Questions and Answers #2 Confidence Intervals

Questions and Answers Sheet 2 Confidence Intervals Question #1 Of 1000 randomly selected cases of lung cancer 843 resulted in death within 10 years. Construct a 95% two-sided confidence interval on the death rate from lung cancer. a) Construct a 95% two-sides confidence interval on the death rate from lung cancer. Round your answer 3 decimal places. ? ≤ 𝑝 ≤? b) Using the point estimate of p obtained from the preliminary sample what sample size is needed to be 95% confident that the error in estimatimating the true value of p is less than 0.00? c) How large must the sample if we wish to be at least 95% confident that the error in estimating p is less than 0.03 regardless of the value of p? Answer: Part (a) Consider the n=1000 and x=843 . Then, the point estimate is, 𝑃̂ = 1 − 0.843 =0.157 Then, at 95% confidence level. 𝛼 = 1 − 0.95 𝛼 = 0.05 𝛼 = 0.025 2 𝑍𝛼 = 𝑍0.025 2 𝑍𝛼 = 1.960 2 Then, the margin of the error is, 𝐸 = 𝑍𝛼 𝑝̂(1−𝑝̂) 2 √ = 1.96√ 𝑛 0.843(0.157) 1000 =0.023 Thus, the 95\% confidence interval for the population p is, 0.843 − 0.023 ≤ 𝑝 ≤ 0.843 + 0.023 0.820 ≤ 𝑝 ≤ 0.866 Part (b) Consider the margin error is, E=0.03 Then, the sample size is, 𝑍𝛼 2 𝑛 = ( 2 ) 𝑃̂(1 − 𝑃̂ ) 𝐸 =( 1.96 2 0.03 ) 0.843 × 0.157 = 564.93 =565 Part(c) For p less than 0.03, the sample size is obtained as, 𝑍𝛼 2 𝑛 = ( 2 ) 𝑃̂ (1 − 𝑃̂ ) 𝐸 =( 1.96 2 0.03 ) 0.5 × 0.5 =1067.11 =1068 Question #2 A poll asked the question, ''What do you think is the most important problem facing this country today?'' Twenty-five percent of the respondents answered ''crime and violence.'' The margin of sampling error was plus or minus 3 percentage points. Following the convention that the margin of error is based on a 95% confidence interval, find a 95% confidence interval for the percentage of the population that would respond ''crime and violence'' to the question asked by the pollsters. Lower limit % Upper limit % Answer: Step 1 The percentage of responndents answered crime and violence = 25% 95% confidence interval margin of error =3% The formula to calculate the confidence interval is shown below Confidence Interval = Sample Statistic ± Margin of Error Step 2 We know sample statistic is given to us as 25% . So substituting the values we get the lower limit and upper limit of 95% confidence interval as shown below. Lower Limit =22% Upper Limit =28% Confidence Interval = Sample Statistic ± Margin of Error Lower Limit = Sample Statistic − Margin of Error Lower Limit = 25 − 3 = 22% Upper Limit = Sample Static + Margin of error Upper Limit = 25 + 3 = 28% Question #3 Express the confidence interval (0.412, 0.789) in the form of 𝑝̂ ± 𝐸. Answer: Step 1 From the confidence interval we get two equations: 𝑝̂ − 𝐸 = 0.412 𝑝̂ + 𝐸 = 0.789 Step 2 dding them we get: 𝑝̂ − 𝐸 + 𝑝̂ + 𝐸 = 0.412 + 0.789 2𝑝̂ = 1.201 𝑝̂ = 0.6005 Using 𝑝̂ − 𝐸 = 0.412 we get: 0.6005 − 𝐸 = 0.412 −𝐸 = 0.412 − 0.6005 −𝐸 = −0.1885 