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Solutions to the problems in Test #3 Math 146 12/4/15 Data Set #1 0.64531 1.94499 1.32485 1.97105 0.86830 0.59840 0.68198 0.91983 1.54369 0.82731 0.36510 0.37426 0.66417 1.26636 0.77177 1.37512 0.48117 1.02087 0.93507 0.71663 Sum 19.2962 Count 20 Sum of squares 22.7967 Data Set #2 Mean 0.79618 Sample Variance 0.02053 Sum 23.8855 Count 30 Descriptive Statistics: Calculate for data set #1 only • Sample mean, sample standard deviation, and sample variance • 1st Quartile, Median, and 3rd Quartile • Minimum, Maximum, and Range Solutions Mean 0.96481 Median 0.84781 Standard Deviation 0.46901 Sample Variance 0.21997 Range 1.60596 Minimum 0.36510 Maximum 1.97105 1 quartile 0.65945 3 quartile 1.28098 Inferential Statistics Confidence Interval for data set #1 Choosing a confidence level, use the previous data for a confidence interval for the “true” mean (the variance is unknown). If you have lots of time left, you may also construct a confidence interval for the second data set. data 1 90% CI for the Mean from 0.78347 to 1.14615 data 1 95% CI for the Mean from 0.74531 to 1.18431 data 1 99% CI for the Mean from 0.66477 to 1.26485 data 2 90% CI for the Mean from 0.75173 to 0.84064 data 2 95% CI for the Mean from 0.74268 to 0.84969 data 2 99% CI for the Mean from 0.67497 to 0.86175 Testing for the mean for data set #2 Test the hypothesis H0: μ = 0.75 H0: μ ≠ 0.75 t Stat 1.76534 p Value 0.04402 The p-value is about 4.4%, which means that you would reject the Null Hypothesis at the 10% (this is a two-tailed test) significance level, but not at any level below 8.8%. Testing for equality of two means (clearly, this refers to both data sets) We now want to check if the two data sets could come from “populations” (i.e., probabilistic models) with the same “true” mean. Set up a test, without assuming that the variances are equal. Hypothesized Mean Difference 0 Observed Mean Difference 0.16863 df 21.3840 t Stat 1.56010 p-value 0.06671 t Critical two-tail 2.07734 The “t Critical two-tailed” refers to a 5% significance, and we see that we would not reject at this level. In fact, this being a 2-tailed test, a p-value of 6.7% means that the lowest significance at which we would reject the Null Hypothesis would be twice this value, or 13.4% – an unusually high choice. If you are curious, the computer used the following means and standard deviations for its simulations: Data Set μ σ Set #1 1 0.4 Set #2 0.8 0.2