How to Find Confidence Interval How to Find Confidence Interval A confidence interval gives an estimated range of values which is likely to include an unknown population parameter, the estimated range being calculated from a given set of sample data. (Definition taken from Valerie J. Easton and John H. McColl's Statistics Glossary v1.1) The common notation for the parameter in question is . Often, this parameter is the population mean , which is estimated through the sample mean . The level C of a confidence interval gives the probability that the interval produced by the method employed includes the true value of the parameter . Example :- Suppose a student measuring the boiling temperature of a certain liquid observes the readings (in degrees Celsius) 102.5, 101.7, 103.1, 100.9, 100.5, and 102.2 on 6 different samples of the liquid. He calculates the sample mean to be 101.82. If he knows that the standard deviation for this procedure is 1.2 degrees, what is the confidence interval for the population mean at a 95% confidence level? In other words, the student wishes to estimate the true mean boiling temperature of the liquid using the results of his measurements. If the measurements follow a normal distribution, then the sample mean will have the distribution N(,). Since the sample size is 6, the standard deviation of the sample mean is equal to 1.2/sqrt(6) = 0.49. Know More About Math Tutor
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The selection of a confidence level for an interval determines the probability that the confidence interval produced will contain the true parameter value. Common choices for the confidence level C are 0.90, 0.95, and 0.99. These levels correspond to percentages of the area of the normal density curve. For example, a 95% confidence interval covers 95% of the normal curve -- the probability of observing a value outside of this area is less than 0.05. Because the normal curve is symmetric, half of the area is in the left tail of the curve, and the other half of the area is in the right tail of the curve. As shown in the diagram to the right, for a confidence interval with level C, the area in each tail of the curve is equal to (1-C)/2. For a 95% confidence interval, the area in each tail is equal to 0.05/2 = 0.025. The value z* representing the point on the standard normal density curve such that the probability of observing a value greater than z* is equal to p is known as the upper p critical value of the standard normal distribution. For example, if p = 0.025, the value z* such that P(Z > z*) = 0.025, or P(Z < z*) = 0.975, is equal to 1.96. For a confidence interval with level C, the value p is equal to (1-C)/2. A 95% confidence interval for the standard normal distribution, then, is the interval (-1.96, 1.96), since 95% of the area under the curve falls within this interval. machine fills cups with margarine, and is supposed to be adjusted so that the content of the cups is 250 g of margarine. As the machine cannot fill every cup with exactly 250 g, the content added to individual cups shows some variation, and is considered a random variable X. This variation is assumed to be normally distributed around the desired average of 250 g, with a standard deviation of 2.5 g. To determine if the machine is adequately calibrated, a sample of n = 25 cups of margarine is chosen at random and the cups are weighed. The resulting measured masses of margarine are X1, ..., X25, a random sample from X. Learn More Solve Math Problems
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Normal Approximation Normal Approximation Computing binomial probabilities using the binomial PDF can be difficult for large n. If tables are used to compute binomial probabilities, calculations typically are only given for selected values of n <= 50 and for selected values of p. The Binomial Applet is much more flexible, since it allows any value of n <= 400 and any valid value of p. However, if n is quite large or if the binomial applet is not available, the normal distribution can be used to approximate the binomial distribution. Binomial/Normal Distribution The following Normal Approximation Applet can be used to experiment with computing approximate and exact binomial probabilities and to assess the conditions for which the normal approximation to the binomial distribution is good. The normal distribution is a good approximation to the binomial when n is sufficiency large and p is not too close to 0 or 1. How large n needs to be depends on the value of p. If p is near 0.5, the approximation can be good for n much less than 20. However, it is better to be conservative and limit the use of the normal distribution as an approximation to the binomial when np > 5 and n(1 - p) > 5.
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Example Suppose that in the biro example (5% defective) we wanted to know the probability of getting less than 40 defectives in a bulk purchase of 100 packets - i.e. 1000 biros. Again, we must assume that this purchase represents a random sample of all biros produced. For the approximation to work we must consider three things: we must match the mean of the Normal distribution to the mean of the Binomial - otherwise the Normal curve will be centred in the wrong place, we must match the standard deviation of the Normal to that of the Binomial - otherwise the Normal curve will not be the correct width we must make an ajustment to take account of the fact that the Binomial variable is discrete while the Normal variable is continuous - the continuity correction. The mean number of defectives in 1000 biros = np = 1000 x 0.05 = 50 The standard deviation = sqrt(npq) = sqrt(1000 x 0.05 x 0.95) = 6.892. Less than 40 defectives means 39 or less on the discrete scale, but 39 extends up as far as 39.5 on the continuous scale. So the approximation is the area under the Normal curve below 39.5. Hence, Pr(defectives < 40) = Pr(defectives <39.5) ~ Pr( Z < [39.5 - 50]/6.892 ) = Pr( Z > -1.524) which, from tables of the Normal distribution, is 0.0638. Using the Binomial formula the corresponding probability would have been 0.0598. You have now completed the recommended route through this unit on the Binomial Distribution. Make sure you visit our Links page to find out about other online resources relevant to this topic. Read More About Simplifying Fractions Math.Tutorvista.com
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