Questions and Answers #3 Confidence Intervals

Questions and Answers Sheet 3 Confidence Intervals Question #1 Assume that we want to construct a confidence interval. Do one of the following, as appropriate: a) find the critical value 𝑡α/2 . b) find the critical value 𝑧α/2 , or c) state that neither the normal distribution nor the t distribution applies. Here are summary statistics for randomly selected weights of newborn girls: n=249, 𝑥 = 31.5ℎ𝑔, 𝑠 = 6.3ℎ𝑔. The confidence level is 95% . A. 𝑡α/2 =? B. 𝑧α/2 =? C. Neither the normal distribution nor the t distribution applies. Answer: Step 1 Introduction: It is known that t test is used to test the hypothesis for a population mean when the sample size is not large enough (< 30) and/or the population variance is not known. In this case the sample size is large enough, however the population variance is not mentioned. The sample standard deviation is used to estimate the population standard deviation. Thus, it would be appropriate to use the t test with degrees of freedom of n-1 . Step 2 Calculation: Here, the level of significance is α = 0.05 . The critical value for the two-tailed test is: 𝑃{𝐻0( |𝑡 𝑛−1 }|≥ 𝑡α,𝑛−1 ) = 0.05, where 𝑡α,𝑛−1 is the critical value. The critical value is such that 𝑃{𝐻0( |𝑡𝑛−1 }≥𝑡α/2,𝑛−1) = 𝑃𝐻0 (𝑡𝑛−1 ≤ −𝑡α/2,𝑛−1 ) = 0.025 The degrees of freedom is 248(=249-1) To obtain the exact critical value we have used Excel. Using Excel formulae: =T.INV.2T(0.05,248) , the critical value is 1.97. Thus, the correct option is A and 𝑡α = 1.98. 2 Question #2 Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts (a) through (c) below. a. Express the confidence interval in the format that uses the ''less than'' symbol. Given that the original listed data use one decimal place, round the confidence intervals limits accordingly. 13.05 𝑀𝑏𝑝𝑠 < μ < 22.15 𝑀𝑏𝑝𝑠 b. Identify the best point estimate of μ and the margin of error. The point estimate of μ 𝑖𝑠 17.60 𝑀𝑏𝑝𝑠. The margin of error is E=? Mbps. Answer: Step 1 Given data, Lower limit =13.05 Upper limit =22.15 Point estimate x=17.60 E=? Step 2 𝐸= = 𝑈𝐿−𝐿𝐿 2 22.15−13.05 2 E=4.55 Question #3 A random sample of 100 automobile owners in the state of Virginia shows that an automobile is driven on average 23,500 kilometers per year with a 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 automobile 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? Answer: a) The 99% confidence interval for the average number of kilometres an automobile is driven is obtained below: The value of mean is 23,500 kilometres, population standard deviation is 3,900 and sample size (n)=100 Critical value: From the standard normal distribution table, for the 99% confidence level the critical values is (𝑧 ∗ )2.58 . The confidence interval formula for the population mean is, σ 𝐶. 𝐼. = 𝑥̅ ± 𝑧 ∗ √𝑛 Substitute mean =23,500 , standard deviation (σ) is 3,900 and sample size (n)=100 . σ 𝐶. 𝐼. = 𝑥̅ ± 𝑧 ∗ √𝑛 = 23,500 ± 2.58 3.900 √100 = 23,500 ± 2.58(390) = 23,500 ± 1006.2 = (23,500 − 1006.2, 23,500 + 1006.2) = (22,493.8, 24,506.2) Thus, the 99% confidence interval for the average number of kilometres an automobile is between 22,493.8 and 24,506.2. b) The required formula is obtained below: σ 𝑀𝐸 = 𝑧α ( ) 2 √𝑛 Substitute 3,900 for σ, 100 for n and 2.58 for 𝑍α . 2 𝑀𝐸 = 2.58 ( = 2.58 ( 3,900 10 3,900 √100 ) ) =2.58(390) =1006.2 With 99% confidence that the possible size of our error will not exceed the 1006.2 kilometres. Question #4 Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. To accomplish this, the records of 200 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.3 seats. 