E=0.1885 Answer: 𝑝̂ + 𝐸 = 0.6005 ± 0.1885 Question #4 A group of scientists studied the adhesion of various biofilms to solid surfaces for possible use in environmental technologies. Adhesion assay is conducted by measuring absorbance at A590. If the scientists want the confidence interval to be no wider than 0.55 𝑑𝑦𝑛𝑒 − 𝑐𝑚2 at 98% confidence interval, how many observations should they take? Assume that the standard deviation is known to be 0.66 dyne𝑐𝑚2 CHOOSE THE RIGHT ANSWER: a.39 b.21 c.16 d.25 e.18 f.32 Answer: Step 1 The width is given to be 0.55. At 98% confidence level, the value of z is 2.326. It is known that 2𝑧𝜎 Width = ⇒ 0.55 = ⇒ 0.55 = ⇒ √𝑛 = ⇒𝑛=[ ⇒𝑛=[ √𝑛 2𝑧𝜎 √𝑛 2(2.326)(0.66) √𝑛 2(2.326)(0.66) 0.55 2(2.326)(0.66) 2 0.55 2 3.07032 2 0.55 ] ] ⇒ 𝑛 = (5.5824)2 ⇒ 𝑛 = 31.1632 ⇒ 𝑛 ≅ 32 Step 2 Therefore, the option F is correct. Question #5 A survey of several 10 to 11 year olds recorded the following amounts spent on a trip to the mall: $19.17,$21.18,$20.38,$25.08 Construct the 90% confidence interval for the average amount spent by 10 to 11 year olds on a trip to the mall. Assume the population is approximately normal. Copy Data Step 4 of 4 : Construct the 90% confidence interval. Round your answer to two decimal places. Answer: Step 1 The mean and standard deviation are obtained as follows: SUM x2 367.4889 448.5924 415.3444 629.0064 1860.432 x 19.17 21.18 20.38 25.08 85.81 Mean = (𝑥) = ∑𝑥 𝑛 = 85.81 4 Standard Deviation = √ = 21.45 (∑ 𝑥) 𝑛 ∑ 𝑥2− 𝑛−1 85.812 4 1860.432− Standard Deviation = √ 3 Standard Deviation =2.55 Step 2 The 90% confidence interval is obtained as follows: Confidence Interval = [𝑥 ± 𝑡(𝑛−1,𝛼) × 2 𝑠 √𝑛 ] Confidence Intarval = [21.45 ± 𝑡(3,0.10) × 2 Confidence Interval = [21.45 ± 2.353 × 2.55 √4 2.55 √4 ] ] Comfidence Interval =[18.45, 24.45] Lower Limit =18.45 Upper Limit =24.45 Question #6 How do you solve? A healthcare provider monitors the number of CAT scans performed each month in each of its clinics. The most recent year of data for a particular clinics follows (the reported variable is the number of CAT scans each month expressed as the number of CAT scans per thousand members of the health plan): 2.31,2.09,2.36,1.95,1.98,2.25,2.16,2.07,1.88,1.94,1.97,2.02. Find a two-sided 95% confidence interval for the standard deviation. What should you do to address any reservations about this confidence interval? Round your answers to two decimal places (e.g. 98.76). ___ ≤ 𝜎 ≤ ___ Answer: Step 1 Given the data, we can calculate the mean and s.d of the data. Sum of all observations 𝑥= Total number of observations 2.31+2.09+2.36+1.95+1.98+2.25+2.16+2.07+1.88+1.94+1.97+2.02 = 12 =2.0816 Mean is 2.0816. Standard deviation is given by; 𝑠=√ ∑𝑥(𝑥 − 𝑥)2 𝑛 =√ =√ (2.31−2.0816)2+(2.09−2.0816)2 …..