Questions and Answers #1 Confidence Intervals

Questions and Answers Sheet 1 Confidence Intervals Question #1 Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist. (𝑡 − 3)𝑦 ′ + (ln 𝑡)𝑦 = 2𝑡, 𝑦(1) = 2 Answer: Divide equation with (t-3) , 𝑦′ + ln 𝑡 2𝑡 𝑦= 𝑡−3 𝑡−3 According to Theorem 2.4.1, 𝑝(𝑡) = ln 𝑡 𝑡−3 and 𝑔(𝑡) = 2𝑡 𝑡−3 Functions p(t), g(t) are continuos on intervals (0, 3) and (3, ∞) because domains are (0, ∞) {3} for p and 𝑅 3 for g. Since 𝑡0 = 1 , solution exists on (0, 3) Question #2 The mean blood cholesterol level for all men aged 20 to 34 years is 𝜇 = 188𝑚𝑔/𝑑𝑙 with a standard deviation of 𝜎 = 41 𝑚𝑔/𝑑𝑙 We suspect that the mean for cross-country runners is lower. A random sample of 20 cross-country runners yielded a mean level of 172 mg/dl . a) Give a 90% confidence interval for the mean blood cholesteral level for cross-country runners. Be sure to change your Critical Value for Z-Score! Lower Bound of Confidence Interval Upper Bound of Confidence Interval b) This confidence interval is shorter than your interval in the previous exercise, even though the intervals come from the same sample. Why does the second interval have a smaller margin of error? Explain c) Does the Population Mean fall inside or outside of the 95% Confidence Interval? The 90% Confidence Interval? What does this tell you about statistical significance? Explain Answer: Step 1 The mean blood cholesterol level for all men aged 20 to 34 years is 𝜇 = 188𝑚𝑔/𝑑𝑙 The standard deviation blood cholesterol level for all men aged 20 to 34 years is 𝜎 = 41𝑚𝑔/𝑑𝑙 Sample of 20 The mean blood cholesterol level for cross country runners aged 20 to 34 years is 𝜇 = 172𝑚𝑔/𝑑𝑙 The standard deviation blood cholesterol level for cross country runners aged 20 to 34 years is 𝜎 = 41𝑚𝑔/𝑑𝑙 Step 2 a) Here: 𝑥̅ = 172 𝜎 = 41 n=20 The 90% Confidence interval for the mean blood cholesterol level for cross-country runners: 𝑥̅ ± 𝑧𝛼 × 2 𝜎 √𝑛 = 172 ± 𝑧5% × 41 √20 = 172 ± 1.6449 × 41 √20 = (156,9198, 187.0802) Question #3 Find the average value fave of the function ff on the given interval. 𝑓(𝑥) = (𝑥 3 𝑥2 +30)2 , 𝑥 ∈ [−3,3] Answer: The average value of the function f on the interval [a,b] is equal to: 𝑏 1 𝑓𝑎𝑣𝑔 = ∫ 𝑓(𝑥)𝑑𝑥 𝑏−𝑎 𝑎 The given function is 𝑓(𝑥) = (𝑥 3 𝑥2 +30)2 , 𝑥 ∈ [−3,3] Thus, the average value of f is given by 𝑓𝑎𝑣𝑔 = 𝑥2 3 1 1 3 𝑥2 𝑑𝑥 = ∫−3 (𝑥 3 𝑑𝑥 ∫ 3−(−3) −3 (𝑥 3 +30)2 6 +30)2 𝑥2 Now first calculate the indefinite integral ∫ (𝑥 3 𝑑𝑥 +30)2 𝑥2 3 1 ∫−3 (𝑥 3 +30)2 𝑑𝑥 = ∫ 3𝑢2 𝑑𝑢 [Apply u - substitution: 𝑢 = 𝑥 3 + 30 ] 1 𝑢−2+1 3 −2+1 1 = ∗ =− =− 3𝑢 1 3(𝑥 3 +30) [Substitute back: 𝑢 = 𝑥 3 + 30 ] Therefore we get 1 𝑥2 3 𝑓𝑎𝑣𝑔 = ∫−3 (𝑥 3 𝑑𝑥 6 +30)2 1 = ([− 6 1 = [− 6 1 171 2 6 1 19 = ∗ = 1 ] 3(𝑥 3 +30) 𝑥=3 1 1 − [− 1 ] 3(𝑥 3 +30) 𝑥=3 ) − (− )] 9 57 Result: 1 3 𝑥2 1 𝑓𝑎𝑣𝑔 = ∫−3 (𝑥 3 𝑑𝑥 = 6 +30)2 57 Question #4 You are interested in finding a 95% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 14 randomly selected physical therapy patients. Round answers to 3 decimal places where possible. 