries, just in case.) Cell phone calculators, smart watches, laptops, and similar items are not permitted on the exam. • You will be provided with a formula sheet. It will be posted on Blackboard prior to the exam. • You will be provided with statistical tables (such as Binomial, Normal, t, and χ 2 ). Topics to Know: • Binomial Distribution • Normal Distribution • Chi-Square Distribution • t -distribution • Chapter 9: Confidence Intervals – 1 Mean; 1 Proportion – Standard Errors / Margin of Errors – How to interpret a confidence interval. – How to draw conclusions using a confidence interval. • Chapter 10: – Errors (Type I, Type II) and the Power of the Test – Hypothesis Testing – stating hypotheses (using symbols), checking assumptions, performing the test, making decisions, writing a conclusion based on the problem setup, and interpreting your results. – Finding Critical Values. Stating Critical Regions. Make decisions and conclusions based on Critical Regions. – Finding p -value’s. Make decisions and conclusions based on p -value’s. – Expect to perform a hypothesis test for each scenario we discussed in class: * 1 Mean; 1 Proportion * Given a confidence interval, perform the hypothesis test. • Concept Questions Notes: • You should also study all of your old homework assignments. Possible exam questions may come from those as well. • This practice problem sheet is not the only thing that you should review. It is an extra resource to help you practice for the final exam. 1

(Some) Previous Material 1. Let X ∼ Bin (9 , 0 . 30). Find the probability. Round your answers to 4 decimal places. (a) Find P ( X ≤ 4). (b) Find P ( X < 6). (c) Find P ( X > 7). (d) Find P ( X ≥ 4). 2. Let Z ∼ N (0 , 1). Find the appropriate z -value. Round your answers to 2 decimal places. (a) Find z 0 . 036 . (b) Find z 0 . 02 . (c) Find - z 0 . 06 . 3. Let T represent the T -Distribution. Find the appropriate t -value, given the degrees of freedom. Round your answers to 3 decimal places. (a) Find t 0 . 04 (19). (b) Find - t 0 . 07 (14). (c) Find t 0 . 03 (8). 4. For a chi-squared distribution, find (a) χ 2 0 . 005 when v = 6; (b) χ 2 0 . 05 when v = 19; (c) χ 2 0 . 01 when v = 12. 2

Chapter 9 5. A random sample of 100 automobile owners in the state of Virginia shows that an automo- bile is driven on average 23,500 kilometers per year with a population standard deviation of 3900 kilometers. Assume the distribution of measurements to be approximately normal. (a) Construct a 99% confidence interval for the average number of kilometers an auto- mobile is driven annually in Virginia. (b) What can we assert with 99% confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year? 6. A random sample of 25 tablets of buffered aspirin contains, on average, 325.05 mg of aspirin per tablet, with a standard deviation of 0.5 mg. Assume that the aspirin content is normally distributed. Construct an 80% confidence interval for the average amount of aspirin per tablet. (Also, what is the value of the standard error and the margin of error?) 7. An efficiency expert wishes to determine the average time that it takes to drill 3 holes in a certain metal clamp. How large a sample will she need to be 95% confident that her sample mean will be within 15 seconds of the true mean? Assume that it is known from previous studies that σ = 40 seconds. 8. Compute a 95% confidence interval for the proportion of defective items in a process when it is found that a sample of size 100 yields 8 defectives. 9. How large a sample is needed in Question 8 if we wish to be 98% confident that our sample proportion will be within 0.05 of the true proportion of defectives? We believe that the value for ˆ p in the previous question seems reasonable. 10. A study is to be made to estimate the proportion of residents of a certain city and its suburbs who favor the construction of a nuclear power plant near the city. How large a sample is needed if one wishes to be at least 95% confident that the estimate is within 0.04 of the true proportion of residents who favor the construction of the nuclear power plant? 3

