GS ECO 302 CH 9 Estimation and Confidence Intervals

Confidence Interval for Population Mean with σ Known

= X̅ +/-z*(σ/√n) To calculate z (1 + CL%) / 2 Note: CL% should be converted to decimal. ex. 97% = .97 Then look up the decimal in the z table and correspond to values for row and column. Step 1: calculate Standard error: = (σ/√n) Step2: calculate Margin of Error = Zα/2*(σ/√n) Step3: calculate Confidence Interval: ± Margin of Error

To compute a confidence interval, we will consider two situations: A confidence interval is computed using two statistics: the sample mean, X̅, and the standard deviation. From previous chapters, you know that the standard deviation is an important statistic because it measures the dispersion, or width, of a population or sample distribution. The results of the central limit theorem allow us to make the following general confidence interval statements using z-statistics: 1. Ninety-five percent of the sample means selected from a population will be within 1.96 standard errors of the population mean. 2. Ninety-nine percent of the sample means will lie within 2.58 standard errors of the population mean.

1. We use sample data to estimate μ with X̅ and the population standard deviation (σ) is known. 2. We use sample data to estimate μ with X̅, and the population standard deviation is unknown. In this case, we substitute the sample standard deviation (s) for the population standard deviation (σ). Remember: μ = population average X̅ = sample average σ = population standard deviation s = sample standard deviation (standard deviation of the sampling distribution of the sample mean) The formula for the population standard deviation is: s = σX̅ = σ/√n (n = sample size) Also recall that this value is called the standard error.

Confidence Intervals for the Sample Mean with σ Not Known More often than not, the population standard deviation is not known. • Then we have to use the sample estimate s in place of σ, where s is estimated from the sample s = sample estimate

= X̅ +/-z*(σ/√n) becomes = X̅ +/-z*(s/√n) BUT you can't just use standard z values. Theoretically, if the population standard deviation is not known but we know that it is a normal distribution, then we have to use something called the "student's t-distribution" or simply the t-distribution.

confidence interval

A range of values constructed from sample data so that the population parameter is likely to occur within that range at a specified probability. The specified probability is called the level of confidence.

Example 3. find the probability that the sample mean will be less than some value or greater than some value. The population mean annual salary for plumbers is $46,700. A random sample of 42 plumbers is drawn from the population. What is the probability that the mean salary of the sample is greater than $44,000. Assume that the σ = $5600

Convert to z of sample. The formula is z = (x-μ) / (σ/√n) x = $44,000 μ = $46,700 σ = $5,600 n = 42 = 44,000 - 46,700 / (5600 / √42) = -2700 / 5600/6.4807 = -2700 / 864.099 = -3.12 z of -3.12 = 0.009 since we are looking for greater than = 1 - .0009 probability that mean of the sample is greater than $44k = 99.91%

Using the formula from professor's notes (1-a)

Example: Suppose the desired confidence level is 90%. This means that we want an interval around the mean which contains 90% (or in decimals 0.90) area or probability of the sampling distribution. In other words, the area or probability outside this range is 10% (or in decimals 0.10). = 1-a = 1-.10 (same as .90) But this alpha is the total of two symmetrical areas or probabilities on the two sides of the mean which are outside the confidence interval. Therefore, on one side only we will have half alpha or α/2 = 1 - a/2 = 1 - .10/2 = 1 - .05 = .95 Looking up .95 z = between 1.64 and 1.65, we go with 1.645 from (1.64 + 1.65) / 2

margin of error formula

From the professor's notes Za/2 * Std. error of X̅ = Za/2 * (σ/√n) Note that the margin of error and the interval get wider (larger) as the desired level of confidence increases.

Point Estimate Example The following examples illustrate point estimates of population means. 1. Tourism is a major source of income for many Caribbean countries, such as Barbados. Suppose the Bureau of Tourism for Barbados wants an estimate of the mean amount spent by tourists visiting the country.

It would not be feasible to contact each tourist. Therefore, 500 tourists are randomly selected as they depart the country and asked in detail about their spending while visiting the island. The mean amount spent by the sample of 500 tourists is an estimate of the unknown population parameter. That is, we let the sample mean serve as a point estimate of the population mean.

Learning Objectives When you have completed this chapter, you will be able to:

LO1 Define point estimate. LO2 Define level of confidence. LO3 Compute a confidence interval for the population mean when the population standard deviation is known. LO4 Compute a confidence interval for the population mean when the population standard deviation is unknown. LO5 Compute a confidence interval for a population proportion. LO6 Calculate the required sample size to estimate a population proportion or population mean. LO7 Adjust a confidence interval for finite populations.