9 6 10 15 19 6 23 26 19 16 11 25 16 11 a) To compute the confidence interval use a t or z distribution. b) With 95% confidence the population mean number of visits per physical therapy patient is between ''?'' and ''?'' visits. c) If many groups of 14 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About ''?'' percent of these confidence intervals will contain the true population mean number of visits per patient and about ''?'' percent will not contain the true population mean number of visits per patient. Answer: Step 1 a) Here, the population standard deviation is unknown, and the sample standard deviation must be used for the calculation of the confidence interval. Thus, to compute the confidence interval, the t-distribution must be used. Step 2 b) The 100(1 − 𝛼)% confidence interval for the population mean, 𝜇 when the population standard deviation, 𝜎 is unknown and the sample standard deviation, s is used, is the t-confidence interval as follows: 𝑠 𝑠 (𝑥̅ − 𝑡{𝛼 /2;𝑛−1 } , 𝑥̅ + 𝑡{𝛼 /2;𝑛−1 √𝑛 such that: 𝑃(𝑥̅ − 𝑡{𝛼 /2;𝑛−1 } Here, 𝐸 = 𝑡{𝛼 /2;𝑛−1 𝑠 √𝑛 √𝑛 𝑠 √𝑛 ≤ 𝜇 ≤ 𝑥̅ + 𝑡{𝛼 /2;𝑛−1 𝑠 √𝑛 = 1− 𝛼 is the margin of error. Here, n is the sample size, 𝑥̅ is the sample mean, and 𝑡{𝛼 /2;𝑛−1 is the critical value of the t-distribution with (n-1) , above which, 100(𝛼/2)% or 𝛼/2 proportion of the observations lie, and below which, 100(1 − 𝛼 + 𝛼/2)% = 100(1 − 𝛼/2)% or (1 − 𝛼/2) proportion of the observations lie. The sample size is, n=14, so that the degrees of freedom is, n-1=13. The means and standard deviations are obtained as follows:  x 9 6 10 15 19 6 23 26 19 16 11 25 16 11 x = 212  x2 81 36 100 225 361 36 529 676 361 256 121 625 256 121 2 x = 3.784 1 𝑥̅ = ∑𝑛𝑖=1 𝑥𝑖 = 𝑛 212 14 ≈ 15.143 𝑠=√ 1 𝑛−1 =√ 1 14−1 ̅̅̅2 ) (∑𝑛𝑖=1 𝑥𝑖2 − 𝑛𝑥 [3,784 − (14) ⋅ (15.143)2 ] ≈ 6.643. The desired confidence level is 95%. Thus, 100(1 − 𝛼) = 95 1 − 𝛼 = 0.95 𝛼 = 0.05 𝛼/2 = 0.025 1 − 𝛼/2 = 0.975. This implies that, 𝑡{𝛼 /2;𝑛−1 = 𝑡0.025;13, such that: 𝑃(𝑡𝑛−1 ≤ t {𝛼 /2;𝑛−1 } = 𝑃(𝑡13 ≤ t 0.05;11 ) = 0.975 = 𝑃(𝑡13 ≤ 2.16) [Using the t-distribution table, or the Excel formula: = 𝑇. 𝐼𝑁𝑉(0.975,13), 𝑃(𝑡13 ≤ 2.16) = 0.975] ⇒ 𝑡0.025;13 = 2.16 First, the margin of error is calculated below: 𝑠 𝐸 = 𝑡{𝛼 /2;𝑛−1 = 𝑡0.025;13 = (2.16) ⋅ 𝑠 √𝑛 √𝑛 6.643 √14 ≈ 3.835 The confidence interval can be obtained as follows: 𝑠 𝑠 (𝑥̅ − 𝑡{𝛼 /2;𝑛−1 } , 𝑥̅ + 𝑡{𝛼 /2;𝑛−1 = (𝑥̅ − 𝐸, 𝑥̅ + 𝐸) √𝑛 √𝑛 ≈ (15.143 − 3.835,15.143 + 3.835) = (11.308,18.978). Hence, with “ 95% confidence, the population mean number of visits per physical therapy patient is between 11.308 (or approximately 11) and 18.978 (or approximately 19) visits.” Step 3 c) As it can be said that 95% of the confidence intervals calculated on from several samples of the same size will contain the true population mean, therefore, the remaining 5%(=100%-95%) of the confidence intervals will not contain the true population mean. Hence, if “many groups of 14 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About 95 percent of these confidence intervals will contain the true population mean number of visits per patient and about 5 percent will not contain the true population mean number of visits per patient.” Question #5 Suppose that the standard deviation of the tube life for a particular brand of TV picture tube is known to be 𝜎 = 400 , but that the mean operating life is not known. Overall, the operating life of the tubes is assumed to be approximately normally distributed. For a sample of n=15 , the mean operating life is 𝑋̅ = 8900ℎ𝑟. Determine (i) the 95 percent and (ii) the 90 percent confidence