board the plane Calculation of 95% confidence interval for the expected mean number of passengers who board the plane: In order to calculate the confidence interval, we used the CONFIDENCE function with the inputs of alpha, mean standard error and sample size of 0.05, 0.34 and 1000 respectively. Therefore based on the calculation conducted on excel spreadsheet, we calculated the upper and lower limit of the interval. Confidence Interval | Confidence Interval | 0.95 | Alpha | 0.05 | +/- Range
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008 | 31.1943 | | 95% Confidence Interval for Mean | Lower Bound | 410.911 | | | | Upper Bound | 537.104 | | | 5% Trimmed Mean | 461.956 | | | Median | 437.000 | | | Variance | 38923.356 | | | Std. Deviation | 197.2900 | | | Minimum | 169.9 | | | Maximum | 975.0 | | | Range | 805.1 | | | Interquartile Range | 219.1 | | | Skewness | 1.096 | .374 | | Kurtosis | 1.011 | .733 | Sale Price | Mean | 454.223 | 30.4397 | | 95% Confidence Interval for Mean | Lower Bound
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Bahria University Bahria Institute of Management & Computer Sciences Karachi Campus Course Title: Statistical Inference Course Code: QTM 220 Credit Hours: Three Semester: 3rd Semester Prerequisite: QTM 160 Aims and Objectives: This main objective of this course is to provide wide application f the statistical tools in business management. It is also aims at to impart in-depth and rigorous knowledge to the business students to inculcate academic excellence in various fields of research and
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statistics, which include measures of central tendency and dispersion. The calculations for these statistics were conducted in an excel worksheet and reported in a summary fashion in this report. The analysis will also include the distribution and confidence intervals for each company, along with an analysis of those findings. A hypothesis test was conducted on all three data sets in order to establish that the mean values were credible. Finally, a regression analysis was done to see if there was a relationship
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estimate of π. Underlying Sample distribution is binomial and can be approximated by normal if: nπ ≥ 5 and n(1- π) ≥ 5. With resulting mean equal to μp=π and standard error equal to σp=π(1-π)n Therefore: WEEK 8 CONFIDENCE INTERVALS: CONFIDENCE INTERVAL FOR μ (σ Known): Assume standard deviation is known, population is normally distributed. If not
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20 HA : µ < 20 A sample of 40 provided a sample mean of 19.4. The population standard deviation is 2. (a) Create a 95% confidence interval for the mean. We know σ, therefore we should use the z − table. This is a one-tailed (lower tail) test, so the 95% confidence interval will be given then by σ x − z.05 √ , ∞ ¯ n 2 19.4 − 1.65 √ , ∞ 40 The 95% confidence interval is µ ∈ [18.878, ∞). (b) What is the p-value? The p-value is the area in the lower tail. First, we calculate the z-value:
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for the six colors vary with Blue being the dominant color of 852, follow by orange 770, green, 616, red 506 yellow 484 and brown 454. Secondly I had to construct the 95% confidence interval for the proportions of all six colors blue, orange, green, yellow, red and the results are the following.95% confidence interval results: | | | | | | p : proportion of successes for population | | | | | Method: Standard-Wald | | | | | | | Proportion | Count | Total | Sample
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Econ 184b Fall 2011 Solutions to Problem Set 1 1. (a) | Y = 0 | Y = 1 | Pr(X) | X = -10 | .225 | .175 | .40 | X = 0 | .0375 | .0125 | .05 | X = 10 | .4875 | .0625 | .55 | Pr(Y) | .75 | .25 | 1.0 | These probabilities are calculated using three facts: Pr(X = x | Y = y) = Pr(X = x, Y = y) Pr(Y = y) n ∑ Pr(X = xj , Y = yi) = Pr(X = xj) i=1 n ∑ Pr(Y = yi) = 1 i=1 (b) E(X) = (-10)(.40) +
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Exam #3 Review | | | | | | |1. |(5 point(s)) | | |A report on health care in the U.S. said that 28% of Americans have experienced
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and is hence not statistically significant. iii) Period 41: Point forecast = 542.9878132; 95% confidence interval = (400.9438205, 685.0318058) Period 42: Point forecast = 456.8909767; 95% confidence interval = (312.3282773, 601.453676) Period 43: Point forecast = 385.2961403; 95% confidence interval = (237.9076139, 532.6846667) Period 44: Point forecast = 531.6563038; 95% confidence interval = (381.1521284, 682.1604783) The assumptions on ε are that it is an independent random variable
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