𝑠𝑥 =? , 𝑛 =? , 𝑛 − 1 =?. Construct a 92% confidence interval for the population average number of unoccupied seats per flight. (State the confidence interval. (Round your answers to two decimal places.) Answer: Step 1 𝑥̅ = 11.6, 𝑠 = 4.3, 𝑛 = 200 𝑠𝑥 = 𝑠 √𝑛 = 4.3 √200 = 0.3041 n=200 n-1=199 Step 2 92% 𝐶. 𝐼. = 𝑥̅ ± 𝑇𝐼𝑁𝑉(0.08,199)(𝑠𝑥̅ ) = 11.6 ± 1.76(0.3041) = 11.6 ± 0.54 = (11.06,12.14) 𝑇𝐼𝑁𝑉(α, 𝑛 − 1) is a Excel function. Question #5 A simple random sample of 60 items resulted in a sample mean of 80. The population standard deviation is σ = 15 a) Compute the 95% confidence interval for the population mean. Round your answers to one decimal place. b) Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean. Round your answers to two decimal places. c) What is the effect of a larger sample size on the interval estimate? Larger sample provides a-Select your answer-larger smaller Item 5 margin of error. Answer: a) The sample size is 60, sample mean is 80, and population standard deviation is 15. Computation of critical value: (1 − α) = 0.95 α = 0.05 α = 0.025 2 α 1 − = 0.975 2 The critical value of z-distribution can be obtained using the excel formula “=NORM.S.INV(0.975)” . The critical value is 1.96. The margin of error is, σ 𝐸 = 𝑧α ( ) 2 √𝑛 15 = 1.96 ( ) √60 =3.80 The margin of error is 3.80. The 95% confidence interval for the population mean is, 𝐶. 𝐼. = 𝑥̅ ± 𝐸 = 80 ± 3.80 = (80 − 3.8, 80 + 3.8) = (76.2, 83.8) Thus, the 95% confidence interval for the population mean is (76.2, 83.8) . b) The margin of error is, σ 𝐸 = 𝑧α ( ) 2 = 1.96 ( √𝑛 15 √120 ) =2.68 The margin of error is 2.68. The 95% confidence interval for the population mean is, 𝐶. 𝐼. = 𝑥̅ ± 𝐸 = 80 ± 2.68 = (80 − 2.68, 80 + 2.68) = (77.32, 82.68) Thus, the 95% confidence interval for the population mean is (77.32, 82.68) . c) The margin of error for sample size 60 is 3.80, and margin of error for sample size 120 is 2.68. It is clear that increasing the sample size causes the decreases the margin of error value Thus, the larger sample provides a smaller margin of error. Question #6 A food safety guideline is that the mercury in fish should be below 1 part per million (ppm). Listed below are the amounts of mercury (ppm) found in tuna sushi sampled at different stores in a major city. Construct a 95% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna sushi? 0.60 0.82 0.09 0.89 1.29 0.49 0.82 What is the confidence interval estimate of the population mean μ? ? 𝑝𝑝𝑚 < μ < ? 𝑝𝑝𝑚 Answer: Step 1 x (x − x ) 0.6 −0.11429 0.82 0.10571 0.09 −0.62429 0.89 0.17571 1.29 0.57571 0.49 −0.22429 0.82 0.10571 Total 5 ( x − x )2 0.01306122 0.01117551 0.38973265 0.03087551 0.33144694 0.05030408 0.01117551 0.83777143 Here n=7 Sampke mean = 𝑥̅ = Sample S.D.= 𝑠 = √ ∑𝑥 𝑛 1 𝑛−1 = 0.714 ∑(𝑥 − 𝑥̅ )2 = 0.3737 Step 2 Confidence Level = 95 Significance Level =  = (100 − 95)% = 0.05 Degrees of freedom = n − 1 = 7 − 1 = 6 Critical value = t * = 2.447[using Excel = TINV (0.05, 6)] Standard Error = 𝑠 √𝑛 = 0.1412 Margin of Error (M.E) = 𝑡 ∗ 𝑠 √𝑛 = 0.3456 Lower Limit = 𝑥̅ − (𝑀. 𝐸) = 0.3687 Upper Limit = 𝑥̅ + (𝑀. 