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Chapter 8. Confidence Intervals. Confidence Intervals. 8.1 z-Based Confidence Intervals for a Population Mean: σ Known 8.2 t-Based Confidence Intervals for a Population Mean: σ Unknown 8.3 Sample Size Determination 8.4 Confidence Intervals for a Population Proportion
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Chapter 8 Confidence Intervals
Confidence Intervals 8.1 z-Based Confidence Intervals for a Population Mean: σ Known 8.2 t-Based Confidence Intervals for a Population Mean: σ Unknown 8.3 Sample Size Determination 8.4 Confidence Intervals for a Population Proportion 8.5 Confidence Intervals for Parameters of Finite Populations (Optional) 8.6 A Comparison of Confidence Intervals and Tolerance Intervals (Optional)
z-Based Confidence Intervals for a Mean: σ Known • The starting point is the sampling distribution of the sample mean • Recall that if a population is normally distributed with mean m and standard deviation σ, then the sampling distribution of x is normal with mean mx= m and standard deviation • Use a normal curve as a model of the sampling distribution of the sample mean • Exactly, because the population is normal • Approximately, by the Central Limit Theorem for large samples
The Empirical Rule • 68.26% of all possible sample means are within one standard deviation of the population mean • 95.44% of all possible observed values of x are within two standard deviations of the population mean • 99.73% of all possible observed values of x are within three standard deviations of the population mean
Example 8.1: The Car Mileage Case • Assume a sample size (n) of 5 • Assume the population of all individual car mileages is normally distributed • Assume the population standard deviation (σ) is 0.8 • The probability that x with be within ±0.7 of µ is 0.9544
Example 8.1 The Car Mileage Case Continued • Assume the sample mean is 31.3 • That gives us an interval of [31.3 ± 0.7] = [30.6, 32.0] • The probability is 0.9544 that the interval [x ± 2σ] contains the population mean µ
Example 8.1: The Car Mileage Case #3 • That the sample mean is within ±0.7155 of µ is equivalent to… • x will be such that the interval [x ± 0.7115] contains µ • Then there is a 0.9544 probability that x will be a value so that interval [x ± 0.7115] contains µ • In other wordsP(x – 0.7155 ≤ µ ≤ x + 0.7155) = 0.9544 • The interval [x ± 0.7115] is referred to as the 95.44% confidence interval for µ
Example 8.1: The Car Mileage Case #4 Three 95.44% Confidence Intervals for m • Thee intervals shown • Two contain µ • One does not
Example 8.1: The Car Mileage Case #5 • According to the 95.44% confidence interval, we know before we sample that of all the possible samples that could be selected … • There is 95.44% probability the sample mean is such the interval [x ± 0.7155] will contain the actual (but unknown) population mean µ • In other words, of all possible sample means, 95.44% of all the resulting intervals will contain the population mean µ • Note that there is a 4.56% probability that the interval does not contain µ • The sample mean is either too high or too low
Generalizing • In the example, we found the probability that m is contained in an interval of integer multiples of sx • More usual to specify the (integer) probability and find the corresponding number of sx • The probability that the confidence interval will not contain the population mean m is denoted by • In the mileage example, a = 0.0456
Generalizing Continued • The probability that the confidence interval will contain the population mean m is denoted by 1 - a • 1 – a is referred to as the confidence coefficient • (1 – a) 100% is called the confidence level • Usual to use two decimal point probabilities for 1 – a • Here, focus on 1 – a = 0.95 or 0.99
General Confidence Interval • In general, the probability is 1 – a that the population mean m is contained in the interval • The normal point za/2 gives a right hand tail area under the standard normal curve equal to a/2 • The normal point - za/2 gives a left hand tail area under the standard normal curve equal to a/2 • The area under the standard normal curve between -za/2 and za/2 is 1 – a
z-Based Confidence Intervals for aMean with σ Known • If a population has standard deviation s (known), • and if the population is normal or if sample size is large (n 30), then … • … a (1-a)100% confidence interval for m is
95% Confidence Level • For a 95% confidence level, 1 – a = 0.95, so a = 0.05, and a/2 = 0.025 • Need the normal point z0.025 • The area under the standard normal curve between -z0.025 and z0.025 is 0.95 • Then the area under the standard normal curve between 0 and z0.025 is 0.475 • From the standard normal table, the area is 0.475 for z = 1.96 • Then z0.025 = 1.96
95% Confidence Interval • The 95% confidence interval is
99% Confidence Interval • For 99% confidence, need the normal point z0.005 • Reading between table entries in the standard normal table, the area is 0.495 for z0.005 = 2.575 • The 99% confidence interval is
