Skip to content
Solved Assignments

Confidence Intervals: Calculations and Applications in Statistical Inference

Explores confidence interval calculations in statistics.

Preview 12 of 37 pages

See the first 12 pages exactly as they appear in the document.

  1. Page 1
    Read page 1 as text

    Confidence Intervals: Calculations and Applications in Statistical Inference Why is a 99% confidence interval wider than a 95% confidence interval? Solution) The definition of a confidence interval is that it contains the true population mean. If I have a 95% confidence interval, that means I am 95% certain that the true population mean is in the interval. If I want to be even more certain, I have to widen the interval. If I can be less certain, I can narrow the interval. So the widest interval will be 99%, and the narrowest would be 90%. Example: you're trying to figure out where in the city Comet Donuts is in, but you really don't know for sure. A desperately hungry person hands you a map and asks you to show him where it is. If someone forces you to be 99% accurate, are you going to draw a wide or narrow circle on the map? You can't afford to be wrong - at 99% you're saying that you'll be wrong one time out of 100! So you draw a big circle. If the person asking doesn't even like donuts, they're just asking for the heck of it, you can be 90% accurate, so you can take a chance and draw a small circle. You'll be wrong 10% of the time . 12. A person claims to be able to predict the outcome of flipping a coin. This person is correct 16/25 times. Compute the 95% confidence interval on the proportion of times this person can predict coin flips correctly. What conclusion can you draw about this test of his ability to predict the future? Solution) WE HAVE GIVEN THAT n = 25 and p = 16/25 And we need to construct the 95% C.I. for the proportion of times this person can predict coins flips correctly as, ṕ ± 1.96 * √ ( ṕ q^/n) =.64± 1.96 * √ (.64*.36/25) = .64 ± .1882

  2. Page 2
  3. Page 3
    Read page 3 as text

    So the 95% C.I. is, (0.4518, 0.8282) So We Are 95 Out Of 100 Attempts are confident that the values of the samples are lies b/w (.4518,.8282) 15. You take a sample of 22 from a population of test scores, and the mean of your sample is 60. (a) You know the standard deviation of the population is 10. What is the 99% confidence interval on the population mean? Solution) We have given that n = 22, sample mean =60 and σ = 10 The 99 % C.I. for the population mean is, = sample mean ± 2.58 * σ / √ n = 60 ± 2.58 * 10 / √ 22 = 60 ± 5.501 So, (54.499, 65.501 ) (b) Now assume that you do not know the population standard deviation, but the standard deviation in your sample is 10. What is the 99% confidence interval on the mean now? Solution) Here we have given that n = 22, sample mean = 60 and S = 10 Here we assume that the population stranded deviation is unknown, and also n<30 so we here use the student t distribution as, = sample mean ± t α /2,v s / √ n - 1 Where v= n - 1 degree of freedom and v= 22 - 1 = 21 = 60± 2.831 * 10/ √ 21 = 60±6.178 So, (53.822, 66.178 ) 18. You were interested in how long the average psychology major at your college studies per night, so you asked 10

  4. Page 4
    Read page 4 as text

    psychology majors to tell you the amount they study. They told you the following times: 2, 1.5, 3, 2, 3.5, 1, 0.5, 3, 2, 4. (a) Find the 95% confidence interval on the population mean. Sol) Here we have given that sample mean = 2.25 n= 10 and σ = 1.0548 We use a standard normal distribution and assume that population standard deviation is known so The 95 % C.I. for the population mean is, = sample mean ± 1.96 * σ / √ n = 2.25 ± 1.96* 1.0548 / √ 10 = 2.25 ± .6538 = (1.596, 2.904) (b) Find the 90% confidence interval on the population mean. THE 90% C.I. IS. = sample mean ± 1.645 * σ / √ n = 2.25 ± 1.645 * 1.0548/ √ 10 = 2.25 ± .549 So, (1.701, 2.799) 100. What is meant by the term “90% confident” when constructing a confidence interval for a mean? a. If we took repeated samples, approximately 90% of the samples would produce the same confidence interval. b. If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the sample mean. c. If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the true value of the population mean. d. If we took repeated samples, the sample mean would equal the population mean in approximately 90% of the samples.

  5. Page 5
    Read page 5 as text

    106. Suppose that a committee is studying whether or not there is waste of time in our judicial system. It is interested in the mean amount of time individuals waste at the courthouse waiting to be called for jury duty. The committee randomly surveyed 81 people who recently served as jurors. The sample mean wait time was eight hours with a sample standard deviation of four hours. Solution) i) x ¯ = __________ 8 hours ii. sx = __________ 4 hours iii. n = __________ 81 iv. n – 1 = __________80 b. Define the random variables X and X ¯ in words. X is the number of hours a person wastes while waiting to be called for s ervice. X bar is the average number of hours wasted for people in the sample. c. c. Which distribution should you use for this problem? Explain your choice. We should use the standard normal distribution because n is large enough. d. Construct a 95% confidence interval for the population mean time wasted. (7.12, 8.88) i. State the confidence interval. Sol)

