Examples for Ch 8

Examples for Ch 8
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  1 Examples for Ch 8 Confidence Intervals for the Sample Mean with Known σ   Example 1:  An auditor takes a random sample of size 36 from a population of 1000 accounts receivable. The mean value of the accounts receivable for the population is known (from a  previous large survey) to be $2600 with the standard deviation of $450. Find the 95% confidence interval for the sample mean. We are not told whether the population is normal or not. But the sample size is large enough to use the normal approximation. So we will use Z-values to construct the confidence interval. The Z-value for 95% confidence interval is 1.96. The standard error of the sample mean is 450/ √   = $75.   Therefore the margin of error is 1.96*75 = $147. Therefore, the confidence interval is: 2600 ± 147 or between $2453 and $2747 You can similarly find the 90% and 99% confidence intervals using the corresponding Z-values.  Now suppose I asked you to find the probability that the sample mean will fall in the interval which is within $150 around the mean? In this case we are given the margin of error in the units of the X values and asked to find the probability of the resulting confidence interval. This is a reverse process relative to above where we are given the confidence level (or probability) and asked to find the interval. You can simply convert this value to the Z-value by dividing by the std. error of $75 for the sample mean to give ± 2 The probability that Z will be between ± 2 from the Z tables = 0.9772  –   0.0228 = 0.9544 or 95.44% chance. Thus the confidence level increased slightly from 95% as the range of estimated interval (or tolerable margin of error) became wider ($150 compared to $147). Similarly, I could ask you to find the probability that the sample mean will be less than some value or greater than some value. But I should not ask the probability that the sample mean will  be exactly equal to some value. Why? Required Sample size for a specified level of error   The (minimum) required sample size for a random sample is n = Z 2 α/2 *σ 2 /E 2   , where Z α/2   is already defined above in the context of confidence Intervals (it is the Z value corresponding to the required confidence level) , σ is the population standard deviation (or the standard deviation of the population from which the sample is taken), and E is the margin of error around the mean expressed in the units of the X variable.  Note that we don’t need to know the mean for determining the sample size. We need only the standard deviation.  2 Example 2: A personnel department analyst wishes to estimate the mean number of training hours needed annually for supervisors in a division of the company within the margin of 3 hours  (that is plus minus 3 hours) with a 95% confidence  level. Based on a large data from other similar companies the analyst estimates the standard deviation  of required training hours to be equal to 20 hours . Find the minimum sample size which will give the required estimate with specified margin of error and level of confidence. Answer: H ere σ = 20 hours, Z α/2  = Z 0.025  = 1.96 (for 95% confidence level), and the margin of error E = 3 hours. Therefore, n = (1.96*20/3) 2  = 170.7 or 171 observations (always rounded up). The required sample size increases as the tolerable margin of error is reduced. Find the required sample size for margin of error of only 1 hour (I bet it will be 9 times the sample size we just obtained). Similarly, we can find the required sample size for other confidence levels, such as 90% (with Z α/2 = Z 0.05 = 1.645) and 99% (with Z α/2 =  Z 0.005 = 2.576). Clearly the sample size increases as the desired confidence level increases and conversely. Similarly you see that the required sample size increases as the standard deviation of the population increases. This makes sense, because you need a larger sample size to have the same level of confidence if the parent  population involves larger variability, other things remaining the same. Example 3 : A small town has 1000 families who make contributions to the only local church. A  poll of 144 randomly selected contributing families reveals that the mean annual family contribution is $500 with a standard deviation of $72. Construct a 95% confidence interval for the mean annual family contribution for this population of families who contribute to this  particular church. Do you see a problem with this question? We are not given the population mean or standard deviation like the previous example. This may be because there was no previous survey done for this population. In such cases the sample results are used as surrogates to the unknown  population parameters in the formulas given above if the sample size is adequate. In this case the sample size 144 is quite large. So