Questions and Answers #4 Confidence Intervals

Questions and Answers Sheet 4 Confidence Intervals Question #1 A study a local high school tried to determine the mean height of females in the US. A study surveyed a random sample of 125 females and found a mean height of 64.5 inches with a standard deviation of 5 inches. Determine a 95% confidence interval for the mean. Confidence Interval 80% 85% 90% 95% 99% 99.5% 99.9% z 1.282 1.440 1.645 1.960 2.576 2.807 3.291 Answer: Step 1 Solution: Given information: n=125 Sample size x=64.5 inches Sample mean s=5 inches sample standard deviation 𝛼 = 0.05 Level of significance Step 2 The 95% confidence interval for the mean is 𝑠 𝑥 ± 𝑡𝛼/2,𝑛−1 × √𝑛 n is large we used 𝑧𝛼/2 instead of 𝑡𝛼/2,𝑛−1 At 𝛼 = 0.05 𝑧𝛼/2 = 𝑧0.05 = 1.96 From Z table (64.5 ± 1.96 × (64.5 ± 1.96 × 5 ) √125 5 11.180339 ) (64.5 ± 1.96 × 0.4472135) (64.5 ± 0.8765384) (64.5 − 0.8765384,64.5 + 0.8765384) (63.62346,65.376538) (63.62,65.38) The 95% confidence interval for the mean is ( 63.62, 65.38) Question #2 margin of error: $130,confidence level: 95%, 𝜎 = $591 what is the sample size? Answer: Step 1 From the provided information, Margin of error (E) = $130 Confidence level = 95% Population standard deviation (𝜎) = 591 Step 2 The z value at 95% confidence level from the standard normal table is 1.96. The required sample size can be obtained as: 𝑛=( =( 𝑧𝛼/2𝜎 2 ) 𝐸 1.96×591 2 ) 130 =79.3963 ≈ 80 Thus, the required sample size is 80. Question #3 a. Locate the critical points of ƒ. b. Use the First Derivative Test to locate the local maximum and minimum values. c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist). 𝑓(𝑥) = 𝑥2 on [-4,4] 𝑥 2 −1 Answer: Step 1 Given: 𝑓(𝑥) = 𝑥2 𝑥 2 −1 on the interval [-4,4] To find critical points, find x such that f'(x)=0 𝑓(𝑥) = 𝑥2 𝑥 2 −1 (𝑥 2 −1)2𝑥−𝑥 2 (2𝑥) 𝑓 ′ (𝑥) = (𝑥 2 −1)2 2𝑥 𝑓 ′ (𝑥) = − (𝑥 2 −1)2 Equate f'(x) equal to 0 2𝑥 − (𝑥 2 2 = 0 −1) -2x=0 x=0 Thus, the critical point is x=0. Step 2 Now the intervals therefore will be [-4, 0] and [0, 4] Assign a test value for each interval to determine if Increasing or Decreasing. Interval x [−4, 0] −2 [0, ] [0, 4] − 1 2 2 f ( x) Remarks 4 Increasing 9 16 Increasing 9 4 − Decreasing 9 The change from increasing to decreasing corresponds to local maximum and the change from decreasing to increasing corresponds to local minimum. Step 3 Therefore, the local maximum is at x=0. (c) Solve for f(x) using the end intervals and critical points x f ( x) −4 1.06 0 0 4 1.06 Remarks AbsoluteMaximum AbsoluteMinimum AbsoluteMaximum Thus, the absolute maximum is at x=-4, 4 and the absolute minimum is at x=0 . Question #4 Customers randomly selected at a grocery store included 172 women and 45 men. 74 of the women and 12 of the men used coupons. Find the 95% confidence interval. Answer: Step 1 Given: Number of women (𝑛1 ) = 172 Number of men (𝑛2 ) = 45 Number of women used coupons (𝑥1 ) = 74 Number of men used coupons (𝑥2 ) = 12 Confidence level = 0.95 So, 𝛼 = 1 − 0.95 = 0.05 Step 2 Sample proportions 𝑥 74 𝑝1 = 1 = ̂ 𝑝 ̂2 = 𝑛1 𝑥2 𝑛2 = 172 12 45 From z tables, 𝑃(𝑧 > 1.96) = 0.025 Step 3 Finding confidence interval (𝑝 ̂1 − 𝑝 ̂) 2 − 𝑧𝛼 × √ 2 𝑝1 × (1 − 𝑝 ̂ ̂) 𝑝 ̂2 × (1 − 𝑝 ̂) 𝑝1 × (1 − ̂) ̂ 𝑝1 𝑝 ̂2 × (1 − 𝑝 ̂) 1 2 2 + , (𝑝 ̂1 − 𝑝 ̂) + 2 + 𝑧𝛼 × √ 𝑛1 𝑛2 𝑛1 𝑛2 2 74 74 12 12 × (1 − ) × (1 − ) 74 12 74 12 √ 172 + 45 45 , ( ( − ) − 1.96 × 172 − ) + 1.96 172 45 