Example: To be sure each cup contains at least the required amount, Del Monte sets the filling operation to dispense 4.01 ounces of peaches and gel in each cup. • 4.01 is the population mean. • The population std deviation is 0.04 ounces. • Now, we select a random sample of 64 cups and determine the sample mean. It is 4.015 ounces of peaches and gel. The 95 percent confidence interval for the population mean of this particular sample is:

Of course not every cup will contain exactly 4.01 ounces of peaches and gel. Some cups will have more and others less. = X̅ +/-1.96*(σ/√n) : MUST KNOW STD. DEV σ = 4.015 +/- 1.96*(.04 / √64) = 4.015 +/-1.96*(.04/8) = 4.015 +/- .0098 95% confidence interval is between 4.0052 and 4.0248 Note the 1.96 is constant for 95% confidence

Example Calculate Confidence Interval 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. Known Sample size : 36 Population Mean: $2600 Population Std. Dev: $450 Find confidence interval level for 95%

Step 1: calculate Standard error: = (σ/√n) = $450 / √36 = $75 Step2: calculate Margin of Error = Zα/2*(σ/√n) = 1.96 (z value of 95%) * $75 = $147 Step3: calc. Confidence Interval: X̅ ± Margin of Error = $2600 - $147 & $2600 + $147 = between $2453 and $2747

Example 2: 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? Basically Step 2 has been completed. Zα/2*(σ/√n) = $150

We already know (σ/√n) = $450 / √36 = $75 so now we have Zα/2 * $75 = $150 (divide both sides by $75) Zα/2 = 2 Convert +/- 2 on z table and subtract z of +2.00 = .9772 z of -2.00 = .0228 .9772 - .0228 = .9544 or 95.44% (slightly higher than 95% when we got $147 as margin of error)

Example statement for confidence level For example, we estimate the mean yearly income for construction workers in the New York-New Jersey area is $85,000. The range of this estimate might be from $81,000 to $89,000.

We can describe how confident we are that the population parameter is in the interval by making a probability statement. We might say, for instance, that we are 90 percent sure that the mean yearly income of construction workers in the New York-New Jersey area is between $81,000 and $89,000.

point estimates

an estimate of the population mean in the form of a single value, usually the sample mean A point estimate is a single value (point) derived from a sample and used to estimate a population value. • A point estimate is a single statistic used to estimate a population parameter.

90/95/99 percent confidence interval For 90% confidence interval, the margin of error = 1.645*(σ/√n) For 95% confidence interval, the margin of error = 1.96*(σ/√n) For 99% confidence interval, the margin of error = 2.576*(σ/√n) 90% confidence interval = ± 1.645*(σ/√n) 95% confidence interval = ± 1.96*(σ/√n) 99% confidence interval = ± 2.576*(σ/√n)

defined as the interval within which x% percent of all possible sample estimates will fall by chance. To calculate confidence level of any other % use this formula to help you locate the z value. (1 + CL%) / 2 Note: CL% should be converted to decimal. ex. 97% = .97 z for 97% = 2.17 Then look up that # in the z table (using std normal table provided). Then look up the decimal in the z table and correspond to values for row and column. If using appendix B in book, subtract .5 The formula will then be X̅ +/- Z value of CL *(σ/√n)

Example - determining sample size 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.

n = [(z*σ)/E]² z = 1.96 (1+.95) / 2 = .975; convert .975 to z value = 1.96 σ = 20 E = +/-3 = [(1.96*20) / 3] ^2 = 170.737 = 171 because we round up to nearest whole # The required sample size increases as the tolerable margin of error is reduced. For example if the margin of error is 1 instead of 3 the sample size required is now: 1,537

The sample mean,X̅ , is not the only point estimate of a population parameter. For example

p, a sample proportion, is a point estimate of π, the population proportion; and s, the sample standard deviation, is a point estimate of σ, the population standard deviation.

Variance formula

s^2 = ∑(x - X)² / (n-1) Thus the σ = √s^2

Determining Sample Size The required sample size increases as the tolerable margin of error is reduced or if the desired confidence level increases. If the standard deviation increases the required sample size will also increase.

the sample size required to estimate the population mean, µ, with a level of confidence (1 - ∝) * 100% with a specified margin of error, E, is given by n = [(z*σ)/E]² where n is rounded *up* to the nearest whole number z is the confidence level σ is the population standard deviation E is the margin of error around the mean expressed in the units of X variable. Note: we do not need to know the mean for determining the sample size only the standard deviation.

The size of the standard error is affected by two values.

• The first is the standard deviation of the population. The larger the population standard deviation, σ, the larger σ/√n. If the population is homogenous, resulting in a small population standard deviation, the standard error will also be small. • However, the standard error is also affected by the number of observations in the sample. A large number of observations in the sample will result in a small standard error of estimate, indicating that there is less variability in the sample means.