Estimating Means From Samples – p.110 What is the average sales increase at the 10 stores? 29% 11.3%44.0% 30.8%16.8% 28.9%31.7% 49.9%26.2% 24.7%25.5% The marketing manager authorized a market test at 10 stores across the country. The lower price was put into effect for 1 month and the resulting increase in sales (as a percent) were recorded as shown in the table. The marketing manger is not willing to risk her job if there is any real chance that decreasing prices for the year will result in lower profits.

Estimating Means From Samples – p.111 What is the required % increase in sales? 27.3% (Break even)

Estimating Means From Samples – p.112 The mean increase from the 10 stores was 29% The required increase to breakeven was 27.3% According to the spreadsheet if we increase sales by 29% the profits will increase by $1,140,000 Should we do it? Our model is based on the mean sales increase for all stores all year. Does our sample provide a good estimate of the population mean?

Estimating Means From Samples – p.119 Consider a normal random variable with mean 50 and standard deviation of 10. What is the probability that a randomly selected value would exceed 52? (That is, P(x > 52) ?) This is a problem we already knew how to do. μ = 50 σ = 10 P(X > 52) = 1- NORMDIST(52, 50,10,true) = =.4207

Estimating Means From Samples – p.120 What is the probability that a sample of size 36 would produce a sample mean of 52 or larger? Find P(x bar > 52) x bar is a normal random variable μ xbar = μ = 50 σ xbar = σ / √ n = 10/√ 36 = 10/6 = 1.67 Note: 3 standard deviations = 3 * 10/6 = 5 (range about to 50 +5) P(x bar > 52) = 1 – NORMDIST(52,50,1.67, true) = =.1155

Estimating Means From Samples – p.121 Consider the same population (μ = 50, σ = 10). Find the probability that the mean of a sample of 81 will fall 48 and 52. μ xbar = μ = 50 σ xbar = σ / √ n = 10/√ 81 = 10/9 = 1.11 Find P(48 < x bar < 52) P(48 < x bar < 52) = NORMDIST(52,50,1.11,true) – NORMDIST (48,50,1.11,true) = =.9284 Or % of sample means should fall between 48 and 52 Note 52 is (52-50)/1.11 = st dev above the mean And 48 is (48-50)/1.11 = st dev below mean

Estimating Means From Samples – p.122 For the same population and the same sample size within what range would 90% of all sample means fall. NORMINV(.95,50,1.11) = NORMINV(.05,50,1.11) = % of all sample means should fall between and % of all sample means should fall within how many standard deviations of the mean? (note we mean stdev of sample means) (51.83 – 50)/ 1.11 = (48.17 – 50)/1.11 = % of all sample means should fall within standard deviations of the population mean What happens if we vary the sample size? the mean? the population standard deviation? Nothing

Estimating Means From Samples – p.123 The same information could have been determined from the standard normal random variable and either the standard normal table or NORMSINV. Find z associated with or NORMSINV(.95) = 1.645