  6. Page 6
    Read page 6 as text

    The 95% C.I. for the population mean worth of coupons is, =x (bar) ± Z α /2 *S/ √ n = 8 ± 1.96(4/ √ 81) = 8 ± 0.8711 So, (7.129, 8.8711) ii) Sketch iii. Calculate the error bound. EBM = Z α /2 *S/ √ n = 1.96(4/ √ 81) = 0.88 e. Explain in a complete sentence what the confidence interval means. If we sampled 81 people 100 times, we would expect 95 of the samples t o have averages in the interval. 112 . In a recent sample of 84 used car sales costs, the sample mean was $6,425 with a standard deviation of $3,156. Assume the underlying distribution is approximately normal. a. Which distribution should you use for this problem? Explain your choice. We should use a standard normal distribution because n>=30

  7. Page 7
    Read page 7 as text

    b. Define the random variable X ¯ in words. X ¯ is the average no of car sales costs c. Construct a 95% confidence interval for the population mean cost of a used car. i. State the confidence interval. We have given that sample maen = 6425 n = 84 and standerd deviation = 3156 The 95% C.I. for the population mean is, = sample mean ± 1.96 * σ / √ n = 6425 ± 1.96 * 3156 / √ 84 = 6425 ± 674.922 S0, (5777.078, 7099.922) ii. Sketch the graph. ii. Calculate the error bound. = EBM = Z α /2 * σ / √ n EBM = 1.96 * 3156 / √ 84 = 674.922 d. Explain what a “95% confidence interval” means for this study. The 95% confidence interval means that w e are 95% confident that the sample mean lie in the interval (5777.078, 7099.922) Use the following information to answer the next exercise: A quality control specialist for a restaurant chain takes a

  8. Page 8
    Read page 8 as text

    random sample of size 12 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with a sample standard deviation of 1.55. Assume the underlying population is normally distributed. 116 . What is the error bound? a. 0.87 b. 1.98 c. 0.99 d. 1.74 120. An article regarding interracial dating and marriage recently appeared in the Washington Post. Of the 1,709 randomly selected adults, 315 identified themselves as Latinos, 323 identified themselves as blacks, 254 identified themselves as Asians, and 779 identified themselves as whites. In this survey, 86% of blacks said that they would welcome a white person into their families. Among Asians, 77% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person. a. We are interested in finding the 95% confidence interval for the percent of all black adults who would welcome a white person into their families. Define the random variables X and P ′ , in words. X ῀ is the no of black adults who are well come a white persons into their families. p ῀ is the probability of the no of black adults who are well come a white persons into their families. b. Which distribution should you use for this problem? Explain your choice.

  9. Page 9
    Read page 9 as text

    We use the standard normal distribution for the population proportion because the sample size is large enough as >=30. c. Construct a 95% confidence interval. i. State the confidence interval. We have given that n = 1709 , p = 323/1709 = .189 The 95% C.I. is. = ṕ ± Z α /2 √ (pq/n = .189 ± 1.96 * √ .189*.811 /1709 = 0.189± 0.0186 So, (0.170, .2076) ii. Sketch the graph. iii. Calculate the error bound. EBM = z α /2 √ (pq/n) = 1.96 * √ .189*.811/1709 = 0.0186 130. On May 23, 2013, Gallup reported that of the 1,005 people surveyed, 76% of U.S. workers believe that they will continue working past retirement age. The confidence level for this study was reported at 95% with a ±3% margin of error. Solution) a. Determine the estimated proportion from the sample. The estimated proportion from the sample, P^ = 76% = .76 b. Determine the sample size. Sample size = n= 1,005 c. Identify CL and α . Confidence level = 95%

  10. Page 10
    Read page 10 as text

    α = 5% = 0.05 d. Calculate the error bound based on the information provided. EBM = z α /2 √ (pq/n) = 1.96 * √ (.76*0.24/1005) = 0.0264 e. Compare the error bound in part d to the margin of error reported by Gallup. Explain any differences between the Values. The stated ±3 represent the MEB. Which means that those working past retirement age. And is equal to 0.0264. And seems that there is no difference b/w EBM and margins of error that is stated in problems. e. Create a confidence interval for the results of this study. The 95% C.I. for the results of study are, = ṕ ± Z α /2 √ (pq/n) = .76 ± 1.96 * √ (.76*.24/1005) = .76 ± 0.0264 (0.7336 , .7864) g. A reporter is covering the release of this study for a local news station. How should she explain the confidence - Interval to her audience? A reporter is converting the release of this study for a local station, this means that if we sampled we would expect that 95 of the samples to have audience in this intervals (0.7336, .7864)

  11. Page 11
    Read page 11 as text

    STAT200 WEEK 4 HOMEWORK Probability Distributions Problem Set Solutions are provided at the end of the document P - 1 The salespeople at Owl Realty sell up to 9 houses per month. The probability distribution of a salesperson selling n houses in a month is as follows: Sales Probability 0 .05 1 .10 2 .15 3 .20 4 .15 5 6 .10 7 .05 8 .05 9 .05 a. What is the probability of selling 5 houses in a month? b. What are the expected value and the standard deviation for the number houses sold by the salespeople per month? P - 2 The following data has been recorded regarding the rental history of a high pressure cleaner at a home improvement store. Demand Frequency 0 20

  12. Page 12
    Read page 12 as text

    1 23 2 27 3 14 4 16 Find the expected demand for the high pressure cleaner.

25 more pages in the full document

Unlock it once and it stays in your library, ready to chat with or turn into flashcards.

Unlock the full document

More Statistics study materials

See all

Study this document with CramX

Once it is in your library, every tool can work from it.