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Chapter 10 You should be able to do hypothesis testing problems using both the critical region method and using the p -value method. 11. A manufacturer has developed a new fishing line, which the company claims has a mean breaking strength of 15 kilograms with a standard deviation of 0.5 kilograms. To test the hypothesis that μ = 15 kilograms against the alternative that μ < 15 kilograms, a random sample of 50 lines will be tested. The critical region is defined to be ¯ x < 14 . 9. Use the normal distribution. (a) Find the probability of committing a type I error when H 0 is true. (b) Evaluate β for the alternatives μ = 14 . 8 and μ = 14 . 9 kilograms. (c) Find the power for the two alternatives stated in part b. (d) How can you decrease α and β simultaneously? 12. A fabric manufacturer believes that the proportion of orders for raw material arriving late is p = 0 . 6. If a random sample of 12 orders shows that 3 or fewer arrived late, the hypothesis that p = 0 . 6 should be rejected in favor of the alternative p < 0 . 6. Use the binomial distribution. (a) Find the probability of committing a type I error if the true proportion is p = 0 . 6. (b) Find the probability of committing a type II error for the alternatives p = 0 . 3 and p = 0 . 4. (c) Find the power for the two alternatives stated in part b. (d) Based on the power, are these tests good ones? 13. An electrical firm manufactures light bulbs that have a lifetime that is approximately normally distributed with a mean of 800 hours and a standard deviation of 40 hours. Test the hypothesis that μ = 800 hours against the alternative, μ 6 = 800 hours, if a random sample of 30 bulbs has an average life of 788 hours. Use a significance level of 0.02. 14. A random sample of 64 bags of white cheddar popcorn weighed, on average, 5.23 ounces with a standard deviation of 0.24 ounces. (a) Test the hypothesis that μ = 5 . 5 ounces against the alternative hypothesis, μ < 5 . 5 ounces, at the 0.05 level of significance. (b) Suppose we talked to the manufacturer of the white cheddar popcorn bags, and found out that the population mean amount of popcorn that should actually be in the bags was 5.5 ounces. Given the result of your hypothesis test in part (a), did we make the correct decision, or was an error made? If an error was made, what type of error did we make? 4

15. It is believed that 60% or more of the residents in a certain area favor an annexation suit by a neighboring city. We test to see if the actual percentage is smaller than this. What conclusion would you draw if only 110 in a sample of 200 voters favored the suit? Use a 0.05 level of significance. 16. Suppose we perform the hypothesis test H 0 : μ = 80 versus H 1 : μ 6 = 80. We do this by constructing a 97% confidence interval. We find the confidence interval for μ to be 75 < μ < 79. What decision should we make for the hypothesis test, assuming a 3% level of significance? 5

Example Concept Questions: 17. When finding sample sizes, how do you round n ? 18. What are we doing when we find a confidence interval? 19. Are confidence intervals guaranteed to contain the population parameter? 20. What is the difference in checking the success / failure condition for proportions when doing a confidence interval or a hypothesis test? 21. What is the difference between a hypothesis test and a confidence interval? 22. What does a p -value actually tell us? How do we calculate a p -value (what is the general formula)? When we calculate a p -value, do we assume that the null hypothesis is true, or that the alternative hypothesis is true? 23. Why do we care about Type I Errors, Type II Errors, and the power of the test? What do they tell us? 24. Is α always equal to 0.05? 25. How are α , β , and power related to each other? 26. How do you decrease α and β simultaneously? 27. How can you increase power? How can you decrease power? 28. What are some of the assumptions that we need to be aware of to perform these tests, or create these confidence intervals? 29. What is the difference in checking the success / failure condition for proportions when doing a confidence interval or a hypothesis test? 30. Can you use the t -distribution without having a normal population? 31. Suppose that a device advertised to increase a car’s gas mileage really does not work. We test it on a small fleet of cars using the null hypothesis H 0 : The devise is not effective. Our data results in a p -value = 0 . 004. What probably happens as a result of our ex- periment? Explain your answer. Possible Answers: (a) We correctly rejected H 0 . (b) We correctly failed to reject H 0 . (c) We rejected H 0 , making a Type I Error. (d) We rejected H 0 , making a Type II Error. 32. What are some of the parameter values that you might use when stating hypotheses? 33. Are you dealing with a 1-sided test, or a 2-sided test? 34. etc. 6

