T-Distributions, One-Sample Confidence Intervals and One-Sample Tests

Unit 7-Inference for Means Chapter 23-Inferences About Means Day 1: t-distributions, one-sample confidence intervals, one-sample tests Test on FRIDAY 12/15 Since we use s to estimate 6, the statistic no longer follows a normal distribution. It follows a t-distribution. Graphs: t 2 Assumptions for T-intervals and T-tests I 1. SRS 2. Sample size constraint...... /not stupid Skewad If n < 15 OK if data are close to normal. If clearly nonnormal or if outliers are present, do not use r Іг 15 < pi <30: OK except in the presence of outliers or strong skewness If n > 30 ОК, even for skewed distributions Graph histogram, box & Whiter Recall. The sampling distribution of a sample mean, . has a mean equal to and a standard deviation of CLT % What happens when we do not know ? The best we can do is use S Standard Error of overline x = 5/(sqrt(n)) There is a different t-distribution for each sample size. A particular distribution is given by the degrees of freedom. df = n - 1 t_{10} = Notation, where k is the degrees of freedom. z = T-distribution Properties N() 1. symmetric 2. bell-shaped 3. more probability in tails than a normal distribution 4. as a increases, the t-distribution approaches the normal curve more closely Ex: Finding t using the t-tables. See WS (a) 90% confidence interval based on 12 observations inut (.95, 11) 1.800 f=12-1 (b) 95% confidence interval thua za 585 30 observatio invT (.975 29) 2.045 90 t ^ prime prime (c) 80% confidence interval from a sample of size 18 m * sqrt(T) * (0.9, 17) = 1.933 We can and hypothesis tests using the t-distribution. overline x plus/minus i' * s/(sqrt(n)) t= overline x - mu x sqrt mu dT = n - 1 On the handout: 26 अ 20 22 22 14 31 1422 22 23 26 31 31 26 31 23 22 11 22 14 31 n = 8 med 225 Q285 225 random state Assumptions/Conditions N one sample & interval rint x * sqrt(g*g_{0}) )^ 2 for the population mean P the mess of vnt. C CSB (mg/100g) 1.49, 28.51 According to my sample, I am 95% confident that the true mean of vit. C in CSB (mg / 100 * g) is between 16.49 and 28.51. 22.5 + 2.365(7191/(sqrt(8))) IS 10 25 30 35 IQRQ-28.5-18=10.5 LE = \mathbb{Q}_{1} - i * 0.5(10R_{r}) = 18 - 1.6(10.5) = 3.2 * 0.5 omega F= 2 + 1.5\{2R\} = 28.5 + 1.5(mu*alpha * 0.5) = 44.25 Pfus the true mens of (mg(100g) it in CSB Η H_{0} / mu + 27 A see problem 2 N one sample t test T overline x = 22.5 O P woluoe =.120 M fail to ryet t = - 1.769914316 According to my sumple, I fail to reject the null hypothesis with a pval= 120 005. There is not evidence to suggest that the true mean amount of vitamin C in CSB is not equal to 27 (mg / 100 * g) 5.6 5.1 4.6 4.8 5.7 64 (a) Find a 90% confidence interval for this patient's phosphate level = true mean of phosphate levels in blood (mg / d * l) ane sample 1 interval random-the sample represents the population (4.819, 5.914) phuphak tund the many pumaly... I am 90% confident that the true mean levels of phosphate in blood (mg / d * l) are between 4.819 and 5.914. The 5.6 5.1 4.6 4.8 5.7 6.4 Thermal trage af las 26 to 4.5 mgill those goed ovidence to the 'falls 43 repastest give and interpret the true mean of phosphate levels in blood (mg / d * l) y = 1 Η mu * 4.6 one sample test >4.8 t = 2 2.086242948 p.0456673133 买-5.366666667 Sx-6653319973 n=6 Assumptions and conditions already verified Based on my sample, I reject the null hypothesis. There is evidence to suggest that the true mean phosphate level in this patient's blood are above 4.8 mg/dl. Classwork/Homework p. 554 #1, 3, 9, 11, 13, 14, 15(omit d), 23, 25, 29-31 mu =4. overline B \mathcal{H}_{2} Q = 18 overline x =5.36 overline 6 Sx = 7.191264542 overline x pm t^ * s/(sqrt(n)) n = 8 overline x = 22.5 mu = 27