+(2.02−2.0816)2 12 0.2689 12 = 0.1497 Step 2 With small samples, the interval of confidence interval is shown in below table; n 2 3 5 10 25 50 95%CI of SD 0.45  SD → 31.9  SD 0.52  SD → 6.29  SD 0.60  SD → 2.87  SD 0.69  SD → 1.83  SD 0.78  SD → 1.39  SD 0.84  SD → 1.25  SD Our sample size is 12, we will go for 0.69 × 𝑆𝐷 → 1.83 × 𝑆𝐷 . Range of 95% CI will given as; 0.69 × 𝑆𝐷 → 1.83 × 𝑆𝐷 0.69 × 0.1497 → 1.83 × 0.1497 0.10 → 0.27 0.10 ≤ 𝜎 ≤ 0.27 Question #7 As part of an annual review of its accounts, a discount brokerage selects a random sample of 28 customers. Their accounts are reviewed for total account valuation, which showed a mean of $35,100, with a sample standard deviation of $8,800. (Use t Distribution Table.) What is a 98% confidence interval for the mean account valuation of the population of customers? (Round your answers to the nearest dollar amount.) 98% confidence interval for the mean account valuation is between $ and $ Answer: Step 1 The provided information are: Sample size (n)=28 Sample mean (𝑥) = 35,100 Sample standard deviation (s)=$8,800 Step 2 The formula to calculate the confidence interval for population mean is: 𝑠 𝑥 ± 𝑡(𝛼,𝑑𝑓) 2 √𝑛 Step 3 Here, the population standard deviation is unknown. So, the t-critical value is used. The degrees of freedom is: df = n - 1 = 28 - 1 = 27 The critical value for 98% confidence level at df =27 can be found using the t-table. That is, 𝑡(𝛼,𝑑𝑓) = 𝑡(0.02,27) 2 =2.473 Step 4 2 Substitute the values in the confidence interval formula. 𝑠 𝐶𝐼 = 𝑥 ± 𝑡(𝛼,𝑑𝑓) √𝑛 2 = 35100 ± 2.473 × 8800 √28 = 35100 ± 4112.141 = (30987.859,39212.141) ≈ (30988,39212) Step 5 Thus, the 98% confidence interval for the mean account valuation is between $30988 and $39212. Question #8 A simple random sample is drawn from a population that is known to be normally distributed. The sample variance, s2, is determined to be 12.6 A. Construct a 90% confidence interval for 𝜎 if the sample size, n, is 20. (Hint: use the result obtained from part (a)) 7.94 (LB) & 23.66 (UB)) B. Construct a 90% confidence interval for 𝜎 if the sample size, n, is 30. (Hint: use the result obtained from part (b)) 8.59 (LB) & 20.63 (UB) Answer: Here 𝜎2 = 12.6 and c=90% which is 0.9. Therefore 1−𝑐 2 = 1−0.9 2 = 0.05 and 1 − ( 1−𝑐 2 ) = 0.95 Part A For n=20 the degrees of freedom are: df=20-1=19 So the 𝜒 squared table values are: 𝑋0.95 = 10.117 and 𝑋0.05 = 30.144 So the upper and lower bounds are given by: 𝑈𝐵 = ( 19 ) (12.6) 10.117 UB=23.66 𝐿𝐵 = ( 19 30.144 ) (12.6) LB=7.94 Part B For n=30 the degrees of freedom are: df=30-1=29 So the 𝜒 squared table values are: 𝑋0.95 = 17.708 and 𝑋0.05 = 42.557 So the upper and lower bounds are given by: 𝑈𝐵 = ( 29 ) (12.6) 17.708 UB=20.63 𝐿𝐵 = ( 29 ) (12.6) 42.557 LB=8.59 Question #9 To monitor complex chemical processes, chemical engineers will consider key process indicators, which may be just yield but most often depend on several quantities. Before trying to improve a process, 9 measurements were made on a key performance indicator. 