172 45 172 45 74 74 12 12 √172 × (1 − 172) 45 × (1 − 45) × + 172 45 0.0147, 0.3125 The confidence interval for difference in women and men proportions of using coupons is 0.0147, 0.3125. Question #5 Consider the following data set below. Find the UPPER INTERVAL of the population mean at 80% confidence interval. (write your answer in two decimal places) Given: 12,10,9,11 Answer: It is given that, 12,10,9,11 Thus, Sample mean, 𝑥 = ∑𝑛 𝑡−1 𝑋𝑡 𝑛 = 10.5 ∑𝑛 𝑡−1(𝑋𝑡 −𝑋) Sample standard deviation, 𝑠 = √ 𝑛−1 2 = 1.29099 Since, Population standard deviation is unknown thus t distribution is used. Sample size, n=4 Degree of freedom =n-1=4-1=3 The value of 𝑡𝛼/2,𝑛−1 at 80% confidence interval is =1.638 [From t table] Now, the 80% confidence interval for the population mean can be calculated as: 𝑠 𝐶𝐼 = 𝑥 ± 𝑡𝛼/2,𝑛−1 × √𝑛 1.29099 = 10.5 ± 1.638 × √4 =(9.44, 11.56) Upper limit =11.56 Question #6 Find the critical value 𝑧𝛼/2 that corresponds to the given confidence level. 96% Answer: Step 1 From the provided information, Confidence level = 96% Level of significance (𝛼) = 1 − 0.96 = 0.04 Step 2 The critical 𝑧𝛼/2 value at 0.04 level of significance from the standard normal table (two tailed) is 2.054. Question #7 Determine the sample size taken from a normal distribution N(M, 25) to get length of interval of 4 using 90% confidence level for the mean. Answer: Step 1 Here we need to find the required sample size. Step 2 Here it is given that the sample is taken from N(M,25). Here we need to determine the sample size. We know 𝑛 = ( 𝑧𝛼/2×6 2 𝐸 ) . At 90% confidence level 𝑧𝛼/2 = 1.645 Also, 𝐸 = ∴𝑛=( 𝑙𝑒𝑛𝑔𝑡ℎ 2 1.645×5 2 2 4 = = 2,6 = 5 2 ) = 16.912 ≈ 17 Sample size n=17 Question #8 The number of rescue calls received by a rescue squad in a city follows a Poisson distribution with 2.83 per day. The squad can handle at most four calls a day. a. What is the probability that the squad will be able to handle all the calls on a particular day? b. The squad wants to have at least 95% confidence of being able to handle all the calls received in a day. At least how many calls a day should the squad be prepared for? c. Assuming that the squad can handle at most four calls a day, what is the largest value of that would yield 95% confidence that the squad can handle all calls? Answer: Step 1 Given Information Number of Rescue calls received follows a Poisson Distribution with 𝜇 = 2.83 per day The squad can handle at most four calls a day P.M.F for Poisson distribution is given by 𝑒 −𝜇 ×𝜇𝑥 𝑃(𝑋 = 𝑥) = 𝑥! Step 2 a) Probability that the squad will be able to handle all the calls on a particular day is 𝑃(𝑋 ≤ 4) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) + 𝑃(𝑋 = 3) + 𝑃(𝑋 = 4) 𝑒 −2.83 ×2.830 0! + 𝑒 −2.83 ×2.831 1! + 𝑒 −2.83 ×2.832 2! + 𝑒 −2.83 ×2.833 3! + 𝑒 −2.83 ×2.834 4! =0.059+0.167+0.2229+0.1577 =0.8430 Step 3 b) The squad wants to have at least 95% confidence of being able to handle all the calls received in a day. So, to find the number of calls to satisfy the above condition Probability that the number of calls to have at least 95% confidence is 𝑃(𝑋 ≤ 𝑥) = ∑𝑥𝑥=0 𝑒 −2.83 ×2.83𝑥 𝑥! ≥ 0.95 Taking x=1 𝑃(𝑋 ≤ 1) = 0.226 (Using Excel = POISSON.DIST(1,2.83,TRUE)) Taking x=2 𝑃(𝑋 ≤ 2) = 0.462 (Using Excel = POISSON.DIST(2,2.83,TRUE)) Taking x=3 𝑃(𝑋 ≤ 3) = 0.685 (Using Excel = POISSON.DIST(3,2.83,TRUE)) Taking x=4 𝑃(𝑋 ≤ 4) = 0.843 (Using Excel = POISSON.DIST(4,2.83,TRUE)) Taking x=5 𝑃(𝑋 ≤ 5) = 0.932 (Using Excel = POISSON.DIST(5,2.83,TRUE)) Taking x=6 𝑃(𝑋 ≤ 6) = 0.97 (Using Excel = POISSON.DIST(6,2.83,TRUE)) The number of calls to home at least 95% confidence is 6 Step 4 c) Assume that the squad can handle at most four calls a day, then finding the largest value of µ that would yield 95% confidence that the squad can handle all calls. 