AGENDA Point Estimation Interval Estimation Tests of Hypotheses Null Hypothesis and Tests of Hypotheses Hypotheses concerning one mean POINT ESTIMATION Have an population of unknown mean Take a sample Point Estimator X bar of sample Estimate of error Known variance… Unknown variance… MAXIMUM ERROR OF ESTIMATE VARIANCE KNOWN For a given sized random sample what will be the maximum size of the estimate error MAXIMUM ERROR OF ESTIMATE VARIANCE UNKNOWN For a given sized random sample what will be the maximum size of the estimate error DETERMINATION OF SAMPLE SIZE For a given level of precision we can also determine how many observations need to be in the sample Rearrage the maximum error of estimate equation Must know alpha, E, and sigma May have to estimate sigma INTERVAL ESTIMATION Point estimates are unlikely to correspond to the actual population parameter Preferable to replace point estimates with interval estimates Need to specify a confidence level that the estimate will cover the true parameter INTERVAL ESTIMATION FOR POP. MEAN WITH SIGMA KNOWN 1-alpha level of certainty that the interval will contain the population mean Alpha is typically 0.01 (99 percent certain) 0.05 (95 percent certain) EXAMPLE Random sample N=100 Sigma=5.1 X bar = 21.6 95% confidence interval? [20.6, 22.6] INTERVAL ESTIMATION FOR POP. MEAN WITH SIGMA UNKNOWN 1-alpha level of certainty that the interval will contain the population mean For “large” (>=30)sample sizes approximate sigma with s Alpha is typically 0.01 (99 percent certain) 0.05 (95 percent certain) EXAMPLE Random sample N=80 Sigma=5.55 X bar = 18.85 99% confidence interval? [17.25, 20.45] INTERVAL ESTIMATION FOR POP. MEAN WITH SIGMA UNKNOWN 1-alpha level of certainty that the interval will contain the population mean For “small” (n<30) sample sizes Alpha is typically 0.01 (99 percent certain) 0.05 (95 percent certain) EXAMPLE Random sample N=16 S=0.68 X bar = 3.42 99% confidence interval? [2.92, 3.92] TESTS OF HYPOTHESES Don’t always want an estimate of a value Sometime want to know if something is true or false Examples Are my suppliers providing the same raw material? Is the product that I produce acceptable? Has my process been changed? NULL HYPOTHESIS AND TESTS OF HYPOTHESES Null Hypotheses Ho Negation of the assertation or claim “No difference” between parameters (mean, variance, etc) Alternate Hypotheses Ha Assertation or claim “Difference” between parameters Can be one or two sided TESTS OF HYPOTHESES Must be specified at some level of significance (alpha level) 1-alpha corresponds to level of confidence of correctness Involves the calculation and comparison of A criterion critical value based on alpha level A test statistic to evaluate the Ho and Ha Result in a decision with respect to the Hypotheses Cannot reject Ho Reject Ho Note that you do not accept nor reject the Ha Decision is based on whether test is one or two sided… MAKING THE DECSION One Sided Hypotheses Test Ha is either greater than some value Cannot reject if test statistic is less than + critical value Reject if test statistic is greater than + critical value Ha is less than some value Cannot reject if test statistic is greater than - critical value Reject if test statistic is less than - critical value MAKING THE DECSION Two Sided Hypotheses Test Cannot reject if: Test statistic is with + or – critical value Reject if Test statistic is less than – critical value Test statistic is greater than + critical value CORRECTNESS OF DECSION When you decide to cannot reject or reject the null hypotheses, you are not guaranteed to make the correct decision Possiblilities Correct decision Wrong decision – Type 1 error Wrong decision – Type 2 error HYPOTHESES POSSIBILITIES TYPE ONE ERROR False positive You say something is “false” but it actually is not Costs you unnecessarily Discard product believing it to be “bad” TYPE TWO ERROR False negative You say something is “true” but it actually is not Costs your customer They use the product believing it to be “ok” GENERAL HYPOTHESES PROCEDURE Identify Ho and Ha Determine level of significance (generally 0.05) Determine “critical value” criterion from level of significance Calculate “test statistic” Make decision Fail to reject Ho Reject Ho EXAMPLE 1 Manufacturer claim that the thermal conductivity of a brick is 0.340 Previous knowledge indicates that the standard deviation is 0.010 A total of 35 observations are taken Mean of 0.343 Level of significance is 0.05 EXAMPLE 1 Approach Interested if the thermal conductivity is not equal to 0.340 This means it can be either less or more then 0.340 Two sided test Known population mean and variance Use Z distribution STEP 1 HYPOTHESES Ho: The mean u is equal to 0.340 Ha: The mean u is not equal to 0.340 STEP 2 DETERMINE LEVEL OF SIGNIFICANCE 0.05 STEP 3 DETERMINE CRITICAL VALUE Use Z table to obtain “critical value” for alpha=0.05 STEP 4 CALCULATIONS Use formula for Z distribution Z=1.77 STEP 5 DECISION Critical values are -1.96 and +1.96 Test statistic is 1.77 -1.96 < 1.77 < +1.96 Cannot reject Ho English There is evidence to support the claim that the bricks have a thermal conductivity of 0.340 at an alpha level of 0.05 EXAMPLE 2 Claimed tire lifetime of at least 28,000 miles is suspect Sample 40 tires Mean=27,463 Std=1,348 Conclusions at an alpha level = 0.05? APPROACH Interested in if the lifetime is <28,000 miles One sided test Known mean, unknown variance Sample size is >30 Use Z test, but substitute sigma with s STEP 1 HYPOTHESES Ho: The mean u is greater than or equal to 28,000 Ha: The mean is less than 28,000 STEP 2 DETERMINE LEVEL OF SIGNIFICANCE 0.01 STEP 3 DETERMINE CRITICAL VALUE Use Z table to obtain “critical value” for alpha=0.01 STEP 4 CALCULATIONS Use formula for Z distribution Z=-2.52 STEP 5 DECISION Critical value =-2.33 Test statistic = -2.52 Test statistic is less than critical value Reject Ho English There is evidence to support the claim that the tires have a lifetime of less than 28,000 at an alpha level of 0.01 EXAMPLE 3 Specs of ribbon require mean breaking strength of 180 pounds Sample 5 observations Mean is 169.5 pounds Std is 5.7 pounds Ho and Ha are given Significance level of 0.01 Assume normality APPROACH Concern about mean breaking strength being less than 180 pounds One sided test Known mean, unknown variance Sample size is < 30 STEP 1 HYPOTHESES Ho: The mean u is greater than or equal to 180 Ha: The mean is less than 180 STEP 2 DETERMINE LEVEL OF SIGNIFICANCE 0.01 STEP 3 DETERMINE CRITICAL VALUE Use t table to obtain “critical value” for alpha=0.01, 4 degrees of freedom STEP 4 CALCULATIONS Use formula for t distribution T= -4.12 STEP 5 DECISION Critical value =-3.747 Test statistic = -4.12 Test statistic is less than critical value Reject Ho English There is evidence to support the claim that the ribbons have a breaking strength of less than 180 pounds at an alpha level of 0.01