123, 106, 114, 128, 113, 109, 120, 102, 111. Assume that the key performance indicator has a normal distribution. (a) Find the sample variance s 2 of the sample. (b) Find a 95% confidence interval for 𝜎 2. (c) Find a 98% confidence interval for 𝜎. Answer: Step 1 (a) to find variance value 1) add them up 123 + 106 + 114 + 128 + 113 + 109 + 120 + 102 + 111 = 1026 2) square your answer 1026 ⋅ 1026 = 1052676 ...and divide by the number of items. we have 9 items , 1052676/9 = 116964 set this number aside for a moment. 3) take your set of original numbers from step 1, and square them individually this time 1232 + 1062 + 1142 + 1282 + 1132 + 1092 + 1202 + 1022 + 1112 = 117520 4) subtract the amount in step 2 from the amount in step 3 117520 - 116964 = 556 5) subtract 1 from the number of items in your data set, 9 - 1 = 8 6) divide the number in step 4 by the number in step 5. this gives you the variance 556/8 = 69.5 ---------------------------------------------------------------------------------------to find standard deviation value take the square root of your answer from step 6. this gives you the standard deviation 8.3367 Step 2 (b) CONFIDENCE INTERVAL FOR VARIANCE 𝑐𝑖 = (𝑛 − 1)𝑠 2 𝜓 2 𝑟𝑖𝑔ℎ𝑡 < 𝜎 2 < (𝑛 − 1)𝑠 2 /𝜓 2 left where, 𝑠 2 = variance 𝜓 2 = (1 − confidence level)/2 𝜓2 = 1 − 𝜓2 n= sample size since 𝛼 = 0.05 𝜓 2 𝑟𝑖𝑔ℎ𝑡 = (1 − confidence level)/2 = (1 − 0.95)/2 = 0.05/2 = 0.025 𝜓 2 = 1 − 𝜓 2 = 1 − 0.025 = 0.975 the two critical values 𝜓 2 left, 𝜓 2 right at 8 df are 17.5345 , 2.18 variance (𝑠 2 ) = 69.5 sample size (n)=9 confidence interval = [8 ⋅ 69.5/17.5345 < 𝜎 2 < 8 ⋅ 69.5/2.18] = [556/17.5345 < 𝜎 2 < 556/2.1797] [31.7089 , 255.081] Step 3 (c) to calculate confidence interval for standard deviation 𝑐𝑖 = (𝑛 − 1)𝑠 2 /𝜓 2 right < 𝜎 2 < (𝑛 − 1)𝑠 2 /𝜓 2 left where, s= standard deviation 𝜓 2 𝑟𝑖𝑔ℎ𝑡 = (1 − 𝑐𝑜𝑛𝑓𝑖𝑑𝑒𝑛𝑐𝑒 𝑙𝑒𝑣𝑒𝑙)/2 𝜓 2 left = 1 − 𝜓 2 right n= sample size since 𝛼 = 0.02 𝜓 2 right =(1 - confidence level)/2 = (1 − 0.98)/2 = 0.02/2 = 0.01 𝜓 2 left = 1 − 𝜓 2 right =1 - 0.01 = 0.99 the two critical values 𝜓 2 left, 𝜓 2 right at 8 df are 20.0902 , 1.646 s.d(s)=8.3367 sample size (n)=9 confidence interval for 𝜎 2 = [8 ⋅ 69.5006/20.0902 < 𝜎 2 < 8 ⋅ 69.5006/1.646] = [556.0045/20.0902 < 𝜎 2 < 556.0045/1.6465] [27.6754 < 𝜎 2 < 337.6888] and confidence interval for 𝜎 = √𝑙𝑜𝑤𝑒𝑟 < 𝜎 < √𝑢𝑝𝑝𝑒𝑟 = [√27.6754 < 𝜎 < √337.6888, ] = [5.2607 < 𝜎 < 18.3763] Question #10 A firm employs 189 junior accountants. In a random sample of 50 of these, the mean number of hours overtime billed in a particular week was 9.7, and the sample standard deviation was 6.2 hours. a. Find a 95% confidence interval for the mean number of hours overtime billed per junior accountant in this firm that week. b. Find a 99% confidence interval for the total number of hours overtime billed by junior accountants in the firm during the week of interest. Answer: Given data, The number of employee is 189. The mean of the overtime is 9.7. The standard deviation of overtime is 6.2 The sample size is 50. Step 1 a) A 95% confidence