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Sample Test #2 - Solutions Descriptive Statistics The following sample was observed, reported sorted with rank and percentile rank: Rank 14.3470313645721 1 13.9511951251287 2 13.2779176069121 3 13.067373009224 4 12.9228692844372 5 12.8741441944157 6 12.6910633361323 7 12.6510189803734 8 12.5264890042628 9 12.4939617464683 10 12.334512558672 11 12.2565739164815 12 12.1923405114508 13 12.1707334449166 14 12.1155856376745 15 12.0387821210453 16 11.9756692824804 17 11.8845835411695 18 11.6922262079605 19 11.5295952141309 20 11.4899749544308 21 11.0564021006897 22 11.0538426539281 23 10.7488343272596 24 10.6074606195043 25 10.2807997575152 26 9.19230499721994 27 9.08190010947565 28 Percentile Rank 100.00% 96.30% 92.59% 88.89% 85.19% 81.48% 77.78% 74.07% 70.37% 66.67% 62.96% 59.26% 55.56% 51.85% 48.15% 44.44% 40.74% 37.04% 33.33% 29.63% 25.93% 22.22% 18.52% 14.81% 11.11% 7.41% 3.70% 0.00% The following are statistics for this sample: Mean 11.9466137717118 Standard Error 0.2315685387286 Median 12.1431595412955 1 quartile 11.3815817409955 3 quartile 12.6610300693131 Standard Deviation 1.22534552988502 Sample Variance 1.50147166760921 Estimating Proportions A simple random sample of 500 individuals was tested for the presence of a specific gene. 150 turned out to be positive. Compute a confidence interval for the proportion of people in the general population that carry that gene. Be sure to specify what method you used in your calculation. Solutions The sample mean is ^p= 150 =0.3 . If we choose a 95% confidence level, applying our formulas we have 500 • Using the “worst case” estimate • Using the “rough” estimate σ= 1 , (0.25617387297117,0.34382612702883) 2 σ=√ ^p ( 1− ^p )≈0.458 … , (0.25983269109473,0.34016730890527) Estimating the Mean Determine a confidence interval, at a level of your choice, for the true mean of the distribution sampled above Confidence intervals, at various levels: 95% CI for the Mean from to 11.4714743772917 12.4217531661319 90% CI for the Mean from to 11.5521857553027 12.341041788121 98% CI for the Mean from to 11.3740235291274 12.5192040142963 99% CI for the Mean from to 11.3050107680508 12.5882167753729 * Estimating the Variance Determine a confidence interval, at a level of your choice, for the true variance of the distribution sampled above We have that the sum of squares of the difference between the data points and the sample mean is equal to 40.5397350254484. Hence, using tables, we have the following confidence intervals 99%: (0.81659390052922,3.43336312652881) 95%: (0.93853904393575,2.78176561847303) 90%: (1.01063146769221,2.5099833725078)