Presentation on theme: "Textbook Chapter 9. Point estimate and level of confidence. Confidence interval for a population mean with known population standard deviation. "— Presentation transcript:
Textbook Chapter 9
Point estimate and level of confidence. Confidence interval for a population mean with known population standard deviation. Confidence interval for a population mean with unknown population standard deviation. Confidence interval for a population proportion. Sample size for attribute and variable sampling.
Point estimate is a statistic, which is used to estimate the population parameter, computed from sample information. Confidence interval estimate is a range of values constructed from sample data, within which the population parameter is likely to fall within that range at a specified probability (level of confidence). Level of confidence is the measure of confidence we have that a confidence interval estimate will include the population parameter. Sample mean, ‘x-bar’, is a point estimate of population mean, μ. Sample proportion, p, is a point estimate of population proportion, π. Sample standard deviation, s, is a point estimate of population standard deviation, σ.
The width of a confidence interval is influenced by: 1. Sample size, n. 2. Population variation (σ 2 ), or, sample variation (s 2 ). 3. Level of confidence.
For a 95% confidence interval, 95% of similarly constructed intervals will contain the parameter being estimated, while the other 5% will not. µxµx µxµx µxµx µxµx 2.5% Sample interval includes population mean. Sample interval does not include population mean.
Characteristics of t-distribution: - continuous distribution - bell-shaped and symmetrical - there is a family of t distributions: All t- distributions have a mean of 0, but standard deviations differ according to the sample size, n. - t-distribution is flatter and more spread out than the standard normal distribution. As the sample size increases, the t distribution approaches the standard normal distribution.
Use z-distribution, when population standard deviation is known or the sample size is greater than 30. Use t-distribution, when population standard deviation is unknown and the sample size is lesser than 30.
Example: A sample of 10 bearings after 50 million revolutions revealed a sample mean of 0.16 mm of diameter wear with a standard deviation of 0.05 mm. (i) Construct a 95 percent confidence interval for the population mean. (ii) Would it be correct for the company to claim that after 50 million revolutions, the population mean amount of diameter wear is 0.15 mm? The company can be 95% confident that the population mean amount of diameter wear is 0.15mm.
Confidence interval for population proportion is: Where p- Sample proportion n- Sample size x – Number of favorable outcomes
Example: In Country Y, a Bill seeking to amend any provision in the Constitution shall not be passed by Parliament unless it has been supported by the votes of not less than two- thirds of the total number of the elected Members of Parliament. MP A is proposing to change the constitution in Country Y, and would like to know if the bill would be passed. A random sample of 200 MP reveals 125 plan to vote in favor of the Bill. What is the estimate of the population proportion? Develop a 95 percent confidence interval for the population proportion. Basing on this sample information, can MP A conclude that the necessary proportion of MP will favor the merger? Why?
A finite population is defined as one that has a fixed upper bound. When sampling without replacement from a finite population, a correction factor is used to reduce the standard error of the mean, according to the relative size of the sample to the size of the population. For a finite population, where the population size is N and the sample size is n, the following adjustment is made to the standard errors of the sample proportion and sample means: If n/N < 0.05, the finite-population correction factor may be ignored.
Note: Finite population correction factor approaches 1, when n/N becomes smaller. When N = 1000
Confidence Interval for estimating mean when population standard deviation is known with finite population correction: Finite-Population Correction Factor Confidence Interval for estimating mean when population standard deviation is unknown with finite population correction: Confidence Interval for estimating proportion when with finite population correction:
Example: There are 300 families in a block of flats. A random sample of 61 of these families revealed the mean monthly contribution to elderly parents is $550 and the standard deviation of this was $80. Develop a 90 percent confidence interval for the population mean, and interpret the confidence interval. - Given: N = 300, n = 61, s = $75 - Check: n/N = 61/300 = 0.203, the finite population correction factor is necessary. - We use the t-distribution as the population standard deviation is unknown. Confidence Interval for estimating mean when population standard deviation is unknown with finite population correction:
Three factors that determine the sample size are as follows: ◦ degree of confidence ◦ maximum allowable error ◦ population variation None of these factors has any direct relationship with the population size.
(1) Degree of confidence -the higher the level of confidence selected, the larger the size of the corresponding sample, vice versa. (2) Maximum allowable error - amount of error the experimenters are willing to tolerate. It is half the width of the confidence interval. A small allowable error will require a larger sample, vice versa. (3) Population variation - if the population is widely dispersed, a large sample is required, vice versa.
To find the sample size for estimating population mean: where: n – sample size E – maximum allowable error Z – z-value corresponding to selected level of confidence σ – population standard deviation *Round up any fractional result, when the calculated value is not a whole number.
Example: An employee in a MNC wants to determine the mean amount managers in the company earn per month. The error in estimating the mean is to be less than $200 with a 95 percent level of confidence. The employee found an annual report by the HR Department that estimated the standard deviation to be $1,500. What is the required sample size? Given: E = $200, z = 1.96 at 95% level of confidence, σ = $1,500.
Example: An government agency would like to estimate the mean monthly household expenditure for a single family house in May within $10 using 99% level of confidence. Based on similar studies the standard deviation is estimated to be $25. How large a sample size is required?
To find the sample size for estimating population proportion: where: n – sample size E – maximum allowable error z – z-value corresponding to selected level of confidence p – estimate from a pilot study or other sources. Otherwise 0.5 is used, because the term p(1 – p) has the largest value when p = 0.5. Sample size calculated will be the largest when p = 0.5. *Round up any fractional result, when the calculated value is not a whole number.
Example: A advertisement firm wanted to estimate the proportion of adults who own a PDA phone. If the firm wanted the estimate to be within 2% of the population proportion, how many adults would they need to contact? Assume a 95% level of confidence and that the firm’s previous studies estimated that 35% of the adults own a PDA phone.
Example: A new bank wants to estimate the proportion of customers that have adopted their services. The manager wants the margin of error to be within 0.05 of the population proportion, the desired level of confidence is 95%. As it is a new bank, no estimate is available for the population proportion. What is the required sample size?