interval for the mean number of hours overtime, 𝜎 Error margin = 𝑧0.05/2 √𝑛 = 𝑧0.0025 6.2 √50 = 1.96 × 0.8768 =1.72 The lower bound of the interval is, 𝐿 = 𝜇 t Error =9.7-1.72 =7.98 The upper bound of the interval is, 𝑈 = 𝜇 + Error =9.7+1.72 =11.42 Step 2 b) A 99% confidence interval for the mean number of hours overtime, 𝜎 Error margin = 𝑧0.01/2 √𝑛 = 𝑧0.005 6.2 √50 = 2.576 × 0.8768 =2.2586 The lower bound of the interval is, 𝐿 = 𝜇 t Error =9.7 - 2.2586 =7.4414 The upper bound of the interval is, 𝑈 = 𝜇 + Error =9.7+2.2586 =11.9586 Question #11 find the value: 𝑡𝑎 2 n=10 for the 90% confidence interval for the mean. Answer: Step 1 d. Given: n=10 𝛼 = 0.1 The degrees of freedom is obtained as below: df=n-1 =10-1 =9 Critical value: The critical value 𝑡𝛼 is obtained from Excel using ''=TINV(probability, deg_freedom)''. 2 Output obtained using Excel is given below: E=1.833113 From the output, the value 𝑡𝛼 is 1.8331 2 Step 2 Answer: 𝑡𝛼 = 1.8331. 2 Question #12 A salesperson goes door-to-door in a residential area to demonstrate the use of a new household appliance to potential customers. At the end of a demonstration,the probability that the potential customer would place an order for the product is aconstant 0.2107. To perform satisfactorily on the job, the salesperson needs at leastfour orders. Assume that each demonstration is a Bernoulli trial. a. If the salesperson wants to be at least 90% confident of getting at least 4 orders, at least how many demonstrations should she make? b. The salesperson has time to make only 22 demonstrations, and she still wants to be at least 90% confident of getting at least 4 orders. She intends to gain this confidence by improving the quality of her demonstration and thereby improving the chances of getting an order at the end of a demonstration. At least to what value should this probability be increased in order to gain the desired confidence? Your answer should be accurate to four decimal places. Answer: Step 1 Independent repeated trials of an experiment with two outcomes only are called Bernoulli trials. Call one of the outcomes ''success'' and the other outcome ''failure''. Let p be the probability of success in a Bernoulli trial. Then the probability of failure q is given by q=1-p n P( k ) = ( ) p k q n − k k Step 2 by calculating the probability of giving 𝑝(0) + 𝑝(1) + 𝑃(2) + 𝑝(3) + 𝑝(4) = .7651 Question #13 Find the value: 𝑡𝑎 if n=18 for the 99% confidence interval for the mean. 2 Answer: Step 1 Given: n=18 and 1 − 𝛼 = 99% Then, 1 − 𝛼 = 0.99 ⇒ 𝛼 = 1 − 0.99 ⇒ 𝛼 = 0.01 Step 2 Now, The degree of freedom =n-1 =18-1 =17 Then, at 𝛼 = 0.01 , from the one tailed t-table𝑡𝑎 = 𝑡0.01 2 2 = 𝑡0.005 = 2.898 Question #14 Assume that 𝜎 is unknown, the lower 100(1 − 𝛼)% confidence bound on 𝜇 is: 𝑠 a) 𝜇 ≤ 𝑥 + 𝑡𝛼,𝑛−1 √𝑛 b) s  n x − t ,n −1 c) \𝑚𝑢 ≤ 𝑥 + 𝑡𝛼/2,𝑛−1 𝑠 √𝑛 s  n d) x − t /2, n −1 Answer: Step 1 To estimate 𝜇, we draw a random sample of 𝑋1 … . . , 𝑋𝑛 of