In general, a confidence interval is given by: [sample statistic Table value like Z/2 * SE], where SE is the standard deviation of the sampling distribution of the statistic. INTERVAL ESTIMATION OF THE POPULATION MEAN If n>30 or if is known and the Population being sampled is normal, a (l) Confidence interval for the population mean is given by x z /2 ( / n) If is unknown and n>30, sample standard de. s can be used as an approximation for n=100 and x =425, =900, 99% Confidence Interval is 425 ± 2.58 * (900/10) or 192.8 to 657.2 or P(192.8 657.2)=0.99 Interval Estimation of the Population Mean for a Normal Population with Unknown If n 30 ,, is unknown and the sample is drawn from a normal population, a (1) confidence interval for the population mean is given by x t /2, n1 [s / n] where t /2, n1 is the critical value for a t- distribution with n-1 degrees of freedom which captures an area of ( / 2) in the right tail of the t distribution. (t has FATTER tails than normal). The sample mean and standard deviation are 29.86 and 7.08, degrees of freedom associated with the problem is d.f.= n1 = 71 = 6. The t-value corresponding to 6 degrees of freedom and 95% confidence is given in the table as (t 0.025, 6 )=2.447. The corresponding confidence interval is 29.86 ± 2.447 * [7.08/ 7 ] 29.86 ± 6.55. Precision and Sample Size The sample size should be selected in relation to the size of the maximum Positive or negative error the decision maker is willing to accept. This can be achieved by setting the error equal to W/2 or one-half the confidence interval width, Error = (z /2 ) / n This equation can be solved for the sample size, n = [ (z /2 ) / Error ] 2 (Remember, always round up when finding n). Estimating Population Attributes (e.g. defectiveness) Estimate the fraction of defectives from a sample of size n=800. If X=5 defectives are ˆ observed, then p = (5/800)= 0.00625 Estimation of an Interval for Population Attribute In order to develop the confidence interval for a population proportion the sampling distribution of the point estimate must be developed. The ˆ random variable p has a binomial distribution, which is approximated with a normal random variable. ˆ The sample proportion, p is distributed normally with mean=p, and variance=[p(1p)/n] The SE or the standard deviation of the sample proportion p is denoted symbolically as p ˆ ˆ p = [ p(1p) / n ] or approx = [ p (1 p ) / n ] ˆ ˆ ˆ If the sample size n is sufficiently large, both np 5 and n(1 p) 5, then 1- confidence interval for the population proportion is given by the expression p z/2 p ˆ ˆ where z/2 is the distance from the point estimate to the end of the interval in standard deviation units, and p is the standard deviation ˆ ˆ of p or standard error (SE). This is a special case of statistic Z/2 * SE] Precision and Sample Size for Population Attributes If W is the width of the interval, W/2 should obviously equal the maximum allowable error. Hence Error = z/2* p = z/2* [ p(1p) /n] ˆ this is solved for n in the formula below. How large a sample would be required to estimate the proportion of buyers with an accuracy of 0.002 and a 95% degree of confidence if the true proportion is approximately 0.008? Use the formula: n = (z/2 )2 p ( 1- p) / Error 2 p =0.008, z/2 = 1.96 for 95% confidence, and error = 0.002. The above formula yields: n=(1.96)2 (0.008)(10.008)/ (0.002)2 =7662 (Remember, always round up when finding n). Thus, to be 95% confident that the proportion is estimated with an error of at most 0.002 requires a sample size of 7,622.
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