we will use $500 and $72 as the surrogates for the unknown  population mean and the standard deviation, respectively. But do we need to use the formula for large population or small population? At first sight the population of 1000 families seems to be large. But remember the rule given above. The population is much smaller than 20 times the sample. So we will use property #3 to find the standard error. We will use the formula given for finite population on page 4 of Instructions for Chapter 7. Therefore,  ̅  = √   √   = (72/ √  )    = 6*0.9257= 5.554  Therefore, a 95% confidence interval would be between 500-1.96*5.554 and 500+ 1.96*5.554 or  between 489 and 511 dollars per year (rounded to whole numbers ignoring cents. We could also use the “t” distribution (discussed below) to build the confidence interval in this case since the  population standard deviation is not given. But the result would be very close to what we  3 obtained using Z distribution because the sample size is very large. I will discuss this issue in the following section. Confidence Intervals for the Sample Mean with σ Not Known   The pdf (probability density function discussed in my previous instructions) for the t-distribution looks like the curve given below. Note that the t-distribution approaches the Z-distribution more and more closely as the df gets larger (or equivalently, the sample size gets larger). In most  practical applications the t-distribution is considered to be close enough to the Z-distribution for d f ≥ 30. Therefore, many authors suggest  as the practical rule of thumb that for df at least 30  just use the Z-distribution (whose values for the three popular confidence levels are well known and can be easily memorized) instead of the theoretically required t-distribution for which we have to look at the table for the corresponding degree of freedom. Also note that for df infinite the t-distribution exactly coincides with the Z-distribution.   Thus we will use t α/2  in place of Z α/2  in our calculation of the margin of error and the confidence interval whenever df is less than 30 and   σ is unknown (assuming ,however, that the parent population is normal). We will follow exactly the same steps (shown above) as the case when σ is known, except that we replace σ by s and Z by t. For df greater than or equal to 30 it is a matter of researcher’s choice. Theoretically t would be more accurate than   Z,  but that would involve reading from the t-table instead of using the popularly known Z-values. So it is up to you which one to use. Example 4: The sample mean operating life for a random sample of 16 light bulbs of a particular  brand is calculated to be 4000 hours with the sample standard deviation of 200 hours. The operating life of bulbs is generally assumed to be approximately normal. Estimate the mean  4 operating life for the population of bulbs from which the sample is taken using a 95% confidence interval. Here n=16, df = 15, the population is normal (approximately) and the population standard deviation is not given. Therefore, we will use the t-distribution instead of the Z-distribution to construct the confidence interval. We are given    = 4000 hours and s = 200 hours. Therefore, the standard deviation (or standard error) of the sample mean denoted by σ ̅  (from my previous Instructions) is given by σ ̅  = √   where we have replaced σ by s.  Or σ ̅  = √   =   50 hours. Now the confidence interval required is 95%. So α = .05 and α/2 = .025 . Therefore, we need to find t .025  from the table for df = 15. This value is 2.131. Next, The margin of error = t α /2 * σ ̅  = 2.131*50 = 106.55 hours. Therefore, The 95% confidence interval for the mean is    ± t α/2 * σ ̅  =    ± t α/2 * √    = 4000 ± 106.55 or between 3893.45 and 4106.55 hours or between 3893 and 4107 hours rounded (because the numbers are very large we can ignore the decimals and round to the nearest whole number). If we had neglected the fact that the population standard deviation is not known and the sample size is quite small (consequently the df is small), then we would be estimating a narrower interval which would be questionable because it would be claiming more precision than warranted by the nature of the sample.  Now can you build 90% and 99% confidence interval estimates of the mean life of bulbs for this sample? (Hint: look for t .050  and t .005 , respectively). Use of Computer For the example of auditor’s sample of accounts receivable ( Example 1  above) 95% confidence level 2600 mean 450 std. dev. 36 n 1.960 z 146.997 half-width 2746.997 upper confidence limit 2453.003 lower confidence limit You can also find the required sample size for a given level of confidence and specified tolerable margin of error. Let us work on the Second example of this instruction using MegaStat.


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