size n from a normal population population 𝑁(𝜇, 𝜎). Let 𝑋 be the mean of the random sample, then 𝑥 ∼ 𝑁 (𝜇, ⇒ 𝑥−𝜇 𝜎 √𝑛 𝜎2 𝑛 ) ∼ 𝑁(0,1) = 𝑍 Since 𝜎 is unknown, the standard deviation σ is replaced by the estimated standard deviation i.e. the standard error S because 𝑆 2 is the unbiased estimator of 𝜎 2 . Since the standard error is an estimate for the true value of the standard deviation, the distribution of the sample mean 𝑋 is no longer normal with mean 𝜇 and standard deviation 𝜎/√𝑛. Instead, the sample mean follows the Student's t distribution with n-1 degrees of freedom and with mean 𝜇 and standard deviation 𝑆/√𝑛. Step 2 From the table of Student’s t distribution with n-1 degrees of freedom where 𝛼 is the confidence coefficient, we can find T such that 𝑃 (−𝑡𝑛−1,𝛼 ≤ 𝑇 ≤ 𝑡𝑛−1,𝛼 ) = 1- 𝛼 2 2 𝑋−𝑢 ⇒ 𝑃 (−𝑡𝑛−1,𝛼 ≤ 𝑠 √𝑛 2 ⇒ 𝑃 (−𝑡𝑛−1,𝛼 𝑆 2 √𝑛 ≤ 𝑡𝑛−1,𝛼 ) = 1 − 𝛼 2 ≤ 𝑋 − 𝜇 ≤ 𝑡𝑛−1,𝛼 ⇒ 𝑃 (−𝑋 − 𝑡𝑛−1,𝛼 𝑆 2 √𝑛 ⇒ 𝑃 (𝑋 + 𝑡𝑛−1,𝛼 𝑆 2 √𝑛 ⇒ 𝑃 (𝑋 − 𝑡𝑛−1,𝛼 𝑆 2 √𝑛 𝑆 )=1−𝛼 ≤ −𝜇 ≤ −𝑋 + 𝑡𝑛−1,𝛼 𝑆 2 √𝑛 ≥ 𝜇 ≥ 𝑋 − 𝑡𝑛−1,𝛼 𝑆 2 √𝑛 )=1−𝛼 𝑆 ) =1−𝛼 2 √𝑛 √𝑛 Thus the 100(1 − 𝛼)% confidence interval for 𝜇 when 𝜎 is unknown is 2  P( X − t n −1,  2 ≤ 𝜇 ≤ 𝑋 + 𝑡𝑛−1,𝛼 )= 1− 𝛼 S S , X +t  n − 1, n n 2 Hence the correct answer for the lower 100(1 − 𝛼)% confidence bound on 𝜇 when 𝜎 is unknown, is option (d). Question #15 Based on a simple random sample of 1300 college students, it is found that 299 students own a car. We wish to construct a 90% confidence interval to estimate the proportions ? of all college students who own a car. A) Read carefully the text and provide each of the following: The sample size ?= from the sample, the number of college students who own a car is ?= the confidence level is CL = %. B) Find the sample proportion ̂? = and ?̂ = 1 −?̂ = Answer: Step 1 (A) The sample size is, n = 1300 The number of college students who own a car is, x = 299 The confidence level is, CL = 90% Step 2 (B) The value of sample proportion is, 𝑥 𝑝̂ = = 𝑛 299 1300 =0.23 The value of 1̂ is, 𝑞̂ = 1 − 𝑝̂ =1-0.23 =0.77 Question #16 Your client from question 4 decides they want a 90% confidence interval instead of a 95% one. To remind you - you sampled 36 students to find out how many hours they spent online last week. Your sample mean is 46. The population standard deviation, 𝜎, is 7.4. What is the 90% confidence interval for this sample mean? Keep at least three decimal places on any intermediate calculations, then give your answer rounded to 1 decimal place, with the lower end first then upper. (_______ , ___________ ) Answer: Step 1 Given, sample size (n)=36 sample mean (𝑥) = 46 standard deviation (𝜎) = 7.4 𝛼 = 1 − 0.90 = 0.1 𝛼 = 0.05 2 𝑍0.05 = 1.645 Step 2 90% confidence interval (𝐶. 𝐼. ): 𝑋 ± 𝑍𝛼 × 𝐶. 𝐼 = 46 ± 1.645 × 7.4 2 𝜎 √𝑛 √36 𝐶. 𝐼 = 46 ± 2.029 𝐶. 𝐼 = (46 − 2.029,46 + 2.029) 𝐶. 𝐼 = (43.971,48.029) 𝐶. 