intervals for estimating the population mean. Answer: Step 1 Confidence interval is obtained as 𝑠 𝑥̅ ± 𝑡𝑛−1;𝛼 /2 × √𝑛 If population standard deviation is known. Confidence interval is obtained as 𝜎 𝑥̅ ± 𝑧{𝛼 /2 × √𝑛 if the population standard deviation is known. i) Here 𝑥̅ = 8900 𝜎 = 400 x=0.05(5%) n=15 So, 𝐶𝐼 = (𝑥̅ ± 𝑧{𝛼 /2 } × = 8900 ± 𝑧0.005 × 2 = 8900 ± 1.96 × = 8900 ± 𝜎 √𝑛 400 √15 400 √15 784 √15 784 3.87 = 8900 ± 202.60 𝐶𝐼 = (8900 − 202.60,8900 + 202.60) = (8697.4,9102.6) i.c. Confidence interval for population mean is (8697.4, 9102.6) = 8900 ± Step 2 ii) n=15 𝑥̅ = 8900 𝜎 = 400 𝛼 = 0.10 𝐶𝐼 = 𝑥̅ ± 𝑧{𝛼 /2 × = 8900 ± 𝑧 0.10 2 × 𝜎 √𝑛 400 √15 400 = 8900 ± 1.645 × = 8900 ± = 8900 ± = 8900 ± √15 658 √15 658 3.87 658 170.02 Step 3 The confidence interval is obtained as 8900 ± 170.02 I.e. 𝐶𝐼 = (8900 − 170.02, 8900 + 170.02) 𝐶𝐼 = (8729.98, 9070.02) Hence the confidence interval is (8729.98, 9070.02) Here the value of 𝑧{0.05 /2 = 1.96 is obtained from z table Also The value of 𝑧{0.10 /2 = 1.645 is obtained from z table. Question #6 On a certain day, a large number of fuses were manufactured, each rated at 15 A. A sample of 75 fuses is drawn from the day’s production, and 17 of them were found to have burnout amperages greater than 15 A. a) Find a 95% confidence interval for the proportion of fuses manufactured that day whose burnout amperage is greater than 15 A. b) Find a 98% confidence interval for the proportion of fuses manufactured that day whose burnout amperage is greater than 15 A. c) Find the sample size needed for a 95% confidence interval to specify the proportion to within ±0.05. Answer: a) Given x=17 and n=75 𝑥 17 𝑝̂ = = = 0.2267 𝑛 75 The critical value corresponding with 95% confidence level is 1.96 95%𝐶𝐼 = 𝑝̂ ± 𝑧 𝑝̂(1−𝑝̂) 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 √ = 0.2267 ± 1.96√ 𝑛 0.2267(1 − 0.2267) 75 =(0.1319, 0.3215) b) The critical value corresponding with 98% confidence level is 2.33 95%𝐶𝐼 = 𝑝̂ ± 𝑧 ̂ (1−𝑝 ̂) 𝑝 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 √ = 0.2267 ± 2.33√ 𝑛 0.2267(1 − 0.2267) 75 =(0.1141, 0.3393) c) The critical value corresponding with 95% confidence level is 1.96 ME=0.05 sample size = ( =( 𝑧𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 2 𝑀𝐸 ) 𝑝(1 − 𝑝) 1.96 2 0.05 ) 0.5(1 − 0.5) Let p=0.5 = 348.16 ≈ 384 Question #7 Let x be a binomial random variable with n=20 and p=42/100. a) Calculate the mean, variance and standard deviation of the random variable x. b) Use the results of part a to calculate the intervals 𝜇 ± 𝜎, 𝜇 ± 2𝜎, and 𝜇 ± 3𝜎. Find the probability that an observation will fall into each of these intervals. c) Are the results of part b consistent with Tchebysheff’s Theorem? With the Empirical Rule? Why or why not? Answer: Step 1 Given that Let x be a binomial random variable with n=20 and p=42/100 The binomial probability distribution function is P[X=x]=P(x) a P[ X = x] = ( ) p x q n− x ; x = 0, 1, b , n, 0  p  1, q = 1 − p = 0; otherwise Mean of binomial distribution is np 𝐸(𝑋) = 𝑛 × 𝑝 42 𝐸(𝑋) = 20 × 100 E(X)=8.4 Variance of binomial distribution function is npq 𝑉𝑎𝑟(𝑋) = 𝛿 2 = 𝑛𝑝𝑞 𝑉𝑎𝑟(𝑋) = 20 × 42 100 × (1 − 42 100 ) 𝑉𝑎𝑟(𝑋) = 20 × 0.42 × 0.58 𝑉𝑎𝑟(𝑋) = 4.872 Standard deviation is (𝑠𝑑) = √𝑉𝑎𝑟(𝑋) 𝑠𝑑 = 𝛿 = √𝑉𝑎𝑟(𝑋) 𝛿 = √4.872 𝛿 = 2.2073 Step 2 Use the results of part a to calculate the intervals 𝜇 ± 𝜎, 𝜇 ± 2𝜎, and 𝜇 ± 3𝜎. The confidence interval for 𝜇 ± 𝜎 𝐶. 