𝑥 𝑃(𝑋 ≤ 4) = ∑ 𝑥=0 𝑒 −2.83 × 2.83𝑥 ≥ 0.95 𝑥! By taking different values of 𝜇, the minimum value of \mu to satisfy the above condition 𝑃(𝑋 ≤ 4) = ∑𝑥𝑥=0 𝑒 −1.97 ×1.97𝑥 𝑥! = 0.95001 The minimum value of 𝜇 = 1.97 Question #9 The lengths of the products produced in a production process are controlled. The standard deviation is 0.45 mm. A quality control specialist keeps checking over 40 units randomly every morning. Average length in one day 35.62mm. when it happens; Which of the following is the confidence interval for the mean population mean at 95% confidence level? A) 𝑃(60 < 𝜇 < 70) = 0,95 B) 𝑃(30 < 𝜇 < 40) = 0,05 C) 𝑃(30 < 𝜇 < 40) = 0,95 D) 𝑃(60 < 𝜇 < 70) = 0,05 E) 𝑃(70 < 𝜇 < 80) = 0,95 Answer: Step 1 It is an important part of statistics. It is widely used. Step 2 Option C is correct Question #10 Survey A asked 1000 people how they liked the new movie Avengers: Endgame and 88% said they did enjoy it. Survey B also concluded that 82% of people liked the move but they asked 1600 total moviegoers. Which of the following is true about this comparison? The margin of error is the same for both. The confidence interval is smaller for Survey B. Increasing the number of people asked does not change the 95% confidence interval. Survey A is more accurate since the percentage is higher. Survey A has more approvals than Survey B. Survey A is better than Survey B since it has a higher percentage. Answer: Step 1 Survey A asked 1000 people how they liked the new movie Avengers: Endgame and 88% said they did enjoy it. Survey B also concluded that 82% of people liked the move but they asked 1600 total moviegoers. Which of the following is true about this comparison? Answer: The confidence interval is smaller for Survey B. Explanation: From the Standard Normal Table, the z* value for 95% level is 1.96. Step 2 The 95% confidence interval for survey A is below: 𝑝̂(1−𝑝̂) Sample statistic ±𝑀𝐸 = 𝑝̂ ± 𝑧 ∗ √ = 0.88 ± 1.96 × √ 0.88×(1−0.88) 1000 = 0.88 ± (1.96 × 0.0103) 𝑛 = 0.88 ± 0.0201 = (0.8599,0.9001) Thus, the 95% confidence interval for survey A is (0.8599, 0.9001) The 95% confidence interval for survey B is below: 𝑝̂(1−𝑝̂) Sample statistic ±𝑀𝐸 = 𝑝̂ ± 𝑧 ∗ √ = 0.82 ± 1.96 × √ 𝑛 0.82×(1−0.82) 1600 = 0.82 ± (1.96 × 0.0096) = 0.82 ± 0.0188 =(0.8012, 0.8388) Thus, the 95% confidence interval for survey B is (0.8012, 0.8388) Answer: The confidence interval is smaller for Survey B. Question #11 Survey A asked 1000 people how they liked the new movie Avengers: Endgame and 88% said they did enjoy it. Survey B also concluded that 82% of people liked the move but they asked 1600 total moviegoers. Which of the following is true about this comparison? 1. The margin of error is the same for both. 2. The confidence interval is smaller for Survey B. 3. Increasing the number of people asked does not change the 95% confidence interval. 4. Survey A is more accurate since the percentage is higher. 