Common use of NORMALITY Clause in Contracts

NORMALITY. At least 10 successes (np ≥ 10) and 10 failures (n(1 − p) ≥ 10). Sta102 / BME102 (▇▇▇▇▇ ▇▇▇▇▇▇) Lec 13 October 15, 2014 9 / 33 Sta102 / BME102 (▇▇▇▇▇ ▇▇▇▇▇▇) Lec 13 October 15, 2014 10 / 33 Back to experimental design...‌‌ Confidence intervals for a proportion Single population proportion Calculating the Confidence Interval Confidence intervals for a proportion Single population proportion The GSS found that 571 out of 670 (85%) of Americans answered the question on experimental design correctly. Estimate (using a 95% confidence interval) the proportion of all Americans who have a good intuition about experimental design? 670 Given: n = 670, pˆ = 571 = 0.85. Are CLT conditions met?

NORMALITY. At least 10 successes ([np ≥ 10) ] and 10 failures ([n(1 − p) ≥ 10)]. Lec 11 March 3, 2014 9 / 32 Sta102 / BME102 (▇▇▇▇▇ ▇▇▇▇▇▇) Lec 13 October 1511 March 3, 2014 9 10 / 33 32 Sta102 / BME102 (▇▇▇▇▇ ▇▇▇▇▇▇) Lec 13 October 15, 2014 10 / 33 Back to experimental design...‌‌ ‌ Confidence intervals for a proportion Single population proportion Calculating the Confidence Interval Confidence intervals for a proportion Single population proportion The GSS found that 571 out of 670 (85%) of Americans answered the question on experimental design correctly. Estimate (using a 95% confidence interval) the proportion of all Americans who have a good intuition about experimental design? 670 Given: n = 670, pˆ = 571 = 0.85. Are CLT conditions met?

NORMALITY. At least 10 successes (np ≥ 10) and 10 failures (n(1 − p) ≥ 10). Sta102 / BME102 (▇▇▇▇▇ ▇▇▇▇▇▇) Lec 13 October 1514 March 4, 2014 2015 9 / 33 Sta102 / BME102 (▇▇▇▇▇ ▇▇▇▇▇▇) Lec 13 October 15, 2014 10 / 33 Back to experimental design...‌‌ 32 Calculating the Confidence Interval‌‌‌ Confidence intervals for a proportion Single population proportion Calculating . −standard error of the Confidence Interval Confidence intervals sample proportion is SE = . What isthe We are given that n = 670, pˆ = 0.85, we also just learned that the p(1 p) n 95% confidence interval for a proportion Single population proportion this proportion? CI = point estimate ± margin of error = point estimate ± critical value × SE The GSS found that 571 out of 670 (85%) of Americans answered the question on experimental design correctly. Estimate (using a 95% confidence interval) the proportion of all Americans who have a good the correct intuition about experimental design? 670 Given: n = 670, pˆ = 571 = 0.85. Are CLT conditions met?