SSF1063: Statistics for Social Sciences LU7: Point and Interval Estimates 3rd March 2008

Point and Interval Estimates An estimate can be a point estimate or an interval estimate Point Estimate – to estimate population parameter

One Situation A researcher wanted to find out the total number of hours people spend watching TV in a week. How can he secure the data? Census or sample survey?

A Point Estimate If we select a sample and compute the value of the sample statistic for this sample, the value gives the point estimate of the studied population parameter Gives the average total number of hours the entire population spend watching TV. Eg: 27.3 hours a week

An Interval Estimate Instead of assigning one value, an interval is constructed around the point estimate, and it is stated that this interval is likely to contain the corresponding population parameter. Eg: 16.5 to 21.3 hours a week

Point and Interval Estimates

Confidence Level & Confidence Interval Each interval is constructed with regard to a given confidence level and is called a confidence interval Given as: Point estimate ±Margin of error x ± z·σx where σx = σ/√n The confidence level associates with how much confidence we have that this interval contains the true population parameter. It is denoted by (1 – α)100%

Estimation of a Population Mean: σ Known

Estimation of a Population Mean: σ Known x ± z·σx where σx = σ/√n The value of z is obtained from the standard normal distribution table. A 95% confidence level means that total area under the normal curve on different sides of μ is 95% or 0.95.

Estimation of a Population Mean: σ Known 1. Find the areas to the left of z1 and z2 respectively Obtained by (1- α)/2 Find the z values from Table IV Appendix C. Respectively the values are -1.96 and 1.96.

Estimation of a Population Mean: σ Known

Estimation of a Population Mean: σ Known

Example 1 A publishing company has just published a new college textbook. Before the company decides the price at which to sell this textbook, it wants to know the average price of all such textbooks in the market. The research department at the company took a sample of 25 comparable textbooks and collected information of their prices. This information produced a mean price of $90.50 for this sample. It is known that the standard deviation of the prices of all such textbooks is $7.50 and the population of such prices is normal. What is the point estimate of the mean price of all such textbooks? Construct a 90% confidence interval for the mean price of all such college textbooks.

Solution 1 Check for the distribution. n = 25, therefore n<30 σ is given and it is mentioned that the population is normally distributed Therefore, n = 25, x = $90.50 and σ = $7.50 Next, calculate the standard deviation of x σx = σ/√n = $7.5/√25 = $1.5

Solution 1 Check for the z value following the confidence level given, and the apply into the formula. Given confidence level 90% Therefore, α = (1 – 0.9) / 2 = 0.05, from the Table IV, the values are z = 1.65 and z = -1.65 x ± z·σx = 90.50 ± 1.65x1.5 = 90.50 ± 2.48 = (90.50 – 2.48) to (90.50 + 2.48) = $88.02 to $92.98 We are 90% confident that the mean price of college textbooks is between $88.02 to $92.98

Confidence Intervals

Confidence Intervals The width of the confidence intervals depends on the size of the margin of error, zσx Which is also depends on the value of z And the value n, as σx = σ/√n Therefore, to decrease the confidence interval, we have two choices: Lower the confidence level Increase the sample size

Example 2 A sample of 1500 homes sold recently in a state gave the mean price of homes equal to $269,720. The population standard deviation of the prices of homes in this state is $68,650. Construct a 99% confidence interval for the mean price of all homes in this state.

Determining The Sample Size for the Estimation of Mean Why waste time testing on 500 batteries if a sample of 40 batteries can give us the confidence interval we are looking for? By knowing the confidence interval, we can determine the sample size we need to produced the required result.

Example 2 An alumni association wants to estimate the mean debt of this year’s college graduates. It is known that the population standard deviation of the debts of this year’s college graduates is $11,800. How large a sample should be selected so that the estimate with a 99% confidence level is within $800 of the population mean?

Solution 2 Use n = z2 ·σ2 E2 The alumni association wants the 99% confidence interval for the mean debt of this year’s college graduate to be x ± 800  E (margin of error)

Estimation of a Population Mean: σ Not Known

Estimation of a Population Mean: σ Not Known t-distribution curve is flatter than the standard normal distribution curve Lower height and wider spread (larger standard deviation) As the sample increases, the t-distribution approaches the standard normal distribution.

Estimation of a Population Mean: σ Not Known

Estimation of a Population Mean: σ Not Known Given by the formula x ± t·sx, where sx = s/√n Since the population standard deviations, σ is not available, we replace it with the sample standard deviation, s – which is an estimator The value of t is obtained from the t-distribution table for (n – 1) degrees of freedom (df) and the given confidence level.

Using the t-distribution Table

Example 3 Sixty four randomly selected adults who buy books for general reading were asked how much they usually spend on books per year. The sample produced a mean of $1450 and a standard deviation of $300 for such annual expenses. Determine a 99% confidence interval for the corresponding population mean.

Solution 3 From the question we know that: The population standard deviation is not given n = 64, x = 1450 and s = 300 df = n – 1 = 64 – 1 = 63 Using s = 300 to replace the population standard deviation, sx = s/√n = 300 /√64 = 37.50

Solution 3 Substitute the values into the formula: x ± t·sx, where sx = s/√n What must you do next? What is your answer?

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