𝐸) = 1.0599 Question #7 The monthly incomes for 12 randomly selected people, each with a bachelor's degree in economics, are shown on the right. Complete parts (a) through (c) below. Assume the population is normally distributed. a) Find the sample mean. b) Find the sample standard deviation c) Construct a 95% confidence interval for the population mean μ. A 95% confidence interval for the population mean is 4450.42 4455.62 4283.26 3946.49 4596.96 4151.52 4527.94 4023.61 4366.46 3727.77 4407.68 4221.73 Answer: Step 1 X 4450.42 4596.96 4366.46 4455.65 4151.52 3727.77 4283.26 4527.94 4407.68 3946.49 4023.61 4221.73 Total = 51159.49 X −X 187.1292 33.6692 103.1692 192.3592 −111.771 −535.521 19.96917 264.6492 144.3892 −316.801 −239.681 −41.5608 Step 2 a) Sample mean: 𝑥̅ = = ∑𝑋 𝑛 51159.49 12 =4263.291 The sample mean is 4263.291 Step 3 b) Sample Standard Deviation 𝑠𝑑 = √ ∑(𝑋 − 𝑋̅)2 𝑛−1 ( X − X )2 35017.33 111335.1 10643.88 37002.05 12492.72 286782.6 398.7676 70039.18 20848.23 100362.8 57446.9 1727.303 Total = 744096.8 =√ 744096.8 12−1 =260.087 The sample standard deviation is 260.087 Step 4 c) 95% Confidence Interval for Population mean: 𝑠𝑑 𝑥̅ ± 𝑡𝑛−1 × √𝑛 260.087 4263.291 ± 2.201 × √12 4263.291 ± 165.253 (4098.038, 4428.544) 95% confidence interval for the population mean is (4098.038, 4428.544) Question #8 It was observed that 30 of the 100 randomly selected students smoked. Find the confidence interval estimate of 95% confidence level for π ratio of smokers in the population. a) P(0.45 65 Here, sample proportion is 0.58, n = 2056 and claimed proportion is 0.65 at significance level α = 0.01 . Hence the standard deviation at 0.65 claimed proportion, 𝑆. 𝐸 = √ =√ =√ claimed proportion(1-claimed proportion) 𝑛 0.65×0.35 2056 0.2275 2056 ≈ 0.0105 So the value of 𝑧𝑐𝑎𝑙𝑐 is: 𝑧𝑐𝑎𝑙𝑐 = = = 𝑝̂−0.65 0.0105 0.58−0.65 0.0105 −0.07 0.0105 =-6.6667 Hence, the value of z is -6.67. Question #12 A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 24 subjects had a mean wake time of 104.0 min. After treatment, the 24 subjects had a mean wake time of 94.5 min and a standard deviation of 23.2 min. Assume that the 24 sample values appear to be from a normally distributed population and construct a 99% confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 104.0 min before the treatment? Does the drug appear to be effective? Construct the 99% confidence interval estimate of the mean wake time for a population with the treatment. ? 𝑚𝑖𝑛 < μ < ? 𝑚𝑖𝑛 Answer: Step 1 The (1 − α)100% confidence interval formula for the population mean when population standard deviation is not known, is defined as follows: 𝑠 𝐶𝐼 = 𝑥 ± 𝑡α,𝑛−1 ( ) √𝑛 2 Here, 𝑡α,𝑛−1 is the critical value of the t-distribution with degrees of freedom of n-1 above which, 2 100(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, 𝑥 is the sample mean, s is the sample standard deviation, and n is the sample size. Step 2 According to the given question s = 23.2 min . The sample mean wake time after treatment is 94.5 min. The sample size is 24. Hence, the degrees of freedom is 23 (= 24 –1) . The level of significance is 0.01. Using the Excel formula =T.INV.2T(0.01,23) the critical value is 2.807. Hence, the confidence interval is, 𝑠 23.2 √𝑛 √24 𝑥 ± 𝑡α,𝑛−1 ( ) = (94.5 ± (2.807) ( 2 )) = (94.5 ± (2.807)(4.736)) = (94.5 ± 13.3) =(81.2, 107.8) Thus, the 99% confidence interval estimate of the mean wake time for a population with drug treatments is 81.2 𝑚𝑖𝑛 < μ < 107.8 𝑚𝑖𝑛 . Here, the value 104.0 does lies within the obtained interval. Thus, the drug does not effectively change the wake time, by reducing it. Hence, the drug appears to be ineffective. Question #13 If n=590 and p=0.93 , construct a 90% confidence interval. Answer: Step 1 Express the expression for the confidence interval. 𝑝(1−𝑝) 𝐶. 