The Effect of a on Confidence Interval Width za/2 = z0.025 = 1.96 za/2 = z0.005 = 2.575
Example 8.2: Car Mileage Case • Given • x = 31.56 • σ = 0.8 • n = 50 • Will calculate • 95 Percent confidence interval • 99 Percent confidence interval
Notes on the Example • The 99% confidence interval is slightly wider than the 95% confidence interval • The higher the confidence level, the wider the interval • Reasoning from the intervals: • The target mean should be at least 31 mpg • Both confidence intervals exceed this target • According to the 95% confidence interval, we can be 95% confident that the mileage is between 31.33 and 31.78 mpg • We can be 95% confident that, on average, the mean exceeds the target by at least 0.33 and at most 0.78 mpg
t-Based Confidence Intervals for aMean: s Unknown • If s is unknown (which is usually the case), we can construct a confidence interval for m based on the sampling distribution of • If the population is normal, then for any sample size n, this sampling distribution is called the t distribution
The t Distribution • The curve of the t distribution is similar to that of the standard normal curve • Symmetrical and bell-shaped • The t distribution is more spread out than the standard normal distribution • The spread of the t is given by the number of degrees of freedom • Denoted by df • For a sample of size n, there are one fewer degrees of freedom, that is, df = n – 1
Degrees of Freedom and thet-Distribution As the number of degrees of freedom increases, the spread of the t distribution decreases and the t curve approaches the standard normal curve
The t Distribution and Degrees of Freedom • As the sample size n increases, the degrees of freedom also increases • As the degrees of freedom increase, the spread of the t curve decreases • As the degrees of freedom increases indefinitely, the t curve approaches the standard normal curve • If n ≥ 30, so df = n – 1 ≥ 29, the t curve is very similar to the standard normal curve
t and Right Hand Tail Areas • Use a t point denoted by ta • ta is the point on the horizontal axis under the t curve that gives a right hand tail equal to a • So the value of ta in a particular situation depends on the right hand tail area a and the number of degrees of freedom • df = n – 1 • a = 1 – a , where 1 – a is the specified confidence coefficient
Using the t Distribution Table • Rows correspond to the different values of df • Columns correspond to different values of a • See Table 8.3, Tables A.4 and A.20 in Appendix A and the table on the inside cover • Table 8.3 and A.4 gives t points for df 1 to 30, then for df = 40, 60, 120 and ∞ • On the row for ∞, the t points are the z points • Table A.20 gives t points for df from 1 to 100 • For df greater than 100, t points can be approximated by the corresponding z points on the bottom row for df = ∞ • Always look at the accompanying figure for guidance on how to use the table
Using the t Distribution • Example: Find t for a sample of size n=15 and right hand tail area of 0.025 • For n = 15, df = 14 • = 0.025 • Note that a = 0.025 corresponds to a confidence level of 0.95 • In Table 8.3, along row labeled 14 and under column labeled 0.025, read a table entry of 2.145 • So t = 2.145
t-Based Confidence Intervals for aMean: s Unknown • If the sampled population is normally distributed with mean , then a (1a)100% confidence interval for is • ta/2 is the t point giving a right-hand tail area of /2 under the t curve having n1 degrees of freedom
Example 8.4 Debt-to-Equity Ratios • Estimate the mean debt-to-equity ratio of the loan portfolio of a bank • Select a random sample of 15 commercial loan accounts • x = 1.34 • s = 0.192 • n = 15 • Want a 95% confidence interval for the ratio • Assume all ratios are normally distributed but σ unknown
Example 8.4 Debt-to-Equity RatiosContinued • Have to use the t distribution • At 95% confidence, 1 – = 0.95 so = 0.05 and /2 = 0.025 • For n = 15, df = 15 – 1 = 14 • Use the t table to find t/2 for df = 14, t/2 = t0.025 = 2.145 • The 95% confidence interval:
Sample Size Determination (z) If s is known, then a sample of sizeso that x is within B units of , with 100(1-)% confidence
Sample Size Determination (t) If σ is unknown and is estimated from s, then a sample of sizeso that x is within B units of , with 100(1-)% confidence. The number of degrees of freedom for the ta/2 point is the size of the preliminary sample minus 1
Example 8.6: The Car Mileage Case • What sample size is needed to make the margin of error for a 95 percent confidence interval equal to 0.3? • Assume we do not know σ • Take prior example as preliminary sample • s = 0.7583 • z/2 = z0.025 = 1.96 • t/2 = t0.025 = 2.776 based on n-1 = 4 d.f.
Example 8.7: The Car Mileage Case • Want to see that the sample of 50 mileages produced a 95 percent confidence interval with a margin or error of ±0.3 • The margin of error is 0.227, which is smaller than the 0.3 desired