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Solutions (Some) Previous Material 1. Let X ∼ Bin (9 , 0 . 30). Find the probability. Round your answers to 4 decimal places. (a) Find P ( X ≤ 4). Solution: Table: look at the table where n = 9. Look at the column where p = 0 . 30. The table already gives areas to the left, so look at the row where r = 4. The probability is 0 . 9012 . Calculator: P ( X ≤ 4) = binomialcdf (9 , 0 . 30 , 4). (b) Find P ( X < 6). Solution: First, convert the value so it has a ≤ in it. P ( X < 6) = P ( X ≤ 5). Table: look at the table where n = 9. Look at the column where p = 0 . 30. Look at the row where r = 5. The probability is 0 . 9747 . Calculator: P ( X < 6) = P ( X ≤ 5) = binomialcdf (9 , 0 . 30 , 5). (c) Find P ( X > 7). Solution: First, convert the value so that it has a ≤ in it. P ( X > 7) = 1 - P ( X ≤ 7). Table: look at the table where n = 9. Look at the column where p = 0 . 30. Look at the row where r = 7. Subtract the probability from 1. The probability is 1 - 0 . 9996 = 0 . 0004 . Calculator: P ( X > 7) = 1 - P ( X ≤ 7) = 1 - binomialcdf (9 , 0 . 30 , 7). (d) Find P ( X ≥ 4). Solution: First, convert the value so it has a ≤ in it. P ( X ≥ 4) = 1 - P ( X < 4) = 1 - P ( X ≤ 3) Table: look at the table where n = 9. Look at the column where p = 0 . 30. Look at the row where r = 3. Subtract the probability from 1. The probability is 1 - 0 . 7297 = 0 . 2703 . Calculator: P ( X ≥ 4) = 1 - P ( X ≤ 3) = 1 - binomialcdf (9 , 0 . 30 , 3). 7

2. Let Z ∼ N (0 , 1). Find the appropriate z -value. Round your answers to 2 decimal places. (a) Find z 0 . 036 . Solution: z 0 . 036 = invnorm (1 - 0 . 036) = 1 . 80 (b) Find z 0 . 02 . Solution: z 0 . 02 = invnorm (1 - 0 . 02) = 2 . 05 (c) Find - z 0 . 06 . Solution: - z 0 . 06 = invnorm (0 . 06) = - 1 . 554773 ≈ - 1 . 55 3. Let T represent the T -Distribution. Find the appropriate t -value, given the degrees of freedom. Round your answers to 3 decimal places. (a) Find t 0 . 04 (19). Solution: t 0 . 04 (19) = invT (1 - 0 . 04 , 19) = 1 . 84953 ≈ 1 . 850 (b) Find - t 0 . 07 (14). Solution: - t 0 . 07 (14) = invT (0 . 07 , 14) = - 1 . 564575 ≈ - 1 . 565 (c) Find t 0 . 03 (8). Solution: t 0 . 03 (8) = invT (1 - 0 . 03 , 8) = 2 . 18915 ≈ 2 . 189 4. For a chi-squared distribution, find (a) χ 2 0 . 005 when v = 6; Solution: 18.548 (b) χ 2 0 . 05 when v = 19; Solution: 30.144 (c) χ 2 0 . 01 when v = 12. Solution: 26.217 8