𝐼 = 𝑝 ± 𝑍 ∗ √ 𝑛 Here, Z* is the critical value . From the Z table, the value corresponding to the region 90.5 is, 𝑍 ∗= 1.31 Step 2 Put 590 for n, 0.93 for p, 1.31 for Z* implies, 𝐶. 𝐼 = 0.93 ± 1.31√ 0.93(1−0.93) 0.93 𝐶. 𝐼 = 0.93 ± 0.346 Thus, the confidence interval is 0.93+0.346 and 0.93-0.346. Question #14 Use the Student's t distribution to find 𝑡𝑐 for a 0.95 confidence level when the sample is 24. (Round your answer to three decimal places.) Answer: Step 1 Obtain the critical value of t using the student's t distribution to find 𝑡𝑐 for a 0.95 confidence level when the sample is 24. The critical value of t is obtained below: From the information, given that n=24 . Obtain the degrees of freedom. The degrees of freedom is obtained below: df=n-1 =24-1 =23 Here, confidence level is 0.95. For (1 − α) = 0.95 α = 0.05 Step 2 Use EXCEL Procedure for finding the critical value of t. Follow the instruction to obtain the critical value of t: 1.Open EXCEL 2.Go to Formula bar. 3.In formula bar enter the function as“=TINV” 4.Enter the probability as 0.05. 5.Enter the degrees of freedom as 23. 6.Click enter EXCEL output: From the EXCEL output, the critical value of t at the 0.95 confidence level with the 23 degrees of freedom is 2.069. Thus, the critical value of t is 2.069. Question #15 A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 13 subjects had a mean wake time of 101.0 min. After treatment, the 13 subjects had a mean wake time of 94.6 min and a standard deviation of 24.9 min. Assume that the 13 sample values appear to be from a normally distributed population and construct a 95% confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 101.0 min before the treatment? Does the drug appear to be effective? Construct the 95% confidence interval estimate of the mean wake time for a population with the treatment. ___ 𝑚𝑖𝑛 < μ <__ min (Round to one decimal place as needed.) What does the result suggest about the mean wake time of 101.0 min before the treatment? Does the drug appear to be effective? The confidence interval ▼ does not include| includes the mean wake time of 101.0 min before the treatment, so the means before and after the treatment ▼ could be the same |are different. This result suggests that the drug treatment ▼ does not have | has a significant effect. Answer: Step 1 Let μ𝑎 denotes the true population mean for after treatment. Let 𝑥𝑏 and 𝑥𝑎 denote the sample mean for before and after treatment data. Then, provided that 𝑥𝑏 = 101.0𝑚𝑖𝑛 and \overline{x_{a}}=94.6 min . The sample size, n=13 and the sample standard deviation for after treatment data is 𝑠𝑎 = 24.9 min . 𝑠 A 95% confidence interval for the true population mean of after treatment data is given by 𝑥𝑎 ± 𝑡𝑐 ⋅ 𝑎 . √𝑛 Where, 𝑡𝑐 is the critical t-score. Step 2 The critical t-score for 95% confidence level with 12 degree of freedom is 𝑡𝑐 = 2.18. Now, substituting, 𝑥𝑎 = 94.6𝑚𝑖𝑛, 𝑠𝑎 = 24.9𝑚𝑖𝑛, and n=13 in the expression, 𝑠 𝑥𝑎 ± 𝑡𝑐 ⋅ 𝑎 √𝑛 𝑠𝑎 24.9 𝑥𝑎 ± 𝑡𝑐 ⋅ = 94.6 ± 2.18 ⋅ 𝑛 √ √13 ≈ 94.6 ± 15.1 =(79.5, 109.7) As, it can be observe that the confidence interval, 79.5𝑚𝑖𝑛 < μ𝑎 < 109.7𝑚𝑖𝑛 includes 𝑥𝑏 = 101.0𝑚𝑖𝑛. Therefore, the true mean for before and after treatment could be the same. This tells that the treatment does not have the significance effect. Question #16 ACTIVITY H: Sampling Theory and Estimation of Parameters 1. A random sample of 50 students in Holy Angel University shows that they spend an average of 3 hours per day on social media apps with a standard deviation of 0.5 hours. Assume a normal distribution. Construct a 90% and 97% confidence interval for the average number of hours spent on social media apps per day. 2. A study was conducted in which two types of fried chicken, Jollibee and KFC, were compared. Number of fried chickens sold, was measured. 