5. Survey A has more approvals than Survey B. 6. Survey A is better than Survey B since it has a higher percentage. Answer: Given data: Survey A asked 1000 people how they liked the new movie Avengers: Endgame and 88% said they did enjoy it. Survey B also concluded that 82% of people liked the move but they asked 1600 total moviegoers. Confidence interval for Survey A 𝑝̂(1−𝑝̂) CI(Proportion) = (𝑝̂ − 𝑧𝑐 √ = (0.88 − 1.96 × √ 𝑛 , 𝑝̂ + 𝑧𝑐 √ 𝑝̂(1−𝑝̂) 𝑛 ) 0.88(1 − 0.88) 0.88(1 − 0.88) , 0.88 + 1.96 × √ ) 1000 1000 = (0.86,0.9) Confidence interval for Survey B 𝐶𝐼(𝑃𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛) = (𝑝̂ − 𝑧𝑐 √ = (0.82 − 1.96 × √ 𝑝̂ (1 − 𝑝̂ ) 𝑝̂ (1 − 𝑝̂ ) , 𝑝̂ + 𝑧𝑐 √ ) 𝑛 𝑛 0.82(1 − 0.82) 0.82(1 − 0.82) , 0.82 + 1.96 × √ ) 1600 1600 =(0.801,0.839) Answer: 2. The confidence interval is smaller for Survey B. Question #12 25.6 23.8 26.5 23.9 24.8 24.7 20.9 26 22.2 27 What Equation to use to construct a 99% lower confidence bound for 𝜎 2 Answer: Step 1 Given observation and calculation of mean and standard deviation is shown below Data 25.6 23.8 26.5 23.9 24.8 24.7 20.9 26 22.2 27 Sum = 245.4 245.4 Mean = x = = 24.540 10 10 Variance s 2 =  (x − x ) 1 t =1 n −1 2 = 32.92 = 3.66 10 − 1 10 s=  (x − x ) i =1 i n −1 2 = 32.92 = 1.91 10 − 1 Step 2 99\% confidence interval for the population variance is given by (𝑛 − 1)𝑠 2 (𝑛 − 1)𝑠 2 𝐶𝐼 = ≤ 𝜎2 ≤ 2 2 𝜒𝛼 𝜒1− 𝛼 2 2 2 2 𝜒𝛼2 = 𝜒0.01 = 𝜒0.005 = 23.5894 (Using Excel = CHISQ.INV.RT(0.005,9)) 2 2 2 2 2 𝜒1− 𝛼 = 𝜒 0.01 = 𝜒0.995 = 1.7349 (Using Excel = CHISQ.INV.RT(0.995,9)) 1− 2 2 (10 − 1)3.66 (10 − 1)3.66 𝐶𝐼 = ≤ 𝜎2 ≤ 23.5894 1.7349 𝐶𝐼 = 1.3964 ≤ 𝜎 2 ≤ 18.9867 99% lower confidence bound for 𝜎 2 is 1.3964 ≤ 𝜎 2 Question #13 Express the confidence interval 53.5%z)=0.025 ∴ 𝑧 = 1.96(∵ 𝑧 − 𝑡𝑎𝑏𝑙𝑒) Therefore for 95% confidence interval you confind z-value=1.96 ∴ 𝑝(−𝑧 < 𝜇 < 𝑧) = 0.95 ∴ 𝑧 = 1.96 Question #15 An online used car company sells second-hand cars. For 30 randomly selected transactions, the mean price is 2800 dollars. a) Assuming a population standard deviation transaction prices of 130 dollars, obtain a 99% confidence interval for the mean price of all transactions. b) Which of the following is the correct interpretation for your answer in part (a)? A) If we repeat the study many times, 99% of the calculated confidence intervals will contain the mean price of all transactions. B) There is a 99% change that the mean price of all transactions lies in the interval C) We can be 99% confident that the mean price for this sample of 30 transactions lies in the interval D) None of the above Answer: Step 1 The provided information are: Sample size (n)=30 Sample mean (𝑥̅ ) = 2800 Population standard deviation (𝜎) = 130 Confidence level =99% Step 2 a) Using the standard normal table, the z-critical value at 99% confidence level is 2.58. The confidence interval is: 𝜎 𝐶𝐼 = 𝑥̅ ± 𝑧 × √𝑛 = 2800 ± 2.58 × 130 √30 =(2738.9, 2861.1) Thus, the 99% confidence interval is (2738.9, 2861.1) Step 3 b) If many confidence interval of same size =(30) is taken and confidence interval is calculated for each of the sample, then about 99% of them will contain the true population parameter. Thus, the correct option is (A). Question #16 In a survey of 2085 adults in a certain country conducted during a period of economic? Uncertainty, 63% thought that wages paid to workers in industry were too low. The margin of error was 8 percentage points with 90% confidence. For parts (1) through (4) below, which represent a reasonable interpretation of the survey results. For those that are not reasonable, explain the flaw. 1) We are 90% confident 63% of adults in the country during the period of economic uncertainty felt wages paid to workers in industry were too low. A) The interpretation is reasonable. B) The interpretation is flawed. The interpretation provides no interval about the population proportion. C) The interpretation is flawed. The interpretation sugguests that this interbal sets the standard for all the other intervals, which is not true. D) The interpretation is flawed. The interpretation indicates that the level of confidence is varying. 2) We are 82% to 98% confident 63% of adults in the country during the period of economic uncertainty felt wages paid to workers in industry were too low. Is the interpretation reasonable? A) The interpretation is reasonable. B) The interpretation is flawed. The interpretation provides no interval about the population proportion. C) The interpretation is flawed. The interpretation sugguests that this interbal sets the standard for all the other intervals, which is not true. D) The interpretation is flawed. The interpretation indicates that the level of confidence is varying. 3) We are 90% confident that the interval from 0.55 to 0.71 contains the true proportion of adults in the country during the period of economic uncertainty who believed wages paid to workers in industry were too low. A) The interpretation is reasonable. B) The interpretation is flawed. The interpretation provides no interval about the population proportion. C) The interpretation is flawed. The interpretation sugguests that this interbal sets the standard for all the other intervals, which is not true. D) The interpretation is flawed. The interpretation indicates that the level of confidence is varying. 4) In 90% of samples of adults in the country during the period of economic uncertainty, the proportion who believed wages paid to workers in industry were too low is between 0.55 and 0.71. A) The interpretation is reasonable. B) The interpretation is flawed. The interpretation provides no interval about the population proportion. C) The interpretation is flawed. The interpretation sugguests that this interbal sets the standard for all the other intervals, which is not true. D) The interpretation is flawed. The interpretation indicates that the level of confidence is varying. Answer: Step 1 Given that in a survey of 2085 adults in a certain country conducted during a period of economic 63% thought that wages paid to workers in industry were too low. The margin of error was 8 percentage points with 90% confidence. Given values are Margin of error (M)=8%=0.08 𝑝̂ = 63% = 0.63 We are 90% confident 63% of adults in the country during the period of economic uncertainty felt wages paid to workers in industry were too low. The confidence interval for 90% confident is 𝐶. 𝐼. = (𝑝̂ − 𝑀, 𝑝̂ + 𝑀) 𝐶. 𝐼. = (0.63 − 0.08, 0.63 + 0.08) 𝐶. 