Chapter 9 5. A random sample of 100 automobile owners in the state of Virginia shows that an automo- bile is driven on average 23,500 kilometers per year with a population standard deviation of 3900 kilometers. Assume the distribution of measurements to be approximately normal. (a) Construct a 99% confidence interval for the average number of kilometers an auto- mobile is driven annually in Virginia. Solution: List what we know. n = 100 ¯ x = 23500 σ = 3900 α = 0 . 01 We know σ so there is no need to estimate it using s . Use z because σ is known. - z α/ 2 = - z 0 . 01 / 2 = - z 0 . 005 = invnorm (0 . 005) = - 2 . 576 z α/ 2 = z 0 . 005 = 2 . 576 The 99% confidence interval for μ is given by ¯ x ± z α/ 2 · σ √ n ⇒ 23500 ± 2 . 576 · 3900 √ 100 The 99% CI for μ is: 22495 . 36 < μ < 24504 . 64 or (22495 . 36 , 24504 . 64) . (b) What can we assert with 99% confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year? Solution: Use the formula for MOE. We have the following values: n = 100 α = 0 . 01 z α/ 2 = 2 . 576 σ = 3900 We can assert that with 99% confidence, MOE ≤ z α/ 2 · σ √ n = 2 . 576 · 3900 √ 100 = 1004 . 64 9

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

6. A random sample of 25 tablets of buffered aspirin contains, on average, 325.05 mg of aspirin per tablet, with a standard deviation of 0.5 mg. Assume that the aspirin content is normally distributed. Construct an 80% confidence interval for the average amount of aspirin per tablet. (Also, what is the value of the standard error and the margin of error?) Solution: List what you know. n = 25 ¯ x = 325 . 05 s = 0 . 5 α = 0 . 20 Use the t distribution with v = 25 - 1 = 24 degrees of freedom because σ is unknown. t α/ 2 (24) = t 0 . 20 / 2 (24) = t 0 . 10 (24) = 1 . 318 The 80% confidence interval for μ is given by ¯ x ± t α/ 2 · s √ n ⇒ 325 . 05 ± 1 . 318 · 0 . 5 √ 25 The 80% CI for μ is: 324 . 92 < μ < 325 . 18 or (324 . 92 , 325 . 18). Note: The margin of error is 1 . 318 · 0 . 5 √ 25 = 0 . 1318 and the standard error is 0 . 5 √ 25 = 0 . 1. 7. An efficiency expert wishes to determine the average time that it takes to drill 3 holes in a certain metal clamp. How large a sample will she need to be 95% confident that her sample mean will be within 15 seconds of the true mean? Assume that it is known from previous studies that σ = 40 seconds. Solution: List what you know. α = 0 . 05 σ = 40 h = 15 We need to calculate n . - z α/ 2 = - z 0 . 05 / 2 = - z 0 . 025 = invnorm (0 . 025) = - 1 . 96 z α/ 2 = z 0 . 025 = 1 . 96 Then we have that h ≥ z α/ 2 · σ √ n ⇒ 15 ≥ 1 . 96 · 40 √ n ⇒ n ≥ (1 . 96)(40) 15 2 ⇒ n ≥ 27 . 318 ⇒ n ≈ 28 10

8. Compute a 95% confidence interval for the proportion of defective items in a process when it is found that a sample of size 100 yields 8 defectives. Solution: ˆ p = 8 / 100 = 0 . 08. α = 0 . 05. z α/ 2 = z 0 . 025 = 1 . 96. A 95% confidence interval for p is given by ˆ p ± z α/ 2 r ˆ p (1 - ˆ p ) n ⇒ 0 . 08 ± 1 . 96 · r 0 . 08(0 . 92) 100 The 95% CI for p is: 0 . 027 < p < 0 . 133 or (0 . 027 , 0 . 133). 9. How large a sample is needed in Question 8 if we wish to be 98% confident that our sample proportion will be within 0.05 of the true proportion of defectives? We believe that the value for ˆ p in the previous question seems reasonable. Solution: α = 0 . 02 so z α/ 2 = z 0 . 01 = 2 . 33. h = 0 . 05. ˆ p = 0 . 08. If ˆ p is used as an estimate of p , we can be 98% confident that the error will be less than a specified MOE when the sample size is approximately n = z 2 α/ 2 ˆ p (1 - ˆ p ) h 2 = (2 . 33 2 )(0 . 08)(0 . 92) (0 . 05) 2 = 159 . 83 ≈ 160 10. A study is to be made to estimate the proportion of residents of a certain city and its suburbs who favor the construction of a nuclear power plant near the city. How large a sample is needed if one wishes to be at least 95% confident that the estimate is within 0.04 of the true proportion of residents who favor the construction of the nuclear power plant? Solution: α = 0 . 05 so z α/ 2 = z 0 . 025 = 1 . 96. h = 0 . 04. We are not given a specific value of ˆ p . We can be at least 95% confident that the error will not exceed a specified MOE when the sample size is n = z 2 α/ 2 4 · h 2 = 1 . 96 2 4(0 . 04) 2 = 600 . 25 ≈ 601 . 11

