Week 2 SIN502S

Estimation of population parameters A point estimate is a value of a single sample statistic which is used to represent the true but unknown population parameter. e.g. sample mean ( ) is a point estimate of the population mean( ). However, a point estimate does not give enough knowledge as to the reliability of this estimate, i.e. in terms of its closeness to its population parameter. Confidence Interval (CI) is a more reliable estimation method than a point estimate.

Confidence Interval Estimation A confidence interval (CI) is a range of values defined around a sample statistic within which the population parameter is expected to lie with a specified level of confidence. Level of confidence is the probability that the interval constructed around a sample statistic will include (cover) the true population parameter. Margin of error is the value added or subtracted from the sample statistic.

Confidence interval estimation for the single population mean When constructing the confidence interval for a single population mean, we use the Z-distribution or the t-distribution. General rule when to use Z or t When the population standard deviation is known, regardless of the sample size we use the Z-distribution When the population standard deviation is unknown (a) Use Z-distribution when (b) Use t-distribution when

Formula or where, = the sample mean or = limits corresponding to a specified level of confidence = the standard error of the sample mean

Examples A tyre manufacturer found that the sample mean tread life of 49 radial tyres tested was 52345 km. The population standard deviation is known to be 12943 km. Estimate the true mean tread life of all tyres manufactured with a 99% confidence level. A clothing store analyzed the value of purchases made on credit card by a sample of 25 credit card customers. The sample mean was found to be N$165,45 with a sample standard deviation of N$38,60. Construct a 95% CI for the actual mean value of credit card purchases at this store.

Homework The marketing manager of Mores Desserts would like to assess the performance of a new flavored pudding that was launched six weeks ago. The result for a sample of 28 supermarkets countrywide indicated average sales of R3140 per week with a population standard deviation of R345. Construct and interpret a 90% CI for the true population mean countrywide.

Cont… Before entering into wage negotiations, the workers representative of the newly formed farm workers trade union wanted to know the average wage of its union members. The average wage of a random sample of 64 members was found to be N$620 per month with a standard deviation of N$160 per month. Estimate the true mean monthly wage paid to farm workers at a 98% confidence level.

The precision of confidence intervals The precision of the confidence interval is determined by the width of the interval: The narrower the CI, the more precise the interval estimate The wider the CI, the less precise the interval estimate The width of the CI is influenced by: The specified level of confidence The sample size The population standard deviation

Homework The average wage of a random sample of 28 employees was found to be N$ 1420 with a sample standard deviation of N$ 160 per month. Assume wages are normally distributed. Construct a 90% confidence interval and a 99% CI for the true mean monthly wages paid to all employees. Compare the two CIs and comment on the results

Confidence interval estimation for the single population proportion For proportion, we assume that the sample size taken is large enough thus we only use the z-distribution. Formula: Where, = the sample proportion Z = limits from the Z-distribution with a specified level of confidence = standard error of the sampling distribution

Example A recent survey amongst 120 street vendors in Johannesburg shows that 42 of them felt that local by-laws still hampered their trading. Estimate the true population proportion of street vendors who believe that local by-laws still hamper their trading with a 95% level of confidence.

Homework In the company’s telephone survey of 400 television viewers in Oshikuku, it was found that 176 persons in this survey watch television daily. Construct a 90% confidence interval for the actual proportion of all persons who watch television daily in Oshikuku.