𝐼. = (0.55, 0.71) Here population proportion is not mention in this interval so The correct option is B) The interpretation is flawed. The interpretation provides no interval about the population proportion. Step 2 We are 82% to 98% confident 63% of adults in the country during the period of economic uncertainty felt wages paid to workers in industry were too low. The correct option is D) The interpretation is flawed. The interpretation indicates that the level of confidence is varying. Step 3 We are 90% confident that the interval from 0.55 to 0.71 contains the true proportion of adults in the country during the period of economic uncertainty who believed wages paid to workers in industry were too low. The correct option is A) The interpretation is reasonable. Step 4 In 90% of samples of adults in the country during the period of economic uncertainty, the proportion who believed wages paid to workers in industry were too low is between 0.55 and 0.71. C) The interpretation is flawed. The interpretation suggests that this interval sets the standard for all the other intervals, which is not true. Question #17 Rock band The Rolling Stones have played scores of concerts in the ltwenty years. For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.12 million dollars. a) Assuming a population standard deviation groos earnings of 0.51 million dollars, obtain a 99% confidence interval for the mean gross earning of all Rolling Stones concerts (in millions). b) Which of the following is the correct interpretation for your answer in part (a)? A) We can be 99% confident that the mean gross earnings for this sample of 30 Rolling Stones concert lies in the interval B) If we repeat the study many times, 99% of the calculated confidence intervals will contain the mean gross earning of all Rolling Stones concerts. C) There is a 99% chance that the mean gross earning of all Rolling Stones concert lies in the interval D) None of the above Answer: Step 1 Sample mean = 𝑋̅ = 2.12 Sample standard deviation s=0.51 Sample size n=30 Level of significance =0.01 Degree of freedom =n-1 =30-1=29 Critical value = 𝑡𝛼/2;𝑑𝑓 = 𝑡0.005, 29 = 2.76 Step 2 a) To find 99% confidence level 𝑠 𝐶. 𝐼. = 𝑋̅ ± 𝑡𝛼/2;𝑑𝑓 × √𝑛 = 2.12 ± 2.76 × 0.51 √30 = 2.12 ± 0.2570 = (1.863, 2.377) Lower limit =1.863 Upper limit =2.377 Step 3 b) Option A is the correct answer We can be 99% confident that the mean gross earnings for this sample of 30 Rolling Stones concerts lies in the interval (1.863, 2.377) Question #18 Using 64 randomly selected phone calls, the average call length was calculated to be 4.2 minutes. It is known from previous studies that the variance of the length of phone calls is 1.44𝑚𝑖𝑛2 . Assuming that the length of calls has a normal distribution a) estimate an interval estimate of the length of a telephone conversation at the 0.95 confidence level b) confidence 0.99 c) Compare the length of the two intervals and explain how the length of the interval depends on the confidence level. Answer: Step 1 As per given by the question, there are 64 randomly selected phone calls, the average call length was calculated to be 4.2 minutes. So, Random value (n) is 64, Mean 𝑋̅ is 4.42 minute, and Variance is 1.44. Then first calculate the standard deviation with the help of means and variance. 