Chapter 10 11. A manufacturer has developed a new fishing line, which the company claims has a mean breaking strength of 15 kilograms with a standard deviation of 0.5 kilograms. To test the hypothesis that μ = 15 kilograms against the alternative that μ < 15 kilograms, a random sample of 50 lines will be tested. The critical region is defined to be ¯ x < 14 . 9. (a) Find the probability of committing a type I error when H 0 is true. Solution: Here, n = 50, μ = 15 and σ = 0 . 5. The critical region is ¯ x < 14 . 9. z = ¯ x - μ σ/ √ n = 14 . 9 - 15 0 . 5 / √ 50 = - 1 . 41 α = P (Type I error) = P (Reject H 0 | H 0 is true) = P ( ¯ X < 14 . 9 | μ = 15 ) = P ( Z < - 1 . 41) = normalcdf ( - 999 , - 1 . 41) = 0 . 0793 . (b) Evaluate β for the alternatives μ = 14 . 8 and μ = 14 . 9 kilograms. Solution: z μ =14 . 8 = ¯ x - μ σ/ √ n = 14 . 9 - 14 . 8 0 . 5 / √ 50 = 1 . 41 β μ =14 . 8 = P (Type II error) = P (Do Not Reject H 0 | H 0 is false) = P ( ¯ X > 14 . 9 | μ = 14 . 8 ) = P ( Z > 1 . 414) = normalcdf (1 . 41 , 999) = 0 . 0793 z μ =14 . 9 = ¯ x - μ σ/ √ n = 14 . 9 - 14 . 9 0 . 5 / √ 50 = 0 β μ =14 . 9 = P (Type II error) = P ( ¯ X > 14 . 9 | μ = 14 . 9 ) = P ( Z > 0) = 0 . 5 (c) Find the power for the two alternatives stated in part b. Solution: Power is calculated as 1 - β . For μ = 14 . 8, the power is 1 - 0 . 0793 = 0 . 9207. For μ = 14 . 9, the power is 1 - 0 . 5 = 0 . 5. You can also think of power as P (Reject H 0 | H 1 is true). (d) How can you decrease α and β simultaneously? Solution: You can increase sample size. 12