90 Jollibee branches were surveyed while 80 were surveyed for KFC. The average fried chicken sold was 7,020 for Jollibee and 8,100 fried chickens for KFC. Find a 94% and 98% confidence interval on .Assume that the population standard deviations are 1000 and 500 for KFC and ????--?????????Jollibee, respectively. 3. The following are the average lengths of today’s top 10 Global Spotify hits, in minutes. 3.02, 2.85, 3.33, 3.43, 2.9, 2.93, 3.23, 2.77, 2.45, and 2.88. Find a 95% confidence interval for the variance of the lengths of this generation’s music, assuming a normal population. Answer: Part 1 As per bartleby guidelines only 1 question is answered, please upload the rest separately. The mean and standard deviation are μ = 3, σ = 0.5 respectively. For 90% confidence, the one tail test is 5% from right end that means it is 95% to right and 5% to the left. 𝑃 (𝑧 ≤ 𝑥−μ σ ) = 0.95 z=1.65 Hence the value of x is given by 𝑥 = μ±𝑧σ 𝑥 = 3 ± 1.65 × 0.5 x=3.825, 2.175 Therefore the interval is 2.175 ≤ 𝑥 ≤ 3.825 for 90% confidence. Part 2 For 97% confidence, the one tail test is 1.5% from right end that means it is 98.5% to right and 1.5% to the left. 𝑃 (𝑧 ≤ 𝑥−μ σ ) = 0.985 z=2.17 Hence the value of x is given by 𝑥 =μ±𝑧σ 𝑥 = 3 ± 2.17 × 0.5 𝑥 = 4.085𝑚 1.915 Therefore the interval is 1.915 ≤ 𝑥 ≤ 4.085 for 97% confidence. Question #17 A certain reported that in a survey of 2006 American adults, 24% said they believed in astrology. a) Calculate a confidence interval at the 99% confidence level for the proportion of all adult Americans who believe in astrology. (Round your answers to three decimal places.) (_______, _______) b) What sample size would be required for the width of a 99% CI to be at most 0.05 irresoective of the value of 𝑝̂ ? (Round your answer up to the nearest integer.) Answer: a) Given data, n=2006 P=24%=0.24 Z-value at 99% confidence is 𝑍𝑐 = 2.576 Margin of error (𝐸) = 𝑍𝑐 × √ 𝑝(1−𝑝) 𝑛 0.24(1−0.24) = 2.576 × √ 2006 = 2.576 × √9.0927218344𝑒 −5 E=0.052 The 99% is 𝑝̂ + 𝐸 = 0.24 ± 0.052 =(0.188, 0.292) b) P=0.5 E=0.05 𝑍𝑐 = 2.576 n=? 7𝑐 2 𝑛 = 𝑃(1 − 𝑝) × ( ) 𝐸 = 0.5(0.5) × ( 2.576 2 0.05 ) n=664 Question #18 If n=600 and 𝑝̂ = 0.95, construct a 95% confidence interval. Give your answer to three decimals. ____ < p < ____ Answer: Given that: 𝑛 = 600, 𝑝̂ = 0.95 The critical value for α = 0.05 is 𝑧𝑐 = 𝑧1−α/2 = 1.96 𝐶𝐼(𝑃𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛) = (𝑝̂ − 𝑧𝑐 √ = (0.95 − 1.96 × √ 𝑝̂ (1 − 𝑝̂ ) 𝑝̂ (1 − 𝑝̂ ) , 𝑝̂ + 𝑧𝑐 √ ) 𝑛 𝑛 0.95(1 − 0.95) 0.95(1 − 0.95) , 0.95 + 1.96 × √ ) 600 600 =(0.933, 0.967) Therefore, the 95% CI for the population proportion is 0.933 < p < 0.967 Question #19 Find the critical value z a/2 that corresponds to a 93% confidence level Answer: Step 1 C.I=93% α = 1 − 0.93 = 0.07 α/2 = 0.035 Step 2 We need to find z such that 𝑃(𝑍 > 𝑧) = 0.035 ⇒ 𝑃(𝑍 < 𝑧) = 1 − 0.035 = 0.965 ⇒ 𝑧 = 1.81 [using Normal table or Excel: NORM.S.INV(0.965)] Question #20 Express the confidence intervel (0.501, 0.609) in the from of p+E Answer: Step 1 Here we need to express the given confidence interval in p+/-E form. Step 2 a) Here the given 𝐶𝐼 = (0.501,0.609) From the given confidence interval; 𝑈+𝐿 0.501+0.609 Point estimate 𝑝 = = 2 2 =0.555 ∴ 𝑝 = 0.555 Now 𝐸 = 𝑈+𝐿 2 = 0.609−0.501 2 ∴ 𝐸 = 0.054 ∴ 𝐶𝐼 = 0.555 ± 0.054 = 0.501 < 𝑝 < 0.609 𝑃 ± 𝐸 = 0.555 ± 0.054 = 0.054