𝐼 ≅ (44.0,48.0) Question #17 The city wishes to estimate the average commute distance for all city employees. They collect a random sample of 97 employees and find a sample mean of 𝑥 = 7.1 𝑥 = 7.1 miles. They also assume a population standard deviation of 𝜎 = 1.5 𝜎 = 1.5 miles. Calculate the z interval to estimate, 𝜇𝜇, the average commute distance for all city employees with 95% confidence. Round your answers to 2 decimal places. Find the Critical Value z*: Point Estimate: Margin of Error: Write the interval as (lower bound, upper bound) Answer: Step 1 The random variable commute distance for all city employees follow normal distribution. The sample size is 97. The sample mean is 7.1. The population standard deviation is 1.5. We have to find the 95% confidence interval for the population mean. We have to use Z-table. Step 2 From Z-table, the 95% critical value is 𝑧 ∗ = 1.96 . Point estimate: 7.1. 𝜎 𝑀𝐸 = 𝑧 ∗× = 1.96 × √𝑛 1.5 √97 =0.30 Margin of error: 0.30. Step 3 𝐶𝐼 = 𝑥 ± 𝑀𝐸 = 7.1 ± 0.30 = (7.1 − 0.30,7.1 + 0.30) +(6.80,7.40) The 95% confidence interval is (6.80, 7.40). Question #18 In a sample of 7 cars, each car was tested for nitrogen-oxide emissions and the following results were obtained: 0.06, 0.11, 0.16, 0.15, 0.14, 0.08, 0.15. Assuming that the sample is representative of all cars in use and that they are normally distributed, construct a 98% confidence interval estimate for the mean amount of nitrogen-oxide emission for all cars. Use the sample data from #4 to construct a 98% confidence interval estimate for the standard deviation of the amount nitrogen-oxide emission for all cars. Answer: Step 1 Determine the t-value by looking in the row starting with degrees of freedom n-1=7-1=6 in table a-3 : 𝑡𝛼/2 = 𝑡0.01 = 3.143 The margin of error is then: 𝑠 0.0389 𝐸 = 𝑡𝛼/2 ⋅ = 3.143 ⋅ ≈ 0.0462 √7 √𝑛 The confidence interval is: 0.0752 = 0.1214 − 0.0462 = 𝑥 − 𝐸 < 𝜇 < 𝑥 + 𝐸 = 0.1214 + 0.0462 = 0.1676 Step 2 Since you have posted a question with multiple sub-parts, we will solve first sub-parts for you. To get remaining sub-part solved please repost the complete question and mention the sub-parts to be solved. Question #19 Find the value: 𝑡𝑎 2 n=23 for the 95% confidence interval for the mean. Answer: Step 1 Given: n=23 95% confidence interval for the mean Step 2 let =23-1 =22 𝛼 = 0.05 𝑡𝛼/2 = ±2 ⋅ 074 Question #20 The probability density function of X is 𝑓(𝑥) = 1 10 for 0 ≤ 𝑥 ≤ 10 . Then 𝑃 (𝑋 + Answer: Step 1 The probability density function is given as, 𝑓(𝑥) = To determine the value of the probability, 𝑃 (𝑋 + 10 𝑃 (𝑋 + ≥ 7) = 𝑃(𝑋 2 + 10 ≥ 7𝑋) 𝑋 𝑃(𝑋 2 − 7𝑋 + 10 ≥ 0) 𝑃(0 ≤ 𝑋 ≤ 2 𝑜𝑟 𝑋 ≥ 5). 10 𝑋 1 10 ≥ 7). Step 2 Solve the probability function for the above intervals, At 0 ≤ 𝑋 ≤ 2 , 2 𝑃(0 ≤ 𝑋 ≤ 2) = ∫0 𝑓(𝑥)𝑑𝑥 1 2 = ∫ 𝑑𝑥 . 10 0 = 1 10 [𝑥]20 for 0 ≤ 𝑥 ≤ 10 . 10 𝑋 ≥ 7) is: = = 1 10 1 ×2 5 Step 3 At 𝑋 ≥ 5 , 10 𝑃(𝑥 ≥ 5) = ∫ 𝑓(𝑥)𝑑𝑥 1 10 = ∫ 𝑑𝑥 . 10 5 = = = = 1 10 1 10 1 10 1 5 [𝑥]10 5 × (10 − 5) ×5 2 Step 4 Thus , the value of the 𝑃 (𝑋 + 𝑃 (𝑋 + 1 1 5 7 2 = + = 10 = 0.7 10 𝑋 10 𝑋 ≥ 7) is, ≥ 7) = 𝑃(0 ≤ 𝑋 ≤ 2 𝑜𝑟 𝑋 ≥ 5)