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

12. A fabric manufacturer believes that the proportion of orders for raw material arriving late is p = 0 . 6. If a random sample of 12 orders shows that 3 or fewer arrived late, the hypothesis that p = 0 . 6 should be rejected in favor of the alternative p < 0 . 6. Use the binomial distribution. (a) Find the probability of committing a type I error if the true proportion is p = 0 . 6. Solution: Here, n = 12, p = 0 . 6. The critical region is x ≤ 3. α = P (Type I error) = P ( X ≤ 3 | p = 0 . 6) = 3 X k =0 12 k (0 . 6) k (0 . 4) 12 - k = binomialcdf (12 , 0 . 6 , 3) = 0 . 0153 . (b) Find the probability of committing a type II error for the alternatives p = 0 . 3 and p = 0 . 4. Solution: Here, n = 12 and the critical region is x ≤ 3. β p =0 . 3 = P (Type II error) = P ( X > 3 | p = 0 . 3) = 12 X k =4 12 k (0 . 3) k (0 . 7) 12 - k = 1 - 3 X k =0 12 k (0 . 3) k (0 . 7) 12 - k = 1 - binomialcdf (12 , 0 . 3 , 3) = 1 - 0 . 4925 = 0 . 5075 . β p =0 . 4 = P ( X > 3 | p = 0 . 4) = 1 - binomialcdf (12 , 0 . 4 , 3) = 1 - 0 . 2253 = 0 . 7747 . (c) Find the power for the two alternatives stated in part b. Solution: Power is calculated as 1 - β . For p = 0 . 3, the power is 1 - 0 . 5075 = 0 . 4925 . For p = 0 . 4, the power is 1 - 0 . 7747 = 0 . 2253 . (d) Based on the power, are these tests good ones? Solution: The power is relatively low for both of these alternatives. It would not be a good test if we want to distinguish between p = 0 . 6 and p = 0 . 4, or between p = 0 . 6 and p = 0 . 3. 13

13. An electrical firm manufactures light bulbs that have a lifetime that is approximately nor- mally distributed with a mean of 800 hours and a standard deviation of 40 hours. Test the hypothesis that μ = 800 hours against the alternative, μ 6 = 800 hours, if a random sample of 30 bulbs has an average life of 788 hours. Use a significance level of 0.02. Solution: Hypotheses H 0 : μ = 800 hours versus H 1 : μ 6 = 800 hours This is a two-sided test. Given Information We have n = 30, μ 0 = 800, ¯ x = 788, σ = 40. Since σ is known, use z . Computations z = ¯ x - μ 0 σ/ √ n = 788 - 800 40 / √ 30 = - 1 . 64 Critical Region α = 0 . 02. α/ 2 = 0 . 01. - z α/ 2 = - z 0 . 01 = invnorm (0 . 01) = - 2 . 33 z α/ 2 = z 0 . 01 = invnorm (1 - 0 . 01) = 2 . 33 Reject H 0 if z < - 2 . 33 or z > 2 . 33 Critical Region Decision and Conclusion Since - 1 . 64 6 < - 2 . 33, Do Not Reject H 0 . There is not enough evidence that the average lifetime of bulbs is different from 800 hours. P -value p -value = 2 · P ( Z < - 1 . 64) = 2 · normalcdf ( - 999 , - 1 . 64) = 0 . 1010 P -value Decision and Conclusion Since 0 . 1010 6 < 0 . 02, Do Not Reject H 0 . This p -value is somewhat “large”. There is not enough evidence that the average lifetime of bulbs is different from 800 hours. 14

14. A random sample of 64 bags of white cheddar popcorn weighed, on average, 5.23 ounces with a standard deviation of 0.24 ounces. Test the hypothesis that μ = 5 . 5 ounces against the alternative hypothesis, μ < 5 . 5 ounces, at the 0.05 level of significance. Solution Part A: Hypotheses H 0 : μ = 5 . 5 versus H 1 : μ < 5 . 5 This is a one-sided test. Given Information We have n = 64, ¯ x = 5 . 23 and s = 0 . 24. Since σ is unknown, use t . (Side note: since n is large, we could also use z , but I’m not requiring you to know this. t is fine, since as n → ∞ , we end up with the normal distribution.) As sump tions Check Assume that we have an approximately normal population. This is required to use the t -distribution. (From the CLT, we are okay because the sample size is “large”.) Test Statistic Calculation t = ¯ x - μ 0 s/ √ n = 5 . 23 - 5 . 5 0 . 24 / √ 64 = - 9 df = n - 1 = 63 Critical Region α = 0 . 05. - t 0 . 05 (63) ≈ - t 0 . 05 (60) = - 1 . 671. Reject H 0 if t < - 1 . 671. Note: If you use the table, we do not have 63 degrees of freedom on it. When we do our degrees of freedom calculations, we always round down to be safe. You could use the value of v = 60 to use the table. Critical Region Decision and Conclusion Since - 9 < - 1 . 671, Reject H 0 . There is enough evidence that the average weight of white cheddar popcorn bags is less than 5.5 ounces. P -value p -value = P ( T < - 9) ( = tcdf ( - 999 , - 9 , 63) ≈ 0 via calculator < 0 . 0005 via table Note: If you use the table, we do not have 63 degrees of freedom on it. When we do our degrees of freedom calculations, we always round down to be safe. You could use the value of v = 60 to use the table. P -value Decision and Conclusion Since 0 < 0 . 05, Reject H 0 . This p -value is “small”. There is enough evidence that the average weight of white cheddar popcorn bags is less than 5.5 ounces. Note: We would end up rejecting the null hypothesis for α levels even smaller than 0.05. Solution Part B: In this case, we Rejected H 0 . However, we found out from the manufacturer that H 0 was actually correct and that μ = 5 . 5. We Rejected H 0 , even though it was true. This means we made a Type I Error . 15

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

15. It is believed that 60% or more of the residents in a certain area favor an annexation suit by a neighboring city. We test to see if the actual percentage is smaller than this. What conclusion would you draw if only 110 in a sample of 200 voters favored the suit? Use a 0.05 level of significance. Solution: This is a one sample proportion. α = 0 . 05. Hypotheses H 0 : p = 0 . 60 H 1 : p < 0 . 60 Note that we could have also written H 0 : p ≥ 0 . 60 to better align with the problem setup. Notes This is a large sample, n = 200. We can use a normal approximation for this. Either use the version with ˆ p = x/n = 110 / 200 = 0 . 55 or use the version with x = 110. As sump tions Check We need to check the success / failure condition. Expected Number of Successes: np 0 = 200(0 . 6) = 120 ≥ 5 X Expected Number of Failures: n (1 - p 0 ) = 200(1 - 0 . 6) = 80 ≥ 5 X Since this condition is met, we can continue with the test. Test Statistic Calculation z = x - np 0 p np 0 (1 - p 0 ) = 110 - 200(0 . 6) p 200(0 . 6)(0 . 4) = - 1 . 44 = ˆ p - p 0 q p 0 (1 - p 0 ) n = 0 . 55 - 0 . 6 q 0 . 6(0 . 4) 200 Critical Region Reject H 0 if z < - z α = - z 0 . 05 = invnorm (0 . 05) = - 1 . 645 Critical Region Decision and Conclusion Since - 1 . 44 6 < - 1 . 645, Do Not Reject H 0 . There is not enough evidence that less than 60% of the residents favor the annexation suit. P -value p -value ≈ P ( Z < - 1 . 44) = normalcdf ( - 999 , - 1 . 44) = 0 . 0749 P -value Decision and Conclusion Since 0 . 0749 6 < 0 . 05, Do Not Reject H 0 . There is not enough evidence that less than 60% of the residents favor the annexation suit. Note: If α were anything bigger than 0.0749, we would Reject H 0 . In this case, “small” is relative and would depend on how much Type I Error we were willing to consider. 16

16. Suppose we perform the hypothesis test H 0 : μ = 80 versus H 1 : μ 6 = 80. We do this by constructing a 97% confidence interval. We find the confidence interval for μ to be 75 < μ < 79. What decision should we make for the hypothesis test, assuming a 3% level of significance? Solution: It is appropriate to look at a confidence interval in this scenario (instead of a 1-sided confi- dence bound) because we have a 2-sided test ( 6 =). Since α = 0 . 03, then we should look at a 97% confidence interval. So far, everything matches up. Next, look to see if 80 is inside the confidence interval or not. Since 80 is NOT inside the interval, we Reject H 0 . There is enough evidence that the mean is different from 80. 17

Stat361_FinalExam_Practice

Related Documents