𝐼. = (𝜇 = 𝛿, 𝜇 + 𝛿) 𝐶. 𝐼. = (8.4 − 2.2073,8.4 + 2.2073) 𝐶. 𝐼. = (6.1927,10.6073) The Confidence interval for 𝜇 ± 2𝜎 𝐶. 𝐼. = (𝜇 − 2𝛿, 𝜇 + 2𝛿) 𝐶. 𝐼. = (8.4 − 2 × 2.2073, 8.4 + 2 × 2.2073) 𝐶. 𝐼. = (3.9854, 12.8146) The confidence interval for 𝜇 ± 3𝜎 is 𝐶. 𝐼. = (𝜇 − 3𝛿, 𝜇 + 3𝛿) 𝐶. 𝐼. = (8.4 − 3 × 2.2073, 8.4 + 3 × 2.2073) 𝐶. 𝐼. = (1.7781, 15.0219) Find the probability that an observation will fall into each of these intervals. X 0 1 2 3 4 P( X  x) 0.000018559 0.0002687 0.001849 0.004 0.014 5 6 7 8 9 10 11 12 13 14 15 0.036 0.075 0.122 0.162 0.177 0.159 0.119 0.073 0.037 0.015 0.005 16 17 18 19 20 0.001 0.000 0.000 0.000 0.000 Sum 2.1261 𝑃(𝑋 ≤ 𝑥) = ∑20 𝑖=1 𝑃[𝑋 = 𝑥] 𝑃(𝑋 ≤ 𝑥) = 2.1261 This is belong to the all the confidence interval that have written in above step. From the chebysheff's theorem 79.48% of the data value will lie within 2.2073 standard deviation from mean. Question #8 Out of 200 people sampled, 116 preferred Candidate A. Based on this, estimate what proportion of the voting population prefers Candidate A with 99% confidence. Out of 500 people sampled, 455 had kids. The best point estimate for pp is. The margin of error for a 99% confidence interval is. Answer: Step 1 We need to construct the 99% confidence interval for the population proportion. Number of favorable cases x=116 Sample size n=200 The sample proportion is calculated as, 𝑥 116 𝑝̂ = = = 0.58 𝑛 200 Step 2 The critical value for significance level 𝛼 = 0.01 is 𝑧𝑐 = 2.5758 The 99% confidence interval is computed as follows, 𝐶. 𝐼. = (𝑝̂ = 𝑧 𝑐√ ̂ (1−𝑝 ̂) 𝑝 𝑛 , 𝑝̂ + 𝑧 𝑐√ ̂ (1−𝑝 ̂) 𝑝 𝑛 0.58(1−0.58) = (0.58 − 2.5758√ 200 ) , 0.58 + 2.5758√ 0.58(1−0.58) 200 ) = (0.58 − 0.0899, 0.58 + 0.0899) = (0.4901, 0.6699) Therefore, the 99% confidence interval for the population proportion is, 0.4901

0.6699 Which indicates that we are 99% confident that the true population proportion is contained by the intervals (0.4901, 0.6699). Question #9 The number of points TEDU b etball team scores against Stanford b etball team is normally distributed. Suppose that over the last 10 games between the two teams TEDU scored the points given below: a) Construct a 95% two-sided confidence interval for the mean of the number of points scored by TEDU. b) The coach of the TEDU team finds out that the variance of the number of points scored is 25. Given this new information, construct a 95% two-sided confidence interval for the mean of the number of points scored by TEDU. c) Compare the confidence intervals found in part (a) and (b). Explain the reason behind the difference. Answer: Part (a) The table containing the mean, variance is shown below. X −X 9 −6 −3 −4 5 −3 1 −3 10 −6 X 74 59 62 61 70 62 66 62 75 59 Total ( X − X )2 81 36 9 16 25 9 1 9 100 36 322 The mean is 65. Use the equation 𝑉(𝑋) = 𝑉(𝑋) = 322 ∑(𝑋−𝑋̅)2 𝑛−1 to find the variance. 9 = 35.78 Use the equation 𝑆. 𝐷 = √𝑉(𝑋) to find the standard deviation. 𝑆. 𝐷 = √35.78 = 5.98 The t value corresponding to 95% confidence interval is 2.26. 𝜎2 Use the equation 𝐶. 𝐼 = 𝑋̅ ± 𝑡 to construct the confidence interval. 𝐶. 𝐼 = 65 ± 2.26 5.982 𝑛 10 = 65 ± 4.28 Therefore, the upper and lower limits are 65+4.28 and 65-4.28 Part (b) Given that, the variance of the distribution is 25. V(X)=25 𝑆. 