Question 3 Continuous Probability Distribution Continuous Probability ...
Question 1. (Topics 1-3) A population consists of all the members of a group about which you want to draw a conclusion (Greek letters (μ, σ, Ν) are used) A sample is the portion of the population selected for analysis (Roman letter (x, s, n) are used for sample data) A parameter is a numerical measure that describes a characteristic of a population A statistic is a numerical measure that describes a characteristic of a sample Class intervals: Width of interval ≅
range no.of desired class groupings
Numerical data is measured on a natural numerical scale (age) Continuous – Data that can take on any real number (time/length) Discrete - Countable number of responses (cannot have 0.5) Categorical data can only be named or categorised Nominal – no order, no response is considered better (gender) Ordinal – There is an order (very good, good, average) Descriptive Statistics - Collect, Present, Characterise data
Arithmetic Mean: 𝑋̅ =
Measures of Dispersion: Variance, Standard Deviation, Coefficient of Variation
Reordered data: 3, 4, 7, 9 Variance: where SXfirstly & SYfind = S.Dev 𝑥 = formula 5.75 𝑛 (𝑥 − 𝑥 )2 𝑖=1 2 𝑠SXSy = = Sample Variance 𝑛−1 [(3 − 7)2 + (4 − 7)2 + (7 − 7)2 + (9 − 7)2 ] 5.75 − 1 [(−4)2 + (−3)2 + (0)2 + (2)2] = 4.75 16 + 9 + 0 + 4 29 = = = 6.10 4.75 4.75 Standard deviation: 𝑠 = 𝑠 2 = 6.1 = 2.46 Coefficient of variation: 𝑠 2.46 𝐶𝑉 = × 100% = × 100% = 61.7% 𝑥 4
𝑛
Median (Position):
Find the following probabilities 1. 𝑃(𝑍 < −1.67) = 0.0475 Read straight from the table. Note: P(Z −2.78) = ? 1 − 𝑃(𝑍 < −2.78) = ? 1 − 0.27 = 0.9973 3.𝑃(0.15 < 𝑍 < 1.99) = ? 𝑃(𝑍 < 1.99) = 0.9767 𝑃(𝑍 < 0.15) = 0.5596 0.9767 − 0.5596 = 0.4171 Solve the following inverse problems for the standard normal distribution 𝑃(𝑍 > ____ ) = 0.01 Look up the Inverse Normal Table 𝑃(𝑍 > 2.3263) = 0.01
2
Range 𝑿𝒎𝒂𝒙 − 𝑿𝒎𝒊𝒏
Z Score: 𝒁 =
̅ 𝑿−𝑿 𝑺
Z Outliers = > 3.0 or 30. Therefore according to the CLT (central limit theorem) at the very least I will end up with approximate normal distribution In other words if we have 30 observations or more, under the CLT we have a ≈ Normal Question 2 If X̅ = 75, S = 24, n = 36, and assuming that the population is normally distributed, construct a 95% confidence interval estimate of the population mean μ.
Question 3 A study conducted by the Australian Stock Exchange found that 46% of 2,405 Australian adults surveyed in 2006 held shares, either directly or indirectly through managed funds or self-managed superannuation funds (2006 Australian Share Ownership Study, ASX). (a)
Construct a 95% confidence interval for the proportion of Australian adults who held shares in 2006. When dealing with populations proportions we always use a Z.
(b) Interpret the interval constructed in (a). As above. I am 95% confident that the true proportion of Australian adults who held shares in 2006 is between 44 and 48% (c)
To construct a follow-up study to estimate the population proportion of adults who currently hold shares to within 0.01 with 95% confidence, how many adults would you interview?
range no.of desired class groupings
Numerical data is measured on a natural numerical scale (age) Continuous – Data that can take on any real number (time/length) Discrete - Countable number of responses (cannot have 0.5) Categorical data can only be named or categorised Nominal – no order, no response is considered better (gender) Ordinal – There is an order (very good, good, average) Descriptive Statistics - Collect, Present, Characterise data
Arithmetic Mean: 𝑋̅ =
Measures of Dispersion: Variance, Standard Deviation, Coefficient of Variation
Reordered data: 3, 4, 7, 9 Variance: where SXfirstly & SYfind = S.Dev 𝑥 = formula 5.75 𝑛 (𝑥 − 𝑥 )2 𝑖=1 2 𝑠SXSy = = Sample Variance 𝑛−1 [(3 − 7)2 + (4 − 7)2 + (7 − 7)2 + (9 − 7)2 ] 5.75 − 1 [(−4)2 + (−3)2 + (0)2 + (2)2] = 4.75 16 + 9 + 0 + 4 29 = = = 6.10 4.75 4.75 Standard deviation: 𝑠 = 𝑠 2 = 6.1 = 2.46 Coefficient of variation: 𝑠 2.46 𝐶𝑉 = × 100% = × 100% = 61.7% 𝑥 4
𝑛
Median (Position):
Find the following probabilities 1. 𝑃(𝑍 < −1.67) = 0.0475 Read straight from the table. Note: P(Z −2.78) = ? 1 − 𝑃(𝑍 < −2.78) = ? 1 − 0.27 = 0.9973 3.𝑃(0.15 < 𝑍 < 1.99) = ? 𝑃(𝑍 < 1.99) = 0.9767 𝑃(𝑍 < 0.15) = 0.5596 0.9767 − 0.5596 = 0.4171 Solve the following inverse problems for the standard normal distribution 𝑃(𝑍 > ____ ) = 0.01 Look up the Inverse Normal Table 𝑃(𝑍 > 2.3263) = 0.01
2
Range 𝑿𝒎𝒂𝒙 − 𝑿𝒎𝒊𝒏
Z Score: 𝒁 =
̅ 𝑿−𝑿 𝑺
Z Outliers = > 3.0 or 30. Therefore according to the CLT (central limit theorem) at the very least I will end up with approximate normal distribution In other words if we have 30 observations or more, under the CLT we have a ≈ Normal Question 2 If X̅ = 75, S = 24, n = 36, and assuming that the population is normally distributed, construct a 95% confidence interval estimate of the population mean μ.
Question 3 A study conducted by the Australian Stock Exchange found that 46% of 2,405 Australian adults surveyed in 2006 held shares, either directly or indirectly through managed funds or self-managed superannuation funds (2006 Australian Share Ownership Study, ASX). (a)
Construct a 95% confidence interval for the proportion of Australian adults who held shares in 2006. When dealing with populations proportions we always use a Z.
(b) Interpret the interval constructed in (a). As above. I am 95% confident that the true proportion of Australian adults who held shares in 2006 is between 44 and 48% (c)
To construct a follow-up study to estimate the population proportion of adults who currently hold shares to within 0.01 with 95% confidence, how many adults would you interview?