𝜎 𝑉= ̅ 𝑋 1.44 = 𝜎 4.42 𝜎 = 6.36 Step 2 Now, a) estimate an interval estimate of the length of a telephone conversation at the 0.95 confidence level. From formula of normal distribution, 𝜎 𝐶𝐼 = 𝑋̅ ± 𝑡𝑧 ⋅ √𝑛 6.36 = 4.42 ± 1.96 ⋅ √64 = 4.42 ± (1.96)(0.79) = 4.42 ± 1.44 Hence, the length of a telephone conversation at the 0.95 confidence level is 5.97, 2.87 Step 3 b). estimate an interval estimate of the length of a telephone conversation at the 0.99 confidence level. From formula of normal distribution, 𝜎 𝐶𝐼 = 𝑋̅ ± 𝑡𝑧 ⋅ √𝑛 = 4.42 ± 2.58 ⋅ 6.36 √64 = 4.42 ± (2.58)(0.79) = 4.42 ± 2.038 Hence, the length of a telephone conversation at the 0.99 confidence level is 6.458, 2.382. Step 4 c). Compare the length of the two intervals and explain how the length of the interval depends on the confidence level. The length of interval at 0.95 confidence level is 5.97, 2.87. and, The length of interval at 0.99 confidence level is 6.458, 2.382. Here, the value of length of interval at 0.99 confidence level is 6.458, 2.382, that is greater than the value of length of interval at 0.95 confidence level is 5.97, 2.87. The length of the interval depends on the confidence level, because if confidence level is increase then length of interval is increase, if confidence level is decrease then length of interval is increase. Question #19 A random sample of 820 households from a city, Stats City, were surveyed. From the statistical technology, we find a 95% confidence interval for the mean number of people in the households was 1.85 < 𝜇 < 3.25 What is the true meaning of the interval presented in the problem about Stats City? Answer: Step 1 Given, A random sample of 820 households from a city, Stats City, were surveyed. From the statistical technology, we find a 95% confidence interval for the mean number of people in the households was 1.85 < 𝜇 < 3.25 Step 2 Interpretation of 95% confidence interval: We are 95% confident that the interval (1.85, 3.25) contains the true mean number of people in the households A confidence interval tries to capture the true value of the parameter it's estimating, which in this case is the true mean number of people in the households. The confidence level tells us the long-term capture rate of these intervals over repeated samples. Question #20 A statistics practitlioner took a random sample of 54 observations form a population whose standard deviation is 29 and computed the sample mean to be 97. Note: For each confidence interval, enter your answer in the forem (LCL, UCL). You must include the parentheses and the comma between the confidence limits. A) Estimate the population mean with 95% confidence. B) Estimate the population mean with 90% confidence. C) Estimate the population mean with 99% confidence. Answer: Step 1 Given: 𝜎 = 29 n=54 𝑥̅ = 97 a) From the Standard Normal Table, the value of 𝑧 ⋅ for 95% level is 1.96 The 95\% confidence interval for the population mean is obtained as below: 𝜎 Sample statistic ±𝑧 ⋅ 𝑆𝐸 = 𝑥̅ ± 𝑧 ⋅ √𝑛 = 97 ± (1.96 × 29 √54 ) 97 ± 7.7349 = (89.2651, 104.7349) Thus, the 95% confidence interval for the population mean is (89.2651, 104.7349). Step 2 b) From the Standard Normal Table, the value of 𝑧 ⋅ for 90% level is 1.645. The 90% confidence interval for the population mean is obtained as below: 𝜎 Sample statistic ±𝑧 ⋅ 𝑆𝐸 = 𝑥̅ ± 𝑧 ⋅ √𝑛 = 97 ± (1.645 × 29 √54 ) = 97 ± 6.4918 = (90.5082, 103.4918) Thus, the 90% confidence interval for the population mean is (90.5082, 103.4918). Step 3 c) From the Standard Normal Table, the value of 𝑧 ⋅ for 99% level is 2.576 The 99% confidence interval for the population mean is obtained as below: 𝜎 Sample statistic ±𝑧 ⋅ 𝑆𝐸 = 𝑥̅ ± 𝑧 ⋅ √𝑛 = 97 ± (2.576 × 29 √54 ) = 97 ± 10.1659 = (86.8341, 107.1659) Thus, the 99% confidence interval for the population mean is (86.8341, 107.1659).