𝐷 = √25 =5 𝜎2 Use the equation 𝐶. 𝐼 = 𝑋̅ ± 𝑡 to construct the interval. 𝐶. 𝐼 = 65 ± 2.26 × 52 𝑛 10 𝐶. 𝐼 = 65 ± 5.65 The upper and lower limits of the confidence interval is 65+5.65 and 65-5.65. Part (c) The standard deviation is defined as the degree of variance of value from mean value. In part (b) the standard deviation is higher compared to part (a). Therefore, the deviation is higher. So, the confidence interval got widened. Question #10 The scores of a certain population on the Wechsler Intelligence Scale for children (WISC) are thought to be normally distributed with a mean and standard deviation of 10. A simple random sample of twenty -five children from this population is taken and each is given the WISC. The mean of the twenty-five score is 𝑥̅ = 104.32. Based on these data a 95% confidence interval for \mu is computed. The 95% confidence interval for 𝜇 is? Answer: Step 1 Given: 𝜎 = 10 𝑥̅ = 104.32 𝑛 = 25 Step 2 The 95% cocfidence interval for the mean is: 𝜎 𝐶. 𝐼. = 𝑥̅ ± 𝑍𝛼/2 × √𝑛 From the Z table 𝑍𝛼/2 = 1.96 = 104.32 ± 1.96 × 10 √25 = 104.32 ± 3.92 = (104.32 − 3.92, 104.32 + 3.92) = (100.40, 108.24) Answer: = (100.40, 108.24) Question #11 Mark each statement TRUE or FALSE. For these problems, assume that to estimate average time spent waiting at a restaurant, you sample 110 people and you get a 95% confidence interval of (4, 6) in minutes. 1. 95% of people wait between 4 and 6 minutes at the restaurant. 2. 95 of the people in your sample waited between 4 and 6 minutes. 3. You are 95% sure your sample resulted in a confidence interval the contains the true mean. Answer: Step 1 Given that there are 110 people in the sample and I get a 95% confidence interval of (4, 6) in minutes. This means I am 95% sure that my sample resulted in a confidence interval the contains the true mean of the population. This does not mean a range that contains 95% of the values. Step 2 Answer: 1. 95% of people wait between 4 and 6 minutes at the restaurant - False 2. 95 of the people in your sample waited between 4 and 6 minutes - False 3. You are 95% sure your sample resulted in a confidence interval the contains the true mean -True Question #12 In a science fair project, Emily conducted an experiment in which she tested professional touch therapists to see if they could sense her energy field. She flipped a coin to select either her right hand or her left hand, and then she asked the therapists to identify the selected hand by placing their hand just under Emily's hand without seeing it and without touching it. Among 358 trials, the touch therapists were correct 172 times. Complete parts (a) through (d).a) Given that Emily used a coin toss to select either her right hand or her left hand, what proportion of correct responses would be expected if the touch therapists made random guesses? (Type an integer or a decimal. Do not round.)b) Using Emily's sample results, what is the best point estimate of the therapists' success rate? (Round to three decimal places as needed.) c) Using Emily's sample results, construct a 90% confidence interval estimate of the proportion of correct